# Class 12 Chemistry MCQ – Chemical Kinetics – Temperature Dependence of the Rate of a Reaction

This set of Class 12 Chemistry Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Chemical Kinetics – Temperature Dependence of the Rate of a Reaction”.

1. The rate of a reaction depends on the temperature.
a) True
b) False

Explanation: For most of the reactions, the rate of reaction becomes nearly double or even more for 10° rise of temperature. The effect of temperature is usually expressed in terms of temperature coefficient.

2. Which of the following is the correct expression for the temperature coefficient (n)?
a) n = Rate constant at T + 10°/Rate constant at T°
b) n = Rate constant at T + 20°/Rate constant at T°
c) n = Rate constant at T + 30°/Rate constant at T°
d)n = Rate constant at T + 40°/Rate constant at T°

Explanation: The effect of temperature on the rate of a reaction is usually expressed in terms of the temperature coefficient which is defined by the equation:
Temperature coefficient (n) = Rate constant at T + 10° (308 K)/Rate constant at T° (298 K).

3. The rate of reaction increases with a rise in temperature.
a) True
b) False

Explanation: The increase in the rate of reaction with a rise in temperature is not due to the increase in the total number of collisions but mainly due to an increase in the total number of effective collisions.

4. What happens to the peak of the curve in the Maxwell-Boltzmann distribution graph if temperature increases?
a) Shifts forward and upward
b) Shifts forward and downward
c) Shifts backward and upward
d) Shifts backward and downward

Explanation: With the increase of temperature, the peak shifts forward but downward. This is because with the increase of temperature, the most probable kinetic energy increases and the fraction of molecules possessing most probable kinetic energy decreases.

5. Which of the following is the correct Arrhenius equation?
a) k = A eEa/RT
b) k = A eEa/T
c) k = A eEa/R
d) k = A e-Ea/RT

Explanation: Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k, was proposed by Arrhenius in 1889. The equation, called Arrhenius equation, is usually written in the form k = A e-Ea/RT.
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6. Which of the following represents the Boltzmann factor?
a) e-Ea
b) eEa
c) e-Ea/RT
d) eEa/RT

Explanation: The factor e-Ea/RT in the Arrhenius equation is called Boltzmann factor. It represents the fraction of the molecules (NE/NT) having energy equal to or greater than E where NE represents the number of molecules with energy E and NT represents the total number of molecules.

7. What is R in the equation k = Ae-Ea/RT?
a) R = 8.314 J K-1 mol-1
b) R = 3.184 J K-1 mol-1
c) R = 4.318 J K-1 mol-1
d) R = 1.438 J K-1 mol-1

Explanation: Saline medium has extra salts such as sodium chloride dissolved in water. It has a greater concentration of electrolyte than ordinary medium. The ions present will favour the formation of more electrochemical cells and favour the transfer of hydrogen ions and will thus promote rusting or corrosion.

8. The activation energy of a reaction is 50 kJ mol-1 and the value of rate constant at 300 K is 2.5×10-5 sec-1. What is the value of the frequency factor, A?
a) 4228.53 s-1
b) 3829.69 s-1
c) 7596.45 s-1
d) 6565.35 s-1

Explanation: Given,
Ea = 50 kJ mol-1
T = 300 K
k = 1.5 × 10-5 sec-1
R = 8.314 J K-1 mol-1
log k = – Ea / (2.303 RT) + log A
OR
log A = log k + Ea / (2.303 RT)
log A = log (1.5 × 10-5) + 50000 / (2.303 × 8.314 × 300)
log A = 3.88061
A = antilog (3.880611)
A = 7596.45 s-1.

9. What is the value of rate constant k if the value of the activation energy Ea and the frequency factor A are 49 kJ / mol and 9 × 1010 s-1 respectively? (T = 313 K)
a) 6 × 102 s-1
b) 9 × 102 s-1
c) 6 × 10-2 s-1
d) 3 × 102 s-1

Explanation: Given,
Ea = 49 kJ / mol-1
T = 313 K
A = 9 × 102 s-1
R = 8.314 J K-1 mol-1
log k = – Ea / (2.303 RT) + log A
log k = – 49000 / (2.303 × 8.314 × 313) + log 9 × 1010
log k = 2.77843
k= antilog (2.77843)
k = 6 × 102 s-1.

10. The rate constant of a reaction is 6 × 10-3 s-1 at 50° and 9 × 10-3 s-1 at 100° C. Calculate the energy of activation of the reaction.
a) 6.123 kJ mol-1
b) 8.124 kJ mol-1
c) 12.357 kJ mol-1
d) 18.256 kJ mol-1

Explanation: Given,
k1=6 × 10-3s-1 T1=50 + 273=323 K
k2=9 × 10-3s-1 T2=100 + 273=373 K
Substituting these values in the equation:
log (k2 / k1 ) = (Ea / 2.303 R) × ((T2 – T1) / T1 T2), we get
log (9 × 10-3 s-1 / 6 × 10-3 s-1 ) = ((Ea / (2.303 × 8.314)) × ((373 – 323) / (373 × 323))
log 9 / 6 = ((Ea / (2.303 × 8.314)) × (50 / (373 × 323))
Ea = 8.124 kJ mol-1.

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