Question

Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl₂produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:

\(2NOCl(g) \rightleftharpoons 2NO(g) + C{l_2}(g)\quad {K_P} = 4.0 \times 1{0^{ - 4}}\)

Short answer

The equilibrium partial pressure of all species are

\(\begin{array}{*{20}{c}}{\,\,{{\rm{P}}_{{\rm{NO}}}} = 0.43{\rm{atm}}}\\{\,\,\,\,\,{{\rm{P}}_{{\rm{C}}{{\rm{l}}_2}}} = 0.215{\rm{atm}}}\\{{{\rm{P}}_{{\rm{NOCl}}}} = 9.57{\rm{atm}}}\end{array}\).

Step-by-step solution

Determine change in partial pressure:

Given information:

\(2{\text{NOCl}}({\text{g}}) \rightleftharpoons 2{\text{NO}}({\text{g}}) + {\text{C}}{{\text{l}}_2}({\text{g}})\)

• The initial partial pressure of NOCl is 10.0 atm
• The equilibrium constant is \({K_p} = 4.0 \times {10^{ - 4}}\).

The equilibrium partial pressure of all species needs to be obtained.

Let the change in pressure be x

We have to determine the value of x,

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_p} = \frac{{{{\left( {{P_{NO}}} \right)}^2} \times \left( {{P_{C{l_2}}}} \right)}}{{{{\left( {{P_{NOCl}}} \right)}^2}}}}\\{4.0 \times {{10}^{ - 4}} = \frac{{{{(2x)}^2} \times x}}{{{{(10.0 - 2x)}^2}}}}\end{array}\)

Since\(\,\,\,{K_p}\) is too small, we will neglect the change in pressure of NOCl, and get

\(\begin{array}{*{20}{c}}{4.0 \times {{10}^{ - 4}} = \frac{{{{(2x)}^2} \times x}}{{{{(10.0)}^2}}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{x^3} = 4.0 \times {{10}^{ - 2}}}\\{\,\,\,\,\,\,{x^3} = 0.01}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0.215{\rm{atm}}}\end{array}\)

Determine equilibrium partial pressure of all species:

The change in the partial pressure obtained \(x = 0.215{\rm{atm}}\)

Therefore, The equilibrium partial pressure of all species

\(\begin{array}{*{20}{c}}{{{\rm{P}}_{{\rm{NO}}}} = 2{\rm{x}} = 0.43{\rm{atm}}}\\{{{\rm{P}}_{{\rm{C}}{{\rm{l}}_2}}} = {\rm{x}} = 0.215{\rm{atm}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\rm{P}}_{{\rm{NOCl}}}} = 10.0{\rm{atm}} - 2{\rm{x}} = 9.57{\rm{atm}}}\end{array}\).