Question:

The experimental data for decomposition of N₂O₅ [2N₂O₅→4NO₂+O₂] in gas phase at 318K are given below: t/s 0 400 800 1200 1600 2000 2400 2800 3200 10²×[N₂O₅]/molL^-1 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 (i) Plot [N₂O₅] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N₂O₅] and t. (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii)

Answer:

(i) Plot [N₂O₅] against t.

(ii) Find the half-life period for the reaction.

Half-life period = t1/2 = 2400 s

(iii) Draw a graph between log[N₂O₅] and t.

(iv) What is the rate law?

Rate law = -d[N₂O₅]/dt = k[N₂O₅]

(v) Calculate the rate constant.

k = -(1/[N₂O₅])*(d[N₂O₅]/dt)

k = -(0.35-1.63)/(3200-0)

k = -0.004375 mol L^-1 s^-1

(vi) Calculate the half-life period from k and compare it with (ii).

Half-life period = ln(2)/k

Half-life period = ln(2)/(-0.004375)

Half-life period = 1580.7 s

The calculated half-life period (1580.7 s) is close to the measured half-life period (2400 s).

Question:

The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Answer:

Step 1: Write down the rate equation for a first order reaction.

Rate = -k[A]

Step 2: Substitute the given rate constant into the rate equation.

Rate = -60 s^-1[A]

Step 3: Calculate the time it will take for the concentration of the reactant to be reduced to 1/16th of its initial value.

Time = ln(1/16)/(-60 s^-1)

Time = 8.86 s

Question:

A reaction is second order with respect to a reaction. How is the rate of reaction affected if the concentration of the reactant is : (a) doubled, (b) reduced to 1/2?

Answer:

A. (a) The rate of reaction will be doubled. B. (b) The rate of reaction will be reduced to 1/4.

Question:

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 2NO(g)→N₂O(g); Rate=k[NO]² (ii) H2₂​O₂(aq)+3I^-(aq)+2H^+→2H2₂O(I)+3I; Rate=k[H2₂​O₂][I^-] (iii) CH₃CHO(g)→CH4​(g)+CO(g); Rate=k[CH₃CHO]^3/2 (iv) C₂H₅Cl(g)→C₂H₄(g)+HCl(g); Rate=k[C₂H₅Cl]

Answer:

(i) Order of reaction: 2 Dimensions of rate constant: M/s

(ii) Order of reaction: 2 Dimensions of rate constant: M^-1 s^-1

(iii) Order of reaction: 3/2 Dimensions of rate constant: M^-3/2 s^-1

(iv) Order of reaction: 1 Dimensions of rate constant: s^-1

Question:

Calculate the half life of a first order reaction from their rate constants given below: (a) 200s^-1; (b) 2 min^-1; (c) 4year^-1.

Answer:

Step 1: Understand the definition of a half-life. The half-life of a reaction is the time it takes for the reactant concentration to decrease by half.

Step 2: Calculate the half-life for each rate constant given.

(a) For a rate constant of 200s^-1, the half-life is 0.5 seconds.

(b) For a rate constant of 2 min^-1, the half-life is 30 seconds.

(c) For a rate constant of 4year^-1, the half-life is 182.5 days.

Question:

In a reaction between A and B, the initial rate of reaction (r0​) was measured for different initial concentrations of A and B as given below: A/molL^-1 0.20 0.20 0.40 B/molL^-1 0.30 0.10 0.05 r0​/molL^-1s^-1 5.07×10^-5 5.07×10^-5 1.43×10^-4 What is the order of the reaction with respect to A and B?

Answer:

1. Calculate the rate of reaction for each set of initial concentrations of A and B.
2. Calculate the change in rate of reaction for each set of initial concentrations of A and B.
3. Calculate the change in concentration of A and B for each set of initial concentrations.
4. Calculate the ratio of the change in rate of reaction to the change in concentration of A and B for each set of initial concentrations.
5. Compare the ratios for each set of initial concentrations and determine the order of the reaction with respect to A and B.

Question:

The rate constant for the first order decomposition of H₂​O₂ is given by the following equation: logk=14.34−1.25×10⁴K/T Calculate Ea​ for this reaction and at what temperature will its half-period be 256 minutes?

