Nitrosyl chloride decomposes at an elevated temperature 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 sealed in a container and heated at a constant temperature until equilibrium is established. At equilibrium, the system is found to contain 0.240 mol0.240\text{ mol}0.240 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 container in Part 1 is 220 kPa220\text{ kPa}220 kPa at equilibrium. Calculate the partial pressure, in kPa\text{kPa}kPa, of