Answer

(c) The role of a catalyst is to change the activation energy of reaction. This is done by either increasing or decreasing activation energy of molecule as catalyst are mainly of two types; +ve catalyst and -ve catalyst.

Note Catalyst are of two types one is positive catalyst which increases rate of reaction by decreasing activation energy and another is negative catalyst which decreases rate of reaction by increasing energy of activation.

• Gibbs energy of reaction: A catalyst does not change the Gibbs energy of a reaction. The Gibbs energy is a state function and depends only on the initial and final states of the reaction, not on the pathway taken or the presence of a catalyst.

• Enthalpy of reaction: A catalyst does not change the enthalpy of a reaction. The enthalpy change is also a state function and is determined by the difference in enthalpy between reactants and products, which remains unaffected by the catalyst.

• Equilibrium constant: A catalyst does not change the equilibrium constant of a reaction. It only speeds up the rate at which equilibrium is reached by lowering the activation energy, but the position of equilibrium itself remains unchanged.

Answer

(c) In the presence of catalyst, the heat absorbed, or evolved during the reaction remains unchanged as there is no change in stability of reactant and product.

• (a) increases: This is incorrect because a catalyst does not alter the enthalpy change of a reaction. It only provides an alternative pathway with a lower activation energy, but the overall heat evolved or absorbed remains the same.

• (b) decreases: This is incorrect because a catalyst does not affect the total energy change of the reaction. The enthalpy change is determined by the difference in energy between reactants and products, which remains constant regardless of the presence of a catalyst.

• (d) may increase or decrease: This is incorrect because the presence of a catalyst does not change the enthalpy change of the reaction. The heat evolved or absorbed is a fixed property of the reaction itself and is not influenced by the catalyst.

Answer

(b) Activation energy of a chemical reaction is related to rate constant of a reaction at two different temperatures i.e., $k_1$ and $k_2$ respectively

$$ \ln (\frac{k_1}{k_2})=\frac{E_{a}}{R}[\frac{1}{T_1}-\frac{1}{T_2}] $$

where,

$$ E_{a}=\text { activation energy } $$

$T_2=$ higher temperature

$T_1=$ lower temperature

$k_1=$ rate constant at temperature $T_1$

$k_2=$ rate constant at temperature $T_2$

This equation is known as Arrhenius equation.

• (a) Determining the rate constant at standard temperature alone does not provide enough information to calculate the activation energy. The activation energy is derived from the change in rate constants at different temperatures, not just a single temperature.

• (c) Determining the probability of collision does not directly give the activation energy. While collision theory is related to reaction rates, the activation energy specifically requires information about how the rate constant changes with temperature.

• (d) Using a catalyst lowers the activation energy of a reaction but does not provide a method to determine the activation energy. Catalysts affect the rate of reaction but do not help in calculating the activation energy from temperature-dependent rate constants.

Answer

(a) Activation energy is the minimum energy required to convert reactant molecules to product molecules. Here, the energy gap between reactants and activated complex is sum of $E_1$ and $E_2$.

$\therefore$ Activation energy $=E_1+E_2$

Product is less stable than reactant as energy of product is greater than the reactant.

• Option (b): This option is incorrect because, although it correctly states that the activation energy of the forward reaction is (E_1 + E_2), it incorrectly states that the product is more stable than the reactant. In the given figure, the energy of the product is higher than that of the reactant, indicating that the product is less stable than the reactant.

• Option (c): This option is incorrect because it states that the activation energy of both the forward and backward reactions is (E_1 + E_2). However, the activation energy of the backward reaction is only (E_1), not (E_1 + E_2). Additionally, it correctly states that the reactant is more stable than the product, but the error in the activation energy calculation makes this option incorrect.

• Option (d): This option is incorrect because it states that the activation energy of the backward reaction is (E_1), which is correct. However, it incorrectly states that the product is more stable than the reactant. In the given figure, the energy of the product is higher than that of the reactant, indicating that the product is less stable than the reactant.

Thinking Process

This problem is based on first order rate of reaction. To solve this question determine the value of total pressure then calculate value of $x$ followed by rate constant. where, $x=$ pressure of gas transform to product

Answer

(b)

$$ \underset{\text{At time t}}{\underset{Initially}{}} \quad \underset{p_i^{-x}}{\underset{p_i}{A(g)}} \longrightarrow \underset{x}{\underset{0}{B(g)}} + \underset{x}{\underset{0}{C(g)}}$$

$$ \text { For first order reaction } \quad x=p_{t}-p_{i} $$

$$ k= \frac{2.303}{t}log \frac{p_i}{p_i-x} $$

$$ = \frac{2.303}{t}log \frac{p_i}{p_i-(p_t-p_i)} $$

$$ = \frac{2.303}{t}log \frac{p_i}{2p_i-p_t} $$

• Option (a): The expression ( k=\frac{2.303}{t} \log \frac{p_{i}}{p_{i}-x} ) is incorrect because it does not account for the relationship between ( x ) and the total pressure ( p_t ). Specifically, ( x ) is defined as ( p_t - p_i ), and substituting this into the expression should yield ( \frac{p_i}{2p_i - p_t} ), not ( \frac{p_i}{p_i - x} ).

• Option (c): The expression ( k=\frac{2.303}{t} \log \frac{p_{i}}{2 p_{i}+p_{t}} ) is incorrect because it incorrectly adds ( p_t ) to ( 2p_i ). The correct relationship involves subtracting ( p_t ) from ( 2p_i ), as derived from the correct substitution of ( x = p_t - p_i ).

• Option (d): The expression ( k=\frac{2.303}{t} \log \frac{p_{i}}{p_{i}+x} ) is incorrect because it incorrectly adds ( x ) to ( p_i ). The correct expression should involve the term ( 2p_i - p_t ), which comes from the correct substitution of ( x = p_t - p_i ).