Answer:

Step 1: Calculate Ea: Ea = -1.25 x 10⁴K/ln(k)

Step 2: Calculate the rate constant (k): k = e^(14.34 - 1.25 x 10⁴K/T)

Step 3: Calculate the temperature (T): T = 1.25 x 10⁴K/ (14.34 - ln(k))

Step 4: Calculate the half-period (t1/2): t1/2 = (ln(2))/k

Step 5: Substitute the value of k into the equation for t1/2: t1/2 = (ln(2))/e^(14.34 - 1.25 x 10⁴K/T)

Step 6: Solve for T: T = 1.25 x 10⁴K/ (14.34 - ln(2/t1/2))

Step 7: Substitute the value of t1/2 (256 minutes): T = 1.25 x 10⁴K/ (14.34 - ln(2/256))

Step 8: Calculate the temperature: T = 897.9 K

Question:

The rate constant for the decomposition of hydrocarbons is 2.418×10^-5s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

Answer:

Answer:

Step 1: Calculate the Arrhenius equation which is given by k = Ae^(-Ea/RT).

Step 2: Substitute the given values in the equation to calculate the pre-exponential factor (A).

A = 2.418 × 10^-5 × e^{(179.9 × 103/8.314 × 546)}

Step 3: Calculate the value of pre-exponential factor (A).

A = 1.30 × 10^13 s^-1

Question:

For the decomposition of azo-isopropane to hexane and nitrogen at 543K, the following data are obtained. t(sec) P(mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant.

Answer:

1. Calculate the pressure change for each time interval:

Time interval 0-360 sec: ΔP = 54.0 - 35.0 = 19.0 mm of Hg Time interval 360-720 sec: ΔP = 63.0 - 54.0 = 9.0 mm of Hg

2. Calculate the rate constant:

k = ΔP/Δt

For the first time interval:

k = 19.0/360 = 0.0527 mm of Hg/sec

For the second time interval:

k = 9.0/360 = 0.0250 mm of Hg/sec

The average rate constant is (0.0527 + 0.0250)/2 = 0.0388 mm of Hg/sec.

Question:

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Answer:

Step 1: Let the time required for completion of 90% of reaction be ’t'.

Step 2: According to the definition of a first order reaction, the rate of reaction is directly proportional to the concentration of the reactant.

Step 3: Therefore, the rate of reaction at 90% completion (t) will be half the rate of reaction at 99% completion (2t).

Step 4: Since the rate of reaction is directly proportional to the time taken for completion, the time required for 99% completion will be twice the time required for 90% completion (2t).

Question:

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4×10¹⁰s^-1. Calculate k at 318K and Ea​.

Answer:

Step 1: Calculate the rate constant (k) at 298K. k = A * e^(-Ea/RT) k = 4×10¹⁰s^-1 * e^{(-Ea/8.314*298)}

Step 2: Calculate the rate constant (k) at 308K. k = 4×10¹⁰s^-1 * e^{(-Ea/8.314*308)}

Step 3: Calculate the ratio of the rate constants at 298K and 308K. k298/k308 = (4×10¹⁰s^-1 * e^{(-Ea/8.314298)}) / (4×10¹⁰s^-1 * e^(-Ea/8.314308))

Step 4: Calculate the ratio of the time required for 10% completion of a first order reaction at 298K and 25% completion of the same reaction at 308K. Time298/Time308 = 0.1/0.25

Step 5: Equate the ratio of the rate constants and the ratio of the times required for completion. (4×10¹⁰s^-1 * e^(-Ea/8.314298)) / (4×10¹⁰s^-1 * e^(-Ea/8.314308)) = 0.1/0.25

Step 6: Solve for Ea. Ea = -8.314 * ln(0.4)

Step 7: Calculate the rate constant (k) at 318K. k = 4×10¹⁰s^-1 * e^{(-Ea/8.314*318)}

Step 8: Calculate the activation energy (Ea) at 318K. Ea = -8.314 * ln(k/4×10¹⁰s^-1)

Question:

The following results have been obtained during the kinetic studies of the reaction: 2A+B→C+D Experiment [A]/molL^-1 [B]/molL^-1 Initial rate of formation of D/molL^-1min^-1 I 0.1 0.1 6.0×10^-3 II 0.3 0.2 7.2×10^-2 III 0.3 0.4 2.88×10^-1 IV 0.4 0.1 2.40×10^-2 Determine the rate law and the rate constant for the reaction.

Answer:

1. Calculate the initial rate of reaction for each experiment using the equation: rate = (change in concentration of product/change in time).

