Question

The following equilibrium pressures at a certain temperature were observed for the reaction $$ \begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{~atm} \\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{~atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{~atm} \end{aligned} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

Short answer

The equilibrium constant, \(K_p\), for the reaction at this temperature is \(9.06 \times 10^{-11}\).

Step-by-step solution

Write the balanced equilibrium reaction

In this problem, we're already given a balanced reaction: \[ 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \]

Write the expression for the equilibrium constant \(K_p\)

Using the balanced chemical equation, write the expression for the equilibrium constant \(K_p\) in terms of the partial pressures of each species: \[K_p = \frac{(P_{NO})^2 \cdot (P_{O_2})}{(P_{NO_2})^2}\]

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

We're given the following partial pressures: \(P_{NO_2} = 0.55~atm\), \(P_{NO} = 6.5 \times 10^{-5}~atm\), and \(P_{O_2} = 4.5 \times 10^{-5}~atm\) Substitute these values into the \(K_p\) expression: \[K_p = \frac{(6.5 \times 10^{-5})^2 \cdot (4.5 \times 10^{-5})}{(0.55)^2}\]

Calculate the value of \(K_p\)

Perform the calculation in the previous step to find the value of \(K_p\): \[K_p = \frac{(6.5 \times 10^{-5})^2 \cdot (4.5 \times 10^{-5})}{(0.55)^2} = 9.06 \times 10^{-11}\] So, the equilibrium constant, \(K_p\), at this temperature is \(9.06 \times 10^{-11}\).