Thinking Process

This problem include graphical representation of Arrhenius equation. To solve this problem transform the Arrhenius equation into equation of straight line taking $\ln k$ on $x$-axis and $\frac{1}{T}$ on $y$-axis

Answer

(a) According to Arrhenius equation, $k=A e^{-E_{a} / R T}$

Taking log on both side $\ln k=\ln (A \cdot e^{-\frac{E_{a}}{R T}})$

$$ \begin{aligned} \ln k & =\ln A-\frac{E_{a}}{R T} \\ \ln k & =-\frac{-E_{a}}{R} \times \frac{1}{T}+\ln A \\ y & =m x+c \end{aligned} $$

This equation can be related to equation of straight line as shown above.

From the graph, it is very clear that slope of the plot $=\frac{-E_{a}}{R}$ and intercept $=\ln A$.

• Option (b): The graph in option (b) shows a positive slope, which contradicts the derived equation $\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$. According to this equation, the slope should be negative, as $-\frac{E_a}{R}$ is negative.

• Option (c): The graph in option (c) is a curve, not a straight line. The derived equation $\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$ represents a linear relationship between $\ln k$ and $\frac{1}{T}$, so the graph should be a straight line.

• Option (d): The graph in option (d) shows a positive slope, which contradicts the derived equation $\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$. According to this equation, the slope should be negative, as $-\frac{E_a}{R}$ is negative.

Answer

(d) According to Arrhenius equation $k=A e^{-E_{a} / R T}$

Here,

$$ \begin{aligned} & k \propto e^{-E_{a}} \\ & k \propto e^{-\frac{1}{T}} \\ & \propto e^{T} \end{aligned} $$

which indicates that as activation energy decreases rate constant increases and as temperature increases rate of reaction increases.

• (a) Rate constant increases exponentially with increasing activation energy and decreasing temperature

  This is incorrect because, according to the Arrhenius equation, the rate constant ( k ) decreases exponentially with increasing activation energy ( E_a ) and decreasing temperature ( T ). Higher activation energy means a larger negative exponent, leading to a smaller ( k ). Similarly, lower temperature also results in a larger negative exponent, reducing ( k ).

• (b) Rate constant decreases exponentially with increasing activation energy and decreasing temperature

  This is partially correct but not entirely accurate. While it is true that the rate constant ( k ) decreases exponentially with increasing activation energy ( E_a ), it also decreases with decreasing temperature ( T ). However, the statement does not fully capture the relationship as it misses the exponential nature of the temperature dependence.

• (c) Rate constant increases exponentially with decreasing activation energy and decreasing temperature

  This is incorrect because, according to the Arrhenius equation, while the rate constant ( k ) does increase exponentially with decreasing activation energy ( E_a ), it actually decreases with decreasing temperature ( T ). Lower temperature results in a larger negative exponent, which reduces ( k ).

Answer

(c) $Zn+$ Dil. $HCl \longrightarrow ZnCl_2+H_2 \uparrow$

Average rate of reaction $=\frac{\text { Change in concentration of } H_2}{\text { Change in time }}=\frac{V_3-0}{40-0}=\frac{V_3}{40}$

• (a) The option $\frac{V_3-V_2}{40}$ is incorrect because it does not account for the initial volume of hydrogen released at time $t=0$. The correct average rate should consider the total volume change from $0$ to $V_3$ over the entire $40$ seconds.

• (b) The option $\frac{V_3-V_2}{40-30}$ is incorrect because it calculates the average rate over a specific interval from $30$ to $40$ seconds, not the entire $40$ seconds. The problem asks for the average rate up to $40$ seconds, not just within the last $10$ seconds.

• (d) The option $\frac{V_3-V_1}{40-20}$ is incorrect because it calculates the average rate over the interval from $20$ to $40$ seconds, not the entire $40$ seconds. The correct average rate should consider the total volume change from $0$ to $V_3$ over the entire $40$ seconds.

Answer

(c) Out of the given four statements, option (c) is not correct.

Order of Reaction

Order of reaction is equal to the sum of powers of concentration of the reactants in rate law expression,

For any chemical reaction

$$ \begin{gathered} x A+y B \longrightarrow \text { Product } \\ \text { Rate }=k[A]^{x}[B]^{y} \\ \text { Order }=x+y \end{gathered} $$

Order of reaction can be a fraction also. Order of reaction is not always equal to sum of the stoichiometric coefficients of reactants in the balanced chemical equation. For a reaction it may or may not be equal to sum of stoichiometric coefficient of reactants.

• (a) The order of a reaction can be a fractional number: This statement is correct. The order of a reaction can indeed be a fractional number, as it is determined by the rate law expression and not necessarily by whole numbers.

• (b) Order of a reaction is experimentally determined quantity: This statement is correct. The order of a reaction is determined through experimental data and cannot be deduced solely from the stoichiometric coefficients of the balanced chemical equation.

• (d) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression: This statement is correct. The order of a reaction is defined as the sum of the exponents of the concentration terms in the rate law expression.

Answer

(b) Reaction occurring at smallest time interval is known as instantaneous rate of reaction e.g., instantaneous rate of reaction at $40 s$ is rate of reaction during a small interval of time close to $40 s$. Volume change during a small time interval close to $40 s$ i.e., $40-30$ s, $50-40 s, 50-30 s, 40-20 s$.

Instantaneous rate of reaction $=\frac{\text { Change in volume }}{\text { Time interval close to } 40 s}$

(a) $r_{\text {inst }}(20 s)=\frac{V_5-V_2}{50-30}$ correct

(b) $r_{\text {inst }}(20 s)=\frac{V_4-V_3}{50-30}$ incorrect, correct is $\frac{V_5-V_3}{50-30}$

(c) $r_{\text {inst }}(10 s)=\frac{V_3-V_2}{40-30}$ correct

(d) $r_{\text {inst }}(20 s)=\frac{V_3-V_1}{40-20}$ correct

• (a) $\frac{V_5-V_2}{50-30}$: This option is incorrect because it calculates the average rate of reaction over a time interval from 30 s to 50 s, which is not close enough to 40 s to be considered an instantaneous rate at 40 s.

• (c) $\frac{V_3-V_2}{40-30}$: This option is incorrect because it calculates the average rate of reaction over a time interval from 30 s to 40 s, which is not close enough to 40 s to be considered an instantaneous rate at 40 s.