2. Plot the initial rate of reaction (y-axis) against the concentration of reactant A (x-axis).

3. Draw a line of best fit through the points on the graph.

4. The slope of the line is equal to the rate constant, k.

5. The rate law for the reaction is rate = k[A]^m[B]n, where m and n are the exponents for the reactants A and B.

6. Calculate the exponents m and n by substituting the rate constants and concentrations obtained from the experiments into the rate law.

7. The rate law and the rate constant for the reaction are: rate = k[A]^m[B]n, where k = slope of the line of best fit, m = exponents for A, and n = exponents for B.

Question:

During nuclear explosion, one of the products is 90Sr with half life of 28.1 yr. If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 year and 60 year if it is not lost metabolically.

Answer:

1. Calculate the decay constant (λ): λ = 0.693 / 28.1 = 0.0247 yr^-1

2. Calculate the amount of 90Sr remaining after 10 years: Amount remaining after 10 years = 1μg x (1/2)^{(λ x 10)} = 0.732μg

3. Calculate the amount of 90Sr remaining after 60 years: Amount remaining after 60 years = 1μg x (1/2)^{(λ x 60)} = 0.039μg

Question:

In a pseudo first order hydrolysis of ester in water the following results were obtained: ^t/s 0 30 60 90 [Ester]/M. 0.55 0.31 0.17 0.085 (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer:

(i) Average rate of reaction between 30 and 60 seconds:

Rate = (Change in [Ester]/Change in time) Rate = (0.31 M - 0.17 M/ 30s - 0.00s) Rate = 0.14 M/s

(ii) Pseudo first order rate constant (k):

k = -(Change in [Ester]/Change in time)/[Ester] k = -(0.31 M - 0.17 M/30s - 0.00s)/0.31 M k = -0.14/0.31 k = 0.45 s^-1

Question:

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Answer:

Answer:

1. The effect of temperature on the rate constant of a reaction is that an increase in temperature generally increases the rate constant of a reaction.

2. This effect of temperature on rate constant can be represented quantitatively by the Arrhenius equation, which states that the rate constant of a reaction is proportional to the exponential of the inverse of the absolute temperature.

Question:

The rate for the decomposition of NH₃​ on platinum surface is zero order. What are the rate of production of N₂ and H₂​ if K=2.5×10^-4mol litre^-1s^-1.

Answer:

Step 1: The rate of the decomposition of NH₃ is zero order, which means that the rate of the reaction is independent of the concentration of NH₃.

Step 2: The rate of production of N₂ and H₂ can be calculated using the rate constant K.

Step 3: The rate of production of N₂ is equal to the rate constant K multiplied by the concentration of NH₃.

Rate of production of N₂ = K x [NH₃] = 2.5 x 10^-4 mol litre^-1s^-1 x [NH₃]

Step 4: The rate of production of H₂ is equal to the rate constant K multiplied by twice the concentration of NH₃.

Rate of production of H₂ = K x 2[NH₃] = 2.5 x 10^-4 mol litre^-1s^-1 x 2[NH₃]

Question:

The decomposition of hydrocarbon follows the equation, k=(4.5×1011s−1)e−28000K/T. Calculate Ea​.

Answer:

Given: k = (4.5 x 10¹¹ s^-1) e^(-28000K/T)

Step 1: Rearrange the equation to solve for e^(-28000K/T).

e^(-28000K/T) = k/(4.5 x 10¹¹ s^-1)

Step 2: Take the natural logarithm of both sides of the equation.

ln(e^(-28000K/T)) = ln(k/(4.5 x 10¹¹ s^-1))

Step 3: Simplify the equation.

-28000K/T = ln(k/(4.5 x 10¹¹ s^-1))

Step 4: Multiply both sides of the equation by T.

-28000K = T*ln(k/(4.5 x 10¹¹ s^-1))

Step 5: Divide both sides of the equation by -28000.

K = -T*ln(k/(4.5 x 10¹¹ s^-1))/28000

Step 6: Multiply both sides of the equation by -1.

Ea = T*ln(k/(4.5 x 10¹¹ s^-1))/28000

Question:

A reaction is first order in A and second order in B : (i) Write differential rate equation. (ii) How is the rate affected when the concentration of B is tripled? (iii) How is the rate affected when the concentration of both A and B is doubled?

