This question is about some gas mixtures at equilibrium.
Nitric oxide can be produced by the direct combination of nitrogen and oxygen gases at high temperatures:
N2(g)+O2(g)⇌2NO(g)ΔH=+180 kJ mol−1\text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{NO}(\text{g}) \quad \Delta H = +180\text{ kJ mol}^{-1}N2(g)+O2(g)⇌2NO(g)ΔH=+180 kJ mol−1
A mixture of 3.20 mol3.20\text{ mol}3.20 mol of $\text{N}_2(\text{g})$ and 3.20 mol3.20\text{ mol}3.20 mol of $\text{O}_2(\text{g})$ is allowed to reach equilibrium at a constant temperature in a 5.00 dm35.00\text{ dm}^35.00 dm3 container.
At equilibrium, there are 1.20 mol1.20\text{ mol}1.20 mol of $\text{NO}(\text{g})$.
Calculate the mole fraction of NO(g)\text{NO}(\text{g})NO(g) in the equilibrium mixture. Give your answer to 3 significant figures.
State why the equilibrium constant (KpK_{\text{p}}Kp) for this reaction has no units.
The temperature of the equilibrium mixture is increased. How does the amount of NO(g)\text{NO}(\text{g})NO(g) change when the new position of equilibrium is reached? Select one:
Sulfur trioxide is produced during the contact process by the oxidation of sulfur dioxide:
2SO2(g)+O2(g)⇌2SO3(g)ΔH=−197 kJ mol−12\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -197\text{ kJ mol}^{-1}2SO2(g)+O2(g)⇌2SO3(g)ΔH=−197 kJ mol−1
The table below shows the mole fractions of each gas in an equilibrium mixture at a total pressure of 150 kPa150\text{ kPa}150 kPa.
| Gas | Mole fraction |
|---|---|
| Sulfur dioxide, SO2\text{SO}_2SO2 | 0.3400.3400.340 |
| Oxygen, O2\text{O}_2O2 | 0.1800.1800.180 |
| Sulfur trioxide, SO3\text{SO}_3SO3 | 0.4800.4800.480 |
Give an expression for KpK_{\text{p}}Kp for this reaction.
Calculate the value of KpK_{\text{p}}Kp at 150 kPa150\text{ kPa}150 kPa. State the units.
State the effect, if any, of an increase in the volume of the container on the value of KpK_{\text{p}}Kp for this reaction at a constant temperature.