Question

*** 8 Parts*** Write the equilibrium equation for the following reactions, and tell whether the reactants or products are favored in each case.

a) 2 CO(g) + O2(g) -------> 2 CO2(g) Keq = 2 x 10^11

b) Cl2O(g) + H2O(g) ------> 2 HOCl(g) Keq = 0.132

c) HF(aq) + H2O(l) ------> H3O+(aq) + F-(aq) Keq = 7.1 x 10^-4 

d) 3 O2(g) -----> 2 O3(g) Keq = 2.5 x 10^-29

e) S2(g) + 2 H2(g) -------> 2 H2S(g) Keq = 2.8 x 10^-21

f) CO(g) + 2 H2(g) -------> CH3OH(g) Keq = 10.5

g) Br2(g) + Cl2(g) ---------> 2 BrCl(g) Keq = 58.0

h) I2(g) -------> 2I(g) Keq = 6.8 x 10^-3

Answer 1

The rates of the forward and reverse reactions are the same when this system is at equilibrium.

At equilibrium:

rateforward

=

ratereverse

If k_f is the rate of the forward reaction and k_r is the rate of the reverse reaction then

Since k_f and k_r are constants, the ratio of k_f divided by k_r must also be a constant. This ratio is the equilibrium constant for the reaction, K_eq. The ratio of the concentrations of the reactants and products is known as the equilibrium constant expression.

Keq = k_f/k_r

To determine the amount of each compound that will be present at equilibrium you must know the Equilibrium Constant. To determine the equilibrium constant you must consider the chemical reaction written in the form:

aA+bB⇌cC+dD

The equilibrium constant is defined as:

K=(C)^c(D)^d/(A)^a(B)^b

Now for the reaction

a) 2 CO(g) + O₂(g) 2 CO₂(g) Keq = 2 x 10¹¹

Keq = [CO₂]²/[CO]²[O₂]

Since the value of Keq is 2 x 10¹¹

The forward reaction is favoured over the reverse so it is a highly products favoured reaction

b) Cl₂O(g) + H₂O(g) 2 HOCl(g) Keq = 0.132

Keq = [HOCl]²/[Cl₂O][H₂O]

Since in this case the Keq is 0.132 which is a fraction the rate constant for the reverse is higher than the rate constant for the forward reaction so it is moderately reactant favoured reaction

c)  HF(aq) + H₂O(l) H₃O⁺(aq) + F^-(aq) Keq = 7.1 x 10^{-4 }

Keq = [H₃O⁺][F^-]/[HF]

Since in this case the Keq is 7.1 x 10^{-4 } which is a fraction the rate constant for the reverse is higher than the rate constant for the forward reaction so it is highly reactant favoured reaction.