Question

Question: Consider the following reactions at some temperature:

For each reaction some quantities of the reactants were placed in separate containers and allowed to come to equilibrium. Describe the relative amounts of reactants and products that are present at equilibrium. At equilibrium, which is faster, the forward or reverse reaction in each case?

$\mathbf{2}\mathbf{NOCl}\left(g\right)\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{2}\mathbf{NO}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Cl}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$

$\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}{\mathbf{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{31}}$

For each reaction some quantities of the reactants were placed in separate containers and allowed to come to equilibrium. Describe the relative amounts of reactants and products that are present at equilibrium. At equilibrium, which is faster, the forward or reverse reaction in each case?

Short answer

Answer

For the first reaction, the amounts of products present at equilibrium are relatively low, and the relative amounts of reactants present are high due to the smaller equilibrium constant value.

For the second reaction, the amounts of products present at equilibrium are relatively high. The amounts of reactants present are relatively small because the equilibrium constant value is very large.

In each case, the forward reaction rate is the same as the rate of the reverse reaction at equilibrium.

Step-by-step solution

Expression for equilibrium constant

The given balanced chemical equation and its K value are shown below:

$2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)\mathrm{K}=1.6\times {10}^{-5}$

The expression for the equilibrium constant, K, in terms of concentration is written below:

$\mathrm{K}=\frac{{\left[\mathrm{NO}\right]}^2\left[{\mathrm{Cl}}_2\right]}{\left[\mathrm{NOCl}\right]}$ …(1)

The given equilibrium constant value is $1.6\times {10}^{-5}$which is less than 1. By using Equation (1), the concentration of nitrosyl chloride (reactant) present is high when compared to the concentrations of nitric oxide and chlorine (products).

Identifying the faster reaction

The given balanced chemical equation and its K value are rewritten below:

$2\mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\mathrm{K}=1\times {10}^{31}$

The expression for the equilibrium constant, K, in terms of concentrations is shown below:

$\mathrm{K}=\frac{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}{{\left[\mathrm{NO}\right]}^2}$ …(2)

The K value for this reaction is $\mathrm{K}=1\times {10}^{31}$which is much higher than 1. According to Equation (2), the amounts of nitrogen and oxygen present at equilibrium are very large, while the concentration of nitric oxide is relatively very low.

We know that at equilibrium, the forward reaction rate is the same as the rate of the reverse reaction. The given two reactions are equilibrium reactions. Hence, both reactions have the same rate in either direction.