Question

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)

Answer 1

1.

K_c = 1.43 ×10¹⁰ and k_f= 6.45×10⁻⁴ M⁻²s⁻¹

k_r = k_f/K_c = (6.45 ×10⁻⁴ M⁻²s⁻¹) / (1.43×10¹⁰) = 4.51 ×10⁻¹⁴ M⁻²s⁻¹

2.

K_c = 1.88×10⁷ and k_f= 1.34×10⁷ s⁻¹

k_r = k_f/K_c = (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. K_c is constant at constant temperature, but they vary as the temperature changes. On increasing temperature in is an exothermic reaction, the value of K_c decreases.