• (d) $\frac{V_3-V_1}{40-20}$: This option is incorrect because it calculates the average rate of reaction over a time interval from 20 s to 40 s, which is not close enough to 40 s to be considered an instantaneous rate at 40 s.

Answer

(a) Rate of reaction is defined as rate of decrease of concentration of any one of reactant with passage of time

$$ \begin{aligned} \text { Rate of reaction } & =\frac{\text { Rate of disappearance of reactant }}{\text { Time taken }} \\ r & =\frac{-d x}{d t} \end{aligned} $$

Thus, as the concentration of reactant decreases with passage of time, rate of reaction decreases.

• (b) The rate of a reaction is same at any time during the reaction

  This statement is incorrect because the rate of a reaction typically changes over time. As the reactants are consumed, their concentrations decrease, which generally leads to a decrease in the reaction rate. The rate of reaction is not constant and varies depending on the concentration of the reactants.

• (c) The rate of a reaction is independent of temperature change

  This statement is incorrect because the rate of a reaction is highly dependent on temperature. According to the Arrhenius equation, an increase in temperature usually increases the reaction rate by providing more energy to the reactant molecules, thereby increasing the frequency and energy of collisions between them.

• (d) The rate of a reaction decreases with increase in concentration of reactant(s)

  This statement is incorrect because, in most cases, the rate of a reaction increases with an increase in the concentration of reactants. According to the rate law, the rate of reaction is directly proportional to the concentration of the reactants raised to a power (which is the order of the reaction). Therefore, higher concentrations of reactants typically lead to a higher reaction rate.

Answer

(c) Given, chemical reaction is

$$ 5 Br^{-}(aq)+BrO_3^{-}(aq)+6 H^{+}(aq) \longrightarrow 3 Br_2(aq)+3 H_2 O(l) $$

Rate law expression for the above equation can be written as

$$ -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t}=-\frac{\Delta[BrO_3^{-}]}{\Delta t}=\frac{-1}{6} \frac{\Delta[H^{+}]}{\Delta t}=\frac{+1}{3} \frac{\Delta[Br_2]}{\Delta t} $$

$$ \Rightarrow \frac{\Delta[Br^{-}]}{\Delta t} =-\frac{\Delta[BrO_3^{-}]}{\Delta t}=\frac{-5}{6} \frac{\Delta[H^{+}]}{\Delta t} $$

$$ \Rightarrow \frac{\Delta[Br^{-}]}{\Delta t} =\frac{5}{6} \frac{\Delta[H^{+}]}{\Delta t} $$

• Option (a): The expression $\frac{\Delta[Br^{-}]}{\Delta t}=5 \frac{\Delta[H^{+}]}{\Delta t}$ is incorrect because it does not account for the stoichiometric coefficients correctly. The correct relationship should involve the ratio of the coefficients, which is $\frac{5}{6}$, not 5.

• Option (b): The expression $\frac{\Delta[Br^{-}]}{\Delta t}=\frac{6}{5} \frac{\Delta[H^{+}]}{\Delta t}$ is incorrect because it inverts the correct ratio. The correct ratio is $\frac{5}{6}$, not $\frac{6}{5}$.

• Option (d): The expression $\frac{\Delta[Br^{-}]}{\Delta t}=6 \frac{\Delta[H^{+}]}{\Delta t}$ is incorrect because it incorrectly multiplies by 6 instead of using the correct ratio of $\frac{5}{6}$.

Answer

(a) The chemical reaction in which energy is evolved during the reaction is known as exothermic reaction i.e., activation energy of product is greater than activation energy of reactants.

Here, only (I) denotes correct picture of exothermic reaction.

• Option (II) is incorrect: This graph represents an endothermic reaction, where the energy of the products is higher than the energy of the reactants. In an endothermic reaction, energy is absorbed from the surroundings, which is the opposite of an exothermic reaction.

• Option (III) is incorrect: This graph also represents an endothermic reaction, where the energy of the products is higher than the energy of the reactants. Similar to option (II), energy is absorbed in this type of reaction, not released.

Answer

(b) Rate law can be written as

$$ \text { Rate }=k[A][B] $$

Rate of reaction w.r.t $B$ is of first order.

$$ R_1=k[A][B] $$

when concentration of reactant ’ $B$ ’ is doubled then rate $(R_2)$

$$ \begin{aligned} & R_2=k[A][2 B] \\ & R_2=2 k[A][B] \\ & R_2=2 R_1 \end{aligned} $$

Therefore; as concentration of $B$ is doubled keeping the concentration of $A$ constant rate of reaction doubles.

• Option (a) the same: This option is incorrect because the rate of the reaction depends on the concentration of reactants. When the concentration of reactant ‘B’ is doubled, the rate of the reaction also doubles, as shown by the rate law (\text{Rate} = k[A][B]). Therefore, the rate constant (k) remains the same, but the rate of the reaction changes.

• Option (c) quadrupled: This option is incorrect because the rate of the reaction is first order with respect to reactant ‘B’. Doubling the concentration of ‘B’ will double the rate of the reaction, not quadruple it. Quadrupling would occur if the reaction were second order with respect to ‘B’, which it is not.

• Option (d) halved: This option is incorrect because doubling the concentration of reactant ‘B’ increases the rate of the reaction. Halving the rate would imply a decrease in the concentration of ‘B’, which is not the case here.

Answer

(c) According the postulates of collision theory there are following necessary conditions for any reaction to be occur

(i) Molecule should collide with sufficient threshold energy.

(ii) Their orientation must be proper.

(iii) The collision must be effective.

• (a) The statement is incorrect because the collision theory does consider the structural features of reacting molecules or atoms, not just treating them as hard spheres.
• (b) The statement is incorrect because while the number of effective collisions does determine the rate of reaction, it is not the only factor; the orientation and energy of the collisions are also crucial.
• (d) The statement is incorrect because it is actually correct according to the collision theory. Molecules must collide with sufficient threshold energy and proper orientation for the collision to be effective.

