Chemical Equilibrium and Chemical Kinetics Learning Goal: To understand the rela

Question

Chemical Equilibrium and Chemical Kinetics

Learning Goal:

To understand the relationship between the equilibrium constant and rate constants.

For a general chemical equation

A+BC+D

the equilibrium constant can be expressed as a ratio of the concentrations:

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

If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows:

forward ratereverse rate==kf[A][B]kr[C][D]

where kf and kr are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal:

kf[A][B]=kr[C][D]

Thus, the rate constants are related to the equilibrium constant in the following manner:

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

Part A: For a certain reaction, Kc= 674 and kf= 8.99 M2s1 . Calculate the value of the reverse rate constant, kr.

Part B:For a different reaction, Kc=2.19×106, kf=7.81×103s1, and kr= 3.57×103 s1 . Adding a catalyst increases the forward rate constant to 2.41×106 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

Explanation / Answer

A) k(r) = k(f) / K(c)

k(r) = 1.33 x 10^-3

B) Catalyst does not change the value of the equilibrium constant K(c)

So the relation
K(c) = k(f)/k(r) with K(c) = 2.19 x 10 ^6
holds for rate constants of catalyzed reaction

Hence:

k(r) = k(f) / K(c)
= 2.41 X10 ^6s¹ / 2.19 X 10 ^ 6 /S
k(r) =1.10

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