Chemistry Questions and Answers – Chemical Kinetics – Temperature Dependence of the Rate of a Reaction

«
»

This set of Chemistry 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
View Answer

Answer: a
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.
advertisement

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°
View Answer

Answer: a
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
View Answer

Answer: a
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.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
advertisement
advertisement

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
View Answer

Answer: b
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
View Answer

Answer: d
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.
advertisement

6. Which of the following represents the Boltzmann factor?
a) e-Ea
b) eEa
c) e-Ea/RT
d) eEa/RT
View Answer

Answer: c
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
View Answer

Answer: a
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.
advertisement

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
View Answer

Answer: c
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
View Answer

Answer: a
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.
advertisement

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
View Answer

Answer: b
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.

Sanfoundry Global Education & Learning Series – Chemistry – Class 12.

To practice all areas of Chemistry, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.