Answer

(d) The time taken for half the reaction to complete. i.e., the time in which the concentration of a reactant is reduced to half of its original value is called half-life period of the reaction.

But it is impossible to perform $100 %$ of the reaction. Whole of the substance never reacts because in every half-life, $50 %$ of the substance reacts. Hence, time taken for $100 %$ completion of a reaction is infinite.

• Option (a): $1.26 \times 10^{15} s$ is incorrect because it suggests a finite time for the complete reaction, which contradicts the nature of a first-order reaction where the reaction never reaches $100 %$ completion.

• Option (b): $2.52 \times 10^{14} s$ is incorrect because it also implies a finite time for the reaction to reach $100 %$ completion. In a first-order reaction, the concentration of the reactant decreases by half in each half-life period, making it impossible to reach complete conversion.

• Option (c): $2.52 \times 10^{28} s$ is incorrect because, although it is an extremely large number, it still represents a finite time. For a first-order reaction, the reaction asymptotically approaches completion but never actually reaches $100 %$.

Answer

(b) Rate of reaction is change in concentration of reactant with respect to time.

$$ \begin{aligned} r & =k[A]^{x}[B]^{y} \\ \frac{\text { Rate of } \operatorname{exp.1}}{\text { Rate of exp.2 }} & =\frac{[0.30]^{x}[0.30]^{y}}{[0.30]^{x}[0.60]^{y}} \\ \frac{0.10}{0.40} & =\frac{[0.30]^{y}}{[0.60]^{y}} \\ \frac{1}{4} & =[\frac{1}{2}]^{y} \\ [\frac{1}{2}]^{2} & =[\frac{1}{2}]^{y} \\ y & =2 \\ \frac{\text { Rate of exp.1 }}{\text { Rate of exp.3 }} & =\frac{[0.30]^{x}[0.30]^{y}}{[0.60]^{x}[0.30]^{y}} \\ \frac{0.10}{0.20} & =[\frac{0.30}{0.60}]^{x}[\frac{0.30}{0.30}]^{y} \\ \frac{1}{2} & =[\frac{1}{2}]^{x}[1]^{y} \\ \frac{1}{2} & =[\frac{1}{2}]^{x} \\ i.e., \quad x & =1 \\ \because \quad \text { Rate } & =k[A]^{x}[B]^{y} \\ \text { Rate } & =k[A]^{y}[B]^{2} \end{aligned} $$

• Option (a) Rate = k[A]²[B]:

  • This option suggests that the rate is second order with respect to A and first order with respect to B. However, from the given data, when the concentration of B is doubled (Experiment 1 to Experiment 2), the rate increases by a factor of 4, indicating that the reaction is second order with respect to B, not A. Therefore, this option is incorrect.

• Option (c) Rate = k[A][B]:

  • This option suggests that the rate is first order with respect to both A and B. However, from the given data, when the concentration of B is doubled (Experiment 1 to Experiment 2), the rate increases by a factor of 4, indicating that the reaction is second order with respect to B. Additionally, when the concentration of A is doubled (Experiment 1 to Experiment 3), the rate doubles, indicating that the reaction is first order with respect to A. Therefore, this option is incorrect.

• Option (d) Rate = k[A]²[B]⁰:

  • This option suggests that the rate is second order with respect to A and zero order with respect to B. However, from the given data, when the concentration of B is doubled (Experiment 1 to Experiment 2), the rate increases by a factor of 4, indicating that the reaction is second order with respect to B. Therefore, this option is incorrect.

Answer

(b) Characteristics of catalyst

(a) It catalyses the forward and backward reaction to the same extent as it decreases energy of activation hence, increases the rate of both the reactions.

(b) Since, reaction quotient is the relation between concentration of reactants and products. Hence, catalyst does not alter Gibbs free energy as it is related to reaction quotient. Thus, Gibbs free energy does not change during the reaction when catalyst is added to it.

$$ \Delta G=-R T \ln Q $$

where, $Q=$ reaction quotient

(c) It doesn’t alter equilibrium of reaction as equilibrium constant is also concentration dependent term.

(d) It provides an alternate mechanism by reducing activation energy between reactants and products.

• (a) It catalyses the forward and backward reaction to the same extent as it decreases energy of activation hence, increases the rate of both the reactions.

• (c) It doesn’t alter equilibrium of reaction as equilibrium constant is also concentration dependent term.

• (d) It provides an alternate mechanism by reducing activation energy between reactants and products.

Answer

$(\boldsymbol{a}, \boldsymbol{b})$ Pseudo first order reaction is a chemical reaction in which rate of reaction depends upon concentration of only one reactant while concentration of another reactant has no effect on rate of reaction.

e.g., hydrolysis of ethyl acetate in presence of excess of water

$$ \begin{gathered} CH_3 COOC_2 H_5+\underset{\text { excess }}{H_2 O} \xrightarrow{H^{+}} CH_3 COOH+C_2 H_5 OH \\ r=k[CH_3 COO C _2 H_5]^{2}[H_2 O]^{0} \end{gathered} $$

Excess $[H_2 O]$ can cause the independency of reaction on $H_2 O$.

Hence, (a) is the correct choice.

• (b) depends on the concentration of reactants present in excess: This option is incorrect because, in a pseudo first order reaction, the rate of reaction is independent of the concentration of the reactant present in excess. The reactant in excess is considered to have a constant concentration, and thus does not affect the rate constant.

• (c) is independent of the concentration of reactants: This option is incorrect because the rate constant of a pseudo first order reaction does depend on the concentration of the reactant that is not in excess. The reaction rate is proportional to the concentration of this reactant.

• (d) depends only on temperature: This option is incorrect because, while the rate constant does depend on temperature, it also depends on the concentration of the reactant that is not in excess. Therefore, it is not solely dependent on temperature.

Answer

(b) $A \longrightarrow B$

Concentration of reactants and products varies exponentially w.r.t time.

(i) Concentration of reactant (here, A) decreases exponentially w.r.t time.

(ii) Concentration of product (here, $B$ ) increases exponentially w.r.t time new line correct graph representing the above reaction is (b).

