Part A For a certain reaction, K c = 7.62×10 10 and k f= 77.9 M 2s1 . Calculate

Question

Part A

For a certain reaction, Kc = 7.62×1010 and kf= 77.9 M2s1 . 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 M2s1 include (multiplication dot) between each measurement.

Part B

For a different reaction, Kc = 2.62×105, kf=65.6s1, and kr= 2.50×104 s1 . Adding a catalyst increases the forward rate constant to 1.85×104 s1 . 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 M2s1 include (multiplication dot) between each measurement.

Explanation / Answer

For the given reaction

Kc = 7.62×1010 and kf= 77.9 M2s1

For a hypothetical equation A + B in equilibrium with C + D

Equilibrium constant

Kc = [C][D]/[A][B]

Or Kc = kf/kr

Given that, kf= 77.9 M2s1

That means the given reaction is second order. Also, given that the reverse reaction is of the same molecularity as the forward reaction. Thus kr is also having same units (M2s1)

Kc = 77.9 /kb

Or kr= kf/Kc = 77.9 M2s1/ 7.62×1010 (Kc doesn’t have a unit)

kr= 10.22 ×1010 M2s1

or

kr= 1.022 ×1011 M2s1

Part 2.

If are adding a catalyst to a reaction in equilibrium, it will enhance the rate of both the forward and backward reactions. Thus equilibrium constant will not vary or we can use the same equilibrium constant.

Given that Kc = 2.62×105 and the forward rate constant is 1.85×104 s1 in the presence of catalyst (From unit, we can understand the rate of the reaction. It is first order reaction)

Kc = kf/kr

Kc = 1.85×104 /kr

Or kr = 1.85×104 /2.62×105 = 0.7061 x 10-1 s1

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