Nitrosyl chloride gas decomposes at high temperatures according to the following equation:
2NOCl(g)⇌2NO(g)+Cl2(g)ΔH=+76 kJ mol−12\text{NOCl}(\text{g}) \rightleftharpoons 2\text{NO}(\text{g}) + \text{Cl}_2(\text{g}) \quad \Delta H = +76\text{ kJ mol}^{-1}2NOCl(g)⇌2NO(g)+Cl2(g)ΔH=+76 kJ mol−1
A 0.800 mol0.800\text{ mol}0.800 mol sample of nitrosyl chloride is placed in a sealed reaction vessel and heated at a constant temperature until equilibrium is established. At equilibrium, the mixture contains 0.220 mol0.220\text{ mol}0.220 mol of chlorine gas. Calculate the mole fraction of each substance at equilibrium. Give your answers to 3 decimal places.
The total pressure in the vessel in Part 1 is 240 kPa240\text{ kPa}240 kPa at equilibrium. Calculate the partial pressure, in kPakPakPa, of NOCl\text{NOCl}NOCl. (If you were unable to answer Part 1, assume that the mole fraction of NOCl\text{NOCl}NOCl is 0.4000.4000.400. This is not the correct answer.)
The table below shows the mole fractions of the three gases in a different equilibrium mixture of the same system:
| Gas | Mole fraction |
|---|---|
| NOCl\text{NOCl}NOCl | 0.4500.4500.450 |
| NO\text{NO}NO | 0.3500.3500.350 |
| Cl2\text{Cl}_2Cl2 | 0.2000.2000.200 |
For this equilibrium mixture, Kp=49.0 kPaK_p = 49.0\text{ kPa}Kp=49.0 kPa.
The equilibrium mixture in Part 3 is compressed into a smaller volume at constant temperature. Deduce the effect, if any, of this change on the equilibrium yield of chlorine and on the value of KpK_pKp.
The equilibrium mixture in Part 3 is allowed to reach equilibrium at a lower temperature. Explain why the equilibrium yield of chlorine decreases.