• Option (a): The graph shows the concentration of the reactant decreasing linearly with time, not exponentially. Similarly, the concentration of the product increases linearly with time, which does not match the exponential behavior described in the question.

• Option (c): The graph shows the concentration of the reactant decreasing exponentially with time, which is correct. However, the concentration of the product remains constant over time, which is incorrect as it should increase exponentially.

• Option (d): The graph shows the concentration of the reactant decreasing exponentially with time, which is correct. However, the concentration of the product also decreases exponentially with time, which is incorrect as it should increase exponentially.

Answer

$(a, c, d)$

Rate law can be determined from balanced chemical equation if it is an elementary reaction.

• Option (b) is incorrect because if the reaction is an elementary reaction, the rate law can be directly determined from the balanced chemical equation. In elementary reactions, the rate law is based on the stoichiometry of the reactants involved in the rate-determining step.

Answer

$(a, d)$

For a balanced chemical equation of an elementary reaction order is same as molecularity and molecularity can never be zero. If molecularity of a reaction is considered to be zero it mean that no reactant is going to transform into product. Consider a chemical reaction.

$$ 2 NO(g)+O_2(g) \longrightarrow 2 NO_2(g) $$

Differential rate law expression can be written as

$$ \frac{d R}{d t}=k[NO]^{2}[O_2] $$

Here, molecularity $=3$, order $=3$

• (b) Order is less than the molecularity: This statement is incorrect for an elementary reaction because, by definition, the order of an elementary reaction is determined directly by its molecularity. In an elementary reaction, the rate law is derived from the stoichiometry of the reactants, meaning the order and molecularity are the same.

• (c) Order is greater than the molecularity: This statement is also incorrect for an elementary reaction. Since the order of an elementary reaction is directly related to the number of molecules participating in the reaction (molecularity), it cannot be greater than the molecularity. The order is a direct reflection of the molecularity in elementary reactions.

Answer

$(a, b)$

Since, the reaction is an unimolecular reaction. Hence, in the slowest step i.e., in the rate determining step the only one reacting species is involved. Therefore, order of reaction and molecularity of reaction is equal to one.

• Option (c) is incorrect because in a unimolecular reaction, the order of the reaction is not zero. The order of the reaction is determined by the concentration of the single reacting species involved in the rate-determining step, which makes the order one.

• Option (d) is incorrect because while both the molecularity and order of the reaction are indeed one in a unimolecular reaction, this option does not provide any new information beyond what is already stated in options (a) and (b). Therefore, it is redundant and does not add any additional value to the explanation.

Answer

$(a, d)$

(a) For a complex reaction, order of overall reaction = molecularity of slowest step As rate of overall reaction depends upon total number of molecules involved in slowest step of the reaction. Hence, molecularity of the slowest step is equal to order of overall reaction.

(d) Since, the completion of any chemical reaction is not possible in the absence of reactants. Hence, slowest step of any chemical reaction must contain at least one reactant. Thus, molecularity of the slowest step is never zero or non-integer.

(b) The order of the overall reaction is not necessarily less than the molecularity of the slowest step. The order of a reaction is determined by the sum of the exponents of the concentration terms in the rate law, which can be equal to or greater than the molecularity of the slowest step, depending on the reaction mechanism.

(c) The order of the overall reaction is not necessarily greater than the molecularity of the slowest step. The order of a reaction is determined by the rate-determining step, which is the slowest step, and can be equal to the molecularity of that step.

Answer

$(a, c, d)$

Given, chemical reaction is

$$ 2 NH_3(g) \frac{1130 K}{\text { Platinum catalyst }} N_2(g)+3 H_2(g) $$

At very high pressure reaction become independent of concentration of ammonia i.e., zero order reaction

Hence,

$$ \begin{aligned} & \text { Rate }=k[p_{NH_3}]^{0} \\ & \text { Rate }=k \end{aligned} $$

(a) Rate of reaction $=$ Rate constant

(b) Rate of decomposition of ammonia will remain constant until ammonia disappears completely.

(c) Since, formation of ammonia is a reversible process further increase in pressure will change the rate of reaction. According to Le-Chatelier principle increase in pressure will favour in backward reaction.

• (b) Rate of the reaction depends on concentration of ammonia

  This option is incorrect because, for a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. In this case, the reaction rate does not depend on the concentration of ammonia ($NH_3$), as indicated by the zero-order kinetics.

Answer

$(a, d)$

When the reactant molecules collide each other they lead to formation of an activated complex. It has highest energy among reactants, products and activated complex. When it decomposes to give product, energy is released and stability of product increases.

Since, the entire concentration of activated complex do not convert into products while, some activated complex may give reactants also.

• Option (b) is incorrect because during the decomposition of an activated complex, energy is released rather than absorbed. The activated complex is at a higher energy state, and its decomposition to form products results in the release of energy.

• Option (c) is incorrect because the energy does change during the decomposition of an activated complex. The activated complex has higher energy, and as it decomposes to form products, energy is released, leading to a decrease in the overall energy.

Answer

(a, c)

Distribution of kinetic energy may be described by plotting a graph of fraction of molecules versus kinetic energy.

Kinetic energy of maximum fraction of molecule is known as most probable kinetic energy. It is important to note that with increase of temperature, peak shifts forward but downward. This means that with increase of temperature,

(i) most probable kinetic energy increases.

(ii) the fractions of molecules possessing most probable kinetic energy decreases.

• Option (b) is incorrect: The fraction of molecules with the most probable kinetic energy does not increase at higher temperatures. Instead, it decreases because the distribution of kinetic energies broadens, leading to a lower peak in the graph.

• Option (d) is incorrect: The most probable kinetic energy does not decrease at higher temperatures. In fact, it increases because higher temperatures provide more energy to the molecules, shifting the peak of the distribution to higher kinetic energies.

Answer

$(a, d)$

According to Maxwell Boltzmann distribution curve, area under the curve must not change with increase in temperature. But with increase in temperature curve broadens and shift towards right hand side due to decrease in fraction of molecules having most probable kinetic energy.