Answer:

(i) Differential rate equation: d[A]/dt = -k[A][B]

(ii) When the concentration of B is tripled, the rate of the reaction will be tripled as well.

(iii) When the concentration of both A and B is doubled, the rate of the reaction will be quadrupled.

Question:

The rate constant for the decomposition of N₂O₅ at various temperatures is given below: T/°C 0 20 40 60 80
10⁵×k/s^-1 0.0787 1.70 25.7 178 2140 Draw a graph between ln k and 1/T and calculate the values of A and Ea​. Predict the rate constant at 30° and 50°C.

Answer:

1. Use the given table to plot a graph of ln k versus 1/T.

2. Use the graph to calculate the values of A and Ea.

3. Calculate the rate constant at 30°C and 50°C using the equation k = A e^(-Ea/RT).

4. Substitute the values of A, Ea, R, and T into the equation to calculate the rate constants at 30°C and 50°C.

Question:

A first order reaction takes 40 min for 30% decomposition. Calculate t1/2​.

Answer:

Step 1: Understand the problem.

We are being asked to calculate the half-life (t1/2) of a first order reaction that takes 40 minutes for 30% decomposition.

Step 2: Calculate the rate constant (k).

The rate constant (k) can be calculated using the following equation: k = -ln(1 - (x/100)) / t where x is the percentage of decomposition and t is the time taken for the reaction.

Plugging in the values from the problem, we get: k = -ln(1 - (30/100)) / 40 k = 0.0045 min^-1

Step 3: Calculate the half-life (t1/2).

The half-life (t1/2) can be calculated using the following equation: t1/2 = ln(2) / k where k is the rate constant.

Plugging in the value for k from Step 2, we get: t1/2 = ln(2) / 0.0045 t1/2 = 154.6 min

Question:

The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume. SO₂Cl₂(g)→SO₂(g)+Cl₂(g) Experiment Time/s^-1 Total pressure/ atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm.

Answer:

Step 1: Calculate the rate of the reaction when total pressure is 0.5 atm.

Rate of reaction = (0.6-0.5)/100 = 0.001 atm/s

Step 2: Calculate the rate of the reaction when total pressure is 0.65 atm.

Rate of reaction = (0.65-0.5)/100 = 0.0015 atm/s

Question:

The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Answer:

1. Calculate the temperature difference between 293K and 313K: 313K - 293K = 20K

2. Calculate the ratio of the rate of the reaction at 313K to the rate of the reaction at 293K: 4:1

3. Calculate the energy of activation of the reaction: The energy of activation of the reaction is the same regardless of the temperature change.

Question:

The half-life for radioactive decay of 14°C is 5730 years. An archaeological artifact containing wood had only 80% of the 14°C found in a living tree. Estimate the age of the sample.

Answer:

1. Determine the half-life of 14°C. Answer: The half-life of 14°C is 5730 years.

2. Calculate the amount of 14°C present in a living tree. Answer: 100% of the 14°C is present in a living tree.

3. Calculate the amount of 14°C present in the archaeological artifact. Answer: 80% of the 14°C is present in the archaeological artifact.

4. Calculate the age of the sample using the half-life of 14°C. Answer: The age of the sample can be calculated by dividing the half-life of 14°C (5730 years) by the amount of 14°C present in the archaeological artifact (80%) to get the age of the sample, which is 7,163 years.

Question:

Consider a certain reaction A→ Products with, k=2.0×10^-2s^-1. Calculate the concentration of A remaining after 100 s, if the initial concentration of A is 1.0molL^-1.

Answer:

Answer: Step 1: The rate equation for this reaction is given by: Rate = -d[A]/dt = k[A]

Step 2: To calculate the concentration of A remaining after 100 s, we need to solve the differential equation: d[A]/dt = k[A]

Step 3: We can solve this equation using the separation of variables technique: Integral of d[A]/[A] = Integral of kdt

Step 4: After integrating both sides, we get: ln[A] = kt + C

Step 5: To determine the constant C, we can use the initial condition given in the problem: At t = 0, [A] = 1.0 mol/L

Step 6: Substituting t = 0 and [A] = 1.0 mol/L in the equation, we get: ln(1.0) = k(0) + C

Step 7: Solving for C, we get: C = ln(1.0)

Step 8: Substituting C and k in the equation, we get: ln[A] = 2.0 × 10^-2s^-1 × 100s + ln(1.0)

Step 9: Solving for [A], we get: [A] = e^{(2.0 × 10-2s-1 × 100s + ln(1.0))}

Step 10: Hence, the concentration of A remaining after 100 s is: [A] = e^{(2.0 × 10-2s-1 × 100s + ln(1.0))} = 0.3678 mol/L

Question:

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2​=3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Answer:

Step 1: Calculate the half-life of the sucrose sample, which is 3.00 hours.

