Answered: . For a certain reaction, Kc = 4.48×106…

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter13: Chemical Equilibrium

Section: Chapter Questions

Problem 2RQ: What is the law of mass action? Is it true that the value of K depends on the amounts of reactants...

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A. For a certain reaction, Kc = 4.48×10⁶ and kf= 4.10×10⁻² M−2⋅s-1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction.

Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1nclude ⋅⋅ (multiplication dot) between each measurement.

B. For a different reaction, Kc = 97.7, kf=3.00×10^3s−1 and kf=3.00×10^3s−1 , and kr= 30.7 s−1. Adding a catalyst increases the forward rate constant to 1.37×10⁶ s−1s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1 include ⋅ (multiplication dot) between each measurement.

Expert Solution

Step 1

Answer:

Equilibrium is a state for a system when rate of forward reaction becomes equal to rate of backward reaction. when we equate the rate of forward reaction to the rate of backward reaction, established by law of mass action, then ratio of forward rate constant and backward rate constant, is considered as a new constant called as equilibrium constant.

Since, equilibrium constant is the ratio of forward rate constant and reverse rate constant, therefore

$K_{C} = \frac{k_{f}}{k_{r}} K_{C} \times k_{r} = k_{f} k_{r} = \frac{k_{f}}{K_{c}} H e r e : K_{c} = e q u i l i b r i u m c o n s \tan t k_{f} = f o r w a r d r a t e c o n s \tan t k_{r} = r e v e r s e r a t e c o n s \tan t$

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