• (b) area under the curve increases with increase in temperature: This is incorrect because the area under the Maxwell-Boltzmann distribution curve represents the total number of molecules, which remains constant regardless of temperature. The distribution of energies changes, but the total number of molecules does not.

• (c) area under the curve decreases with increase in temperature: This is incorrect for the same reason as option (b). The area under the curve represents the total number of molecules, which does not change with temperature. The shape of the curve changes, but the area remains constant.

Answer

$(a, b)$

Arrhenius equation can be written as $k=A \cdot e^{\frac{-E_{a}}{R T}}$

$k \propto e^{-E_{a}}$ i.e., rate of reaction increases with decrease in activation energy.

$$ k \propto e^{-\frac{1}{T}} $$

$k \propto e^{T}$ i.e., rate of reaction increases with increase in temperature.

• (c) Rate constant decreases exponentially with increase in temperature: This is incorrect because according to the Arrhenius equation, the rate constant ( k ) increases exponentially with an increase in temperature, not decreases.

• (d) Rate of reaction decreases with decrease in activation energy: This is incorrect because according to the Arrhenius equation, the rate of reaction increases with a decrease in activation energy, not decreases.

Answer

$(b, d)$

Function of Catalyst As the catalyst is added to the reaction medium rate of reaction increases by decreasing activation energy of molecule. Hence, it follows an alternative pathway.

Catalyst does not change the enthalpy change of reaction. Energy of reactant and product remain same in both catalysed and uncatalysed reaction.

Hence, (a) and (d) are incorrect statements.

• Catalyst raises the activation energy: This is incorrect because a catalyst actually lowers the activation energy required for a reaction to proceed, thereby increasing the reaction rate.

• Catalyst alters enthalpy change of the reaction: This is incorrect because a catalyst does not change the enthalpy change of the reaction. The energy of the reactants and products remains the same in both catalyzed and uncatalyzed reactions.

Thinking Process

This problem includes graphical representation of zero order reaction. To solve this problem.

(i) Write rate equation of zero order reaction.

(ii) Transform it into equation of straight line.

(iii) Transform it into a curve representing rate versus time.

Answer

$(a, d)$

For a zero order reaction

$$ \begin{gathered} {[R]=(-k) t+[R]_0} \quad …(i) \\ \uparrow \uparrow \uparrow \\ y=m \times x+c \end{gathered} $$

On comparing with Eq. of straight line

$y=[R]$ concentration

$x=t$ time

Slope $(m)=-k$ rate constant

Intercept $(c)=[R]_0$ initial concentration

On rearranging Eq. (i)

$$ \begin{aligned} & \frac{[R]-[R]_0}{t}=-k \\ & \frac{[R]-[R]_0}{t}=-k t^{0} \\ & \quad \text { Rate } \propto t^{0} \end{aligned} $$

• Option (b): This graph shows a plot of concentration ([R]) versus time (t) that is not linear. For a zero-order reaction, the plot of ([R]) versus (t) should be a straight line with a negative slope. Since this graph is not linear, it is incorrect for a zero-order reaction.

• Option (c): This graph shows a plot of the rate of reaction versus time (t). For a zero-order reaction, the rate of reaction is constant and does not depend on time. Therefore, the graph should be a horizontal line. Since this graph is not a horizontal line, it is incorrect for a zero-order reaction.

Answer

(a, $d)$

For the first order reaction

$$ \begin{aligned} & k=\frac{2.303}{t} \log \frac{[R]_0}{[R]} \\ & \frac{k t}{2.303}=\log \frac{[R]_0}{[R]} \\ & \begin{matrix} \log \frac{[R]_0}{[R]} & =(\frac{k}{2.303}) &\times t&+0 \\ \uparrow & \uparrow & \uparrow & \uparrow \\ y= &\quad m & \quad x &+c \end{matrix} \\ \end{aligned} $$

Correct plot of $\log \frac{[R]_0}{[R]}$ can be represented by (d)

where,

$$ \text { slope }=\frac{k}{2.303} $$

The time taken for any fraction of the reaction to complete is independent of the initial concentration. Let, us consider it for half of the reaction to complete.

$$ \begin{aligned} t & =\frac{2.303}{k} \log \frac{a}{a-x} \\ \text{For half-life} \quad t & =t_{1 \backslash 2} \text { and } x=\frac{a}{2} \\ t_{1 \backslash 2} & =\frac{2.303}{k} \log \frac{a}{a-\frac{a}{2}} \\ t_{1 \backslash 2} & =\frac{2.303}{k} \log 2 \end{aligned} $$

Half-life time

$$ t_{1 \backslash 2}=\frac{0.693}{k} $$

$t_{1 / 2}$ is independent of initial concentration. Hence, correct plot of $t_{1 / 2}$ and $[R]_0$ can be represented by a.

• Option (b): This graph is incorrect because it shows a linear relationship between the concentration of the reactant ([R]) and time (t). For a first-order reaction, the concentration of the reactant decreases exponentially with time, not linearly. The correct plot should show a logarithmic relationship.

• Option (c): This graph is incorrect because it shows a linear relationship between the concentration of the reactant ([R]) and the logarithm of time (\log t). For a first-order reaction, the correct relationship is between the logarithm of the concentration (\log [R]) and time (t), not the other way around.

Answer

Presence of one of the reactants in excess, as in such a condition, its concentration remains constant and rate of such reaction depends upon concentration of one reactant only and reaction is known as pseudo first order reaction e.g., acid catalysed hydrolysis of ethyl acetate.

$$ \underset{\text { Ethyl acetate }}{CH_3 COOC_2 H_5}+H_2 O \xrightarrow[\text { Acetic acid }]{H^{+}} \underset{\text { Ethyl alcohol }}{CH_3 COOH}+\underset{2}{C_2 H_5 OH} $$

This reaction is bimolecular but is found to be of first order as experimentally it is observed that rate of reaction depends upon the concentration of ethyl acetate not on water as it is present in excess.

Answer

For reaction $2 A+B \longrightarrow C$ if the rate of reaction is zero then it can be represented as

$$ \text { Rate }=k[A]^{0}[B]^{0}=k $$

i.e., rate of reaction is independent of concentration of $A$ and $B$.

