Question: Calculate the pressures of NO, Cl₂, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl₂. (Hint: K_P is small; assume the reverse reaction goes to completion then comes back to equilibrium.)

Given information:

• The partial pressure of NO is 4.0 atm
• The partial pressure of Cl₂ is 2.0 atm
• The partial pressure of\({K_p} = 2.5 \cdot {10^3}\)

We have to find the pressure of NO, Cl₂, and NOCl is an equilibrium mixture.

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

Since K_p is small, we will assume that the reverse reaction goes to completion then comes back to equilibrium.

Therefore, the initial partial pressure of NOCl is 0 atm.

We have to determine the value of x,

\(\begin{array}{*{20}{c}}{{K_p} = \frac{{{{\left( {{P_{NOCl}}} \right)}^2}}}{{{{\left( {{P_{NO}}} \right)}^2} \times \left( {{P_{C{l_2}}}} \right)}}}\\{2.5 \cdot {{10}^3} = \frac{{{{(2x)}^2}}}{{{{(4 - 2x)}^2} \times (2 - x)}}}\\{2.5 \cdot {{10}^3} = \frac{{4{x^2}}}{{\left( {16 - 16x + 4{x^2}} \right) \times (2 - x)}}}\\{2500 = \frac{{4{x^2}}}{{32 - 48x + 24{x^2} - 4{x^3}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{4{x^2} = - 10000{x^3} + 60000{x^2} - 120000x + 80000}\\{0 = - 10000{x^3} + 60004{x^2} - 120000x + 80000}\end{array}\)

Using equation solver, we get

\(x \approx 1.887{\rm{atm}}\)

The change in the partial pressure obtained\(x \approx 1.887{\rm{atm}}\)

Therefore, the pressures of NO, Cl₂, and NOCl in an equilibrium mixture is\({{\rm{P}}_{{\rm{NO}}}} = 4{\rm{atm}} - 2{\rm{x}} = 0.226{\rm{atm}}\)

\({{\rm{P}}_{{\rm{C}}{{\rm{l}}_2}}} = 2{\rm{atm}} - {\rm{x}} = 0.113{\rm{atm}}\)

\({{\rm{P}}_{{\rm{NOCl}}}} = 2{\rm{x}} = 3.774{\rm{atm}}\).