Question

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ When equilibrium is established, the partial pressures of the gases $$ \text { are: } P_{\mathrm{N}_{2}}=1.200 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.800 \mathrm{~atm}, P_{\mathrm{NO}}=0.0220 \mathrm{~atm} . $$ (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure 1.200 atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

Short answer

Answer: The equilibrium constant (\(K\)) at this temperature is \(5.07 \times 10^{-4}\). When equilibrium is reestablished, the partial pressures of the gases are \(P_{\mathrm{N}_{2}} = 1.19556 \mathrm{~atm}\), \(P_{\mathrm{O}_{2}} = 1.19556 \mathrm{~atm}\), and \(P_{\mathrm{NO}} = 0.03088 \mathrm{~atm}\).

Step-by-step solution

Write the expression for the equilibrium constant K

For a given reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ The equilibrium constant, \(K\), is given by the following expression: $$K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}] \cdot [\mathrm{O}_{2}]}$$ In this case, the partial pressures of the gases at equilibrium are given instead of concentrations. We can use the partial pressures to write the expression for K in terms of pressure: $$K_p = \frac{P_{\mathrm{NO}}^2}{P_{\mathrm{N}_{2}} \cdot P_{\mathrm{O}_{2}}}$$

Calculate the equilibrium constant K

We have the initial partial pressures for the gases at equilibrium: $$P_{\mathrm{N}_{2}}=1.200 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.800 \mathrm{~atm}, P_{\mathrm{NO}}=0.0220 \mathrm{~atm}$$ We can now plug these values into the expression for \(K_p\) to calculate the equilibrium constant: $$K_p = \frac{(0.0220)^2}{(1.200) \cdot (0.800)}$$ Calculate \(K_p\): $$K_p = 5.07 \times 10^{-4}$$ (a) The equilibrium constant \(K\) at this temperature is \(5.07 \times 10^{-4}\).

Calculate the new partial pressures after adding oxygen

When more oxygen is added to make its partial pressure 1.200 atm, we need to determine the new partial pressures when equilibrium is reestablished. Let \(x\) be the change in the partial pressure due to the addition of oxygen. Then, the new equilibrium partial pressures can be expressed as: $$P'_{\mathrm{N}_{2}}=1.200 - x \mathrm{~atm}$$ $$P'_{\mathrm{O}_{2}}=1.200 - x \mathrm{~atm}$$ $$P'_{\mathrm{NO}}=0.0220 + 2x \mathrm{~atm}$$ At the new equilibrium, the reaction quotient (\(Q\)) will be equal to \(K_p\). We can write the expression for \(Q\) using the new partial pressures: $$Q_p = \frac{(0.0220 + 2x)^2}{(1.200 - x)(1.200 - x)}$$ Now we can set \(Q_p = K_p\) $$5.07 \times 10^{-4} = \frac{(0.0220 + 2x)^2}{(1.200 - x)(1.200 - x)}$$ Solve the equation for \(x\). After solving, we get: $$x = 0.00444$$ Now, we can calculate the new partial pressures at equilibrium using the value of \(x\): $$P'_{\mathrm{N}_{2}}=1.200 - 0.00444 = 1.19556 \mathrm{~atm}$$ $$P'_{\mathrm{O}_{2}}=1.200 - 0.00444 = 1.19556 \mathrm{~atm}$$ $$P'_{\mathrm{NO}}=0.0220 + 2(0.00444) = 0.03088 \mathrm{~atm}$$ (b) When equilibrium is reestablished after adding more oxygen, the partial pressures of the gases are: $$P'_{\mathrm{N}_{2}} = 1.19556 \mathrm{~atm}$$ $$P'_{\mathrm{O}_{2}} = 1.19556 \mathrm{~atm}$$ $$P'_{\mathrm{NO}} = 0.03088 \mathrm{~atm}$$