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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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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

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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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