Answer

We can determine the rate of this reaction as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. e.g., for the given reaction,

(i) Keeping $[O_2]$ constant, if the concentration of $NO$ is doubled, rate is found to become four times. This shows that,

Rate $\propto[NO]^{2}$

(ii) Keeping $[NO]$ constant, if the concentration of $[O_2]$ is doubled, rate is also found to become double. This shows that,

$$ \text { Rate } \propto[O_2]^{2} $$

Hence, overall rate law will be

$$ \text { Rate }=k[NO]^{2}[O_2] $$

Rate law expression $-\frac{1}{2} \frac{\Delta[NO]}{\Delta t}=-\frac{\Delta[O_2]}{\Delta t}$

$$ =\frac{1}{2} \frac{\Delta[NO_2]}{\Delta t} $$

Answer

If the reaction is elementary reaction then order and molecularity have same value because elementary reaction proceeds in a single step.

Answer

Rate of any elementary reaction can be represented as

$$ r=k[A]^{n} $$

After changing concentration to its triple value $A=3 A, r$ becomes $27 r$

$$ \begin{aligned} & 27 r=k[3 A]^{n} \\ & \frac{r}{27 r}=\frac{k[A]^{n}}{k[3 A]^{n}} \\ & \frac{1}{27}=\left[\frac{1}{3}\right]^{n} \Rightarrow \left[\frac{1}{3}\right]^{3}= \left[\frac{1}{3}\right]^{n} \end{aligned} $$

Hence, $n=3$

Order of reaction is three.

Answer

For zero order reaction $[R]=[R]_0-k t$

For completion of the reaction $[R]=0$

$$ \therefore \quad t=\frac{[R]_0}{k} $$

Answer

During an elementary reaction, the number or atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction, the order of reaction with respect to $B$ would have been 1 , but in the given rate law it is $\frac{3}{2}$. This indicates that the reaction is not an elementary reaction. Hence, this reaction must be a complex reaction.

Answer

According to collision theory apart from the energy considerations, the colliding molecules should also have proper orientation for effective collision.

This condition might not be getting fulfilled in the reaction as it shows the number of reactants taking part in a reaction, which can never be zero.

Answer

No, the molecularity can never be zero or a fractional number as it shows the number of reactants taking part in a reaction which can never be zero.

Answer

(i) For $A \longrightarrow B$ the given graph shows a zero order reaction. Mathematically represented as

$$ [R]=-k t+[R]_0 $$

Which is equation of straight line. Hence, reaction is a zero order.

(ii) Slope $=-k$

(iii) Unit of zero order reaction is mole $L^{-1} S^{-1}$

Answer

Because activation energy of the reaction is very high at room temperature but at high temperature $H-H$ and $O-O$ bond break and colliding particles cross the energy barrier. This is why reaction between $H_2(g)$ and $O_2(g)$ does not lead to formation of water at room temperature while keeping in the same vessel.

Answer

At higher temperatures, larger fraction of colliding particles can cross the energy barrier (i.e., the activation energy) which leads to faster rate.

Answer

For combustion reactions, activation energy of fuels is very high at room temperature. So, fuels do not burn by themselves at room temperature.

Answer

According to collision theory, we know that to complete any chemical reaction there must be effective collision between reactant particles and they must have minimum sufficient energy. The probability of more than three molecules colliding simultaneously is very small. Hence, possibility of molecularity being three is very low.

Answer

The rate of a reaction depends on the concentration of the reactants. As the reaction proceeds in forward direction, concentration of reactant decreases and that of products increases. So, the rate of reaction generally decreases during the course of reaction.

Answer

Thermodynamically the conversion or diamond to graphite is highly feasible but this reaction is very slow because its activation energy is high.

Hence, thermodynamic feasibility of the reaction alone cannot decide the rate of reaction.

Answer

As we know with increase in temperature rate of reaction increases, Hence, we heat oxalic acid solution before starting of titration to increase the rate of decolourisation.

Answer

Molecularity of the reaction is the number of molecules taking part in an elementary step. For this we require at least a single molecule leading to the value of minimum molecularity of one. Hence, molecularity of any reaction can never be equal to zero.

Answer

A complex reaction occurs through a number of steps i.e., elementary reactions. Number of molecules involved in each elementary reaction may be different, i.e., the molecularity of each step may be different. Therefore, it is meaningless to talk of molecularity of the overall complex reaction.

On the other hand, order of complex reaction depends upon the molecularity of the slowest step. Hence, it is not meaningless to talk of the order of a complex reaction.

Answer

Balanced chemical equation often leads to incorrect order or rate law. e.g., the following reaction seems to be a tenth order reaction

$$ KClO_3+6 FeSO_4+3 H_2 SO_4 \longrightarrow KCl+3 H_2 O+3 Fe_2(SO_4)_3 $$

This is actually a second order reaction. Actually the reaction is complex and occurs in several steps. The order of such reaction is determined by the slowest step in the reaction mechanism.

Order is determined experimentally and is confined to the dependence of observed rate of reaction on the concentration of reactants.

Answer

A. $\rightarrow(1)$

B. $\rightarrow(2)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$ For zero order reaction rate equation may be written as

$$ [R]=-k t+[R_0] $$

Which denotes a straight line equation similar to $y=m x+c$

On transforming (i) $\quad \frac{[R]-[R_0]}{t}=-k$

$$ k=\frac{[R_0]-[R]}{t} $$

Rate $=k \cdot[t]^{0}$

Rate $\propto[t]^{0}$

For a first order reaction $\frac{d x}{d t} \propto$ [concentration]

$\therefore$ Graph between rate and concentration may be drawn as

$$ \begin{aligned} k & =\frac{2.303}{t} \log \frac{[R]_0}{[R]} \\ \frac{k t}{2.303} & =\log [\frac{R^{\circ}}{R}] \\ \frac{k t}{2.303}= & \log [R]_0-\log [R] \\ \log [R] & =\underset{\text { slope }}{(\frac{-k}{2.303}) t}+\underset{\text { intercept }}{\log [R]_0} \end{aligned} $$

Answer

A. $\rightarrow(3) \quad$

B. $\rightarrow(1) \quad$

C. $\rightarrow(4) \quad$

D. $\rightarrow(6) \quad$

E. $\rightarrow(2) \quad$

F. $\rightarrow(5)$

1. Catalyst alters the rate of reaction by lowering activation energy.

2. Molecularity can’t be fraction or zero. If molecularity is zero, then reaction is not possible.

3. Second half-life of first order reaction is same as first because half-life time is temperature independent.

4. $e^{-E_{a} / R T}$ refers to the fraction of molecules with kinetic energy equal to or greater than activation energy.

5. Energetically favourable reactions are sometimes slow due to improper orientation of molecule cause some ineffective collision of molecules.

6. Area under the Maxwell, Boltzmann curve is constant because total probability of molecule taking part in a chemical reaction is equal to one.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(1)$

C. $\rightarrow(3)$

1. Diamond can’t be converted into graphite under ordinary condition.

2. Instantaneous rate of reaction completes at very short span of time.

3. Average rate of reaction occurs to a long duration of time.

Answer

A. $\rightarrow(2) $

B $\rightarrow(1) \quad$

C $\rightarrow(4) \quad$

D $\rightarrow(3)$

1. Mathematical expression for rate of reaction is known as rate law.

2. Rate of reaction for zero order reaction is equal to rate constant

$$ \begin{aligned} & r & =k[A]^{0} \\ \therefore \quad & r & =k \end{aligned} $$

3. Unit of rate of reaction is same as that of rate of reaction.

4. Order of complex reaction is determined by rate of a reaction, which is slowest.

Answer

(b) Both assertion and reason are correct, but the reason is not the correct explanation of assertion.

Order of reaction can be zero or fractional as order of reaction is directly related to sum of power of reactants. Reason is a correct statement but not correct explanation.

Answer

(e) Assertion is incorrect and reason is correct.

Order and molecularity may or may not be same as order of reaction is sum of power of reactant which can be determined experimentally. But molecularity is sum of stoichiometric coefficient of rate determining elementary step.

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

Enthalpy of reaction i.e., difference of total enthalpy of reactants and product remains constant in the presence of a catalyst. As a catalyst participating in the reaction forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains same.

Answer

(e) Assertion is incorrect, but reason is correct.

Correct assertion is “only effective collision lead to formation of product.” Reason defines correct meaning of effective collision, and criterion of collision theory for completion of reaction.

Only those collisions in which molecules have correct orientation and sufficient energy lead to formation of product.

Answer

(c) Assertion is correct, but reason is incorrect.

Rate constant determined from Arrhenius equation are fairly accurate for simple and complex molecules because only those molecules which have proper orientation during collision (i.e., effective collision) and sufficient kinetic energy lead the chemical change.

Answer

Only effective collision lead to the formation of products. It means that collisions in which molecules collide with sufficient kinetic energy (called threshold energy = activation energy + energy possessed by reacting species).

And proper orientation lead to a chemical change because it facilitates the breaking of old bonds between (reactant) molecules and formation of the new ones i.e., in products.

e.g., formation of methanol from bromomethane depends upon the orientation of the reactant molecules.

The proper orientation of reactant molecules leads to bond formation whereas improper orientation makes them simply back and no products are formed.

To account for effective collisions, another factor $P$ (probability or steric factor) is introduced $K=P z_{A B} e^{-E a / R T}$.

Answer

Kinetic energy is directly proportional to the absolute temperature and the number of molecules possessing higher energies increases with increase in temperature, i.e., most probable kinetic energy increases with increase in temperature.

Energy of activation is related to temperature by the following Arrhenius equation

$$ k=A e^{-E_{a} / R T} $$

Thus, it also shows an increase with rise in temperature.

Answer

A catalyst is a s’ubstance which increases the speed of a reaction without itself undergoing any chemical change.

According to “intermediate complex formation theory” reactants first combine with the catalyst to form an intermediate complex which is short-lived and decomposes to form the products and regenerating the catalyst.

The intermediate formed has much lower potential energy than the intermediate complex formed between the reactants in the absence of the catalyst.

Thus, the presence of catalyst lowers the potential energy barrier and the reaction follows a new alternate pathway which require less activation energy.

We know that, lower the activation energy, faster is the reaction because more reactant molecules can cross the energy barrier and change into products.

Enthalpy, $\Delta H$ is a state function. Enthalpy of reaction, i.e., difference in energy between reactants and product is constant, which is clear from potential energy diagram.

Potential energy diagram of catalysed reaction is given as

Answer

The difference between instantaneous rate of reaction and averatge rate of a reaction are as below

Answer

A reaction in which one reactant is present in large amount and its concentration does not get altered during the course of the reaction, behaves as first order reaction. Such reaction is called pseudo first order reaction.

e.g., (i) hydrolysis of ethyl acetate

$ \begin{array}{llllllll} \text{ conc. }& \mathrm{CH}_3 \mathrm{COOC_2H_5} & + &\mathrm{H}_2 \mathrm{O} & \xrightarrow{\mathrm{H}^{+}} & \mathrm{CH}_3 \mathrm{COOH} & + & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \\ \text{Conc.} t= 0 & 0.01 \mathrm{~mol} & & 10 \mathrm{~mol} && 0 \mathrm{~mol} & & 0 \mathrm{~mol} \\ \text{Conc. at } t & 0 \mathrm{~mol} & & 9.99 \mathrm{~mol} && 0.01 \mathrm{~mol} & & 0.01 \mathrm{~mol} \end{array} $

where, $k=k^{\prime}[H_2 O]$

e.g., (ii) inversion of cane sugar

$$ C_{12} H_{22} O_{11}+H_2 O \xrightarrow{H^{+}} C_6 H_{12} O_6+C_6 H_{12} O_6 $$

Rate of reaction $=k[C_{12} H_{22} O_{11}]$

where $\quad k=k^{\prime}[H_2 O]$