Question

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of \(0.095 \mathrm{~atm}, 0.171 \mathrm{~atm}\), and \(0.28\) atm for \(\mathrm{NO}, \mathrm{Cl}_{2}\), and \(\mathrm{NOCl}\), respectively. Calculate \(K_{p}\) for this reaction at \(500 \mathrm{~K}\).

Short answer

The equilibrium constant, \(K_p\), for the reaction \(2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\) at 500 K is approximately 51.10.

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation for the given reaction is: \[2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\]

Write the expression for \(K_p\)

The expression for \(K_p\) for the given reaction is: \[K_p = \frac{[NOCl]^2}{[NO]^2 \times [Cl_2]}\]

Find the equilibrium partial pressures

We are given the equilibrium partial pressures of each gas: \[P_{NO} = 0.095~atm\] \[P_{Cl_2} = 0.171~atm\] \[P_{NOCl} = 0.280~atm\]

Substitute the partial pressures into the expression for \(K_p\)

We will plug the given partial pressures into the expression for \(K_p\): \[K_p = \frac{(0.280)^2}{(0.095)^2 \times (0.171)}\]

Calculate \(K_p\)

Performing the calculations, we get: \[K_p = \frac{0.0784}{0.00153375}\] \[K_p ≈ 51.10\] Therefore, the equilibrium constant, \(K_p\), for this reaction at 500 K is approximately 51.10.