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.

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.

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.

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.

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)

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

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

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

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

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

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 $=\ln A$.

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.

Answer

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

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

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.

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

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.

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

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

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

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

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.

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.

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.

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.

Answer

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

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

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.

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.

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).

Answer

$(a, c, d)$

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

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 $=3$

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.

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.

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.

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.

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.

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.

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.

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.

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)

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

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.

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