Question

# For a general chemical equation A+B⇌C+D the equilibrium constant can be expressed as a ratio of...

For a general chemical equation

A+B⇌C+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 rate=kf[A][B]

reverse rate=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=kf/kr=[C][D]/[A][B]

1.For a certain reaction, Kc = 1.43×1010 and kf= 6.45×10−4 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.

2.For a different reaction, Kc = 1.88×107, kf=8.92×105s−1, and kr= 4.75×10−2 s−1 . Adding a catalyst increases the forward rate constant to 1.34×107 s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

3.Yet another reaction has an equilibrium constant Kc=4.32×105 at 25 ∘C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to200 ∘C , what will happen to the equilibrium constant? (increase/decrease/not change)

1.

Kc = 1.43 ×1010 and kf= 6.45×10−4 M−2s−1

kr = kf/Kc = (6.45 ×10−4 M−2s−1) / (1.43×1010) = 4.51 ×10−14 M−2s−1

2.

Kc = 1.88×107 and kf= 1.34×107 s−1

kr = kf/Kc = (1.34 ×107 s−1) / (1.88×107) = 7.12 ×10−1 s−1

3.  Raising the temperature to 200 C favors the reverse reaction (endothermic).

Equilibrium constant changes if we change the temperature of the system. Kc is constant at constant temperature, but they vary as the temperature changes. On increasing temperature in is an exothermic reaction, the value of Kc decreases.