Step 2: Calculate the fraction of the sample that remains after 8 hours by subtracting 8 hours from the half-life and then dividing the result by the half-life.

Fraction remaining = (3.00 - 8.00) / 3.00

Fraction remaining = -5.00 / 3.00

Fraction remaining = -1.67

Since the fraction remaining is negative, this means that all of the sucrose has decomposed after 8 hours.

Question:

Mention the factors that affect the rate of a chemical reaction.

Answer:

1. Concentration of reactants: The higher the concentration of reactants, the higher the rate of reaction.

2. Temperature: Increasing the temperature of a reaction increases the kinetic energy of the reactants, which increases the rate of reaction.

3. Presence of a catalyst: A catalyst provides an alternative pathway for the reaction to occur, which increases the rate of reaction.

4. Surface area: Increasing the surface area of the reactants increases the rate of reaction, as more particles are available for reaction.

5. Pressure: Increasing the pressure of a reaction increases the rate of reaction.

Question:

For the reaction: 2A+B→A2​B, the rate =k[A][B]² with k=2.0×10^-6mol^-2L²s^-1.
Calculate the initial rate of the reaction when [A]=0.1molL^-1,[B]=0.2 molL^-1. Calculate the rate of reaction after [A] is reduced to 0.06 molL^-1.

Answer:

Answer: Initial rate of reaction when [A]=0.1molL^-1 and [B]=0.2molL^-1: Rate = k[A][B]² = 2.0×10^-6mol^-2L²s^-1 × 0.1molL^-1 × 0.2molL^-1 = 4.0×10^-8molL^-1s^-1

Rate of reaction after [A] is reduced to 0.06molL^-1: Rate = k[A][B]² = 2.0×10^-6mol^-2L²s^-1 × 0.06molL^-1 × 0.2molL^-1 = 2.4×10^-8molL^-1s^-1

Question:

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/molL^-1 B]/molL^-1 Initial rate/ molL^-1min^-1 I 0.1 0.1 2.0×10^-2 II - 0.2 4.0×10^-2 III 0.4 0.4 - IV - 0.2 2.0×10^-2

Answer:

Experiment [A]/molL^-1 B]/molL^-1 Initial rate/ molL^-1min^-1 I 0.1 0.1 2.0×10^-2 II 0.2 0.2 4.0×10^-2 III 0.4 0.4 4.0×10^-2 IV 0.2 0.2 2.0×10^-2

Question:

The decomposition of A into product has value of k as 4.5×10³s^-1 at 10°C and energy of activation 60 kJmol^-1. At what temperature would k be 1.5×10⁴s^-1?

Answer:

1. Rearrange the given equation to isolate the variable ’temperature’: k = 4.5×10³s^-1 * e^(-60kJmol^-1/R * (1/T - 1/T_0))

2. Substitute the given values into the equation: k = 4.5×10³s^-1 * e^(-60kJmol^-1/R * (1/T - 1/293.15K))

3. Rearrange the equation to solve for ‘T’: T = 293.15K * (1 + (60kJmol^-1/R) * ln(4.5×10³s^-1/1.5×10⁴s^-1))^-1

4. Substitute the given values into the equation: T = 293.15K * (1 + (60kJmol^-1/(8.314JK^-1mol^-1)) * ln(4.5×10³s^-1/1.5×10⁴s^-1))^-1

5. Calculate the value of T: T = 293.15K * (1 + (7.235K^-1) * ln(3))^-1

6. Calculate the value of T: T = 293.15K * (1 + 21.705K^-1)^-1

7. Calculate the value of T: T = 293.15K * 0.0462

8. Calculate the value of T: T = 13.55K

Question:

The decomposition of dimethyl ether leads to the formation of CH4​,H₂​ and CO and the reaction rate is given by: Rate=k[CH₃OCH₃]^3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether. i.e., Rate=k[PCH₃OCH₃​]^3/2 If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Answer:

Units of rate: bar^3/2 min^-1

Units of rate constant: bar^3/2 min^-1 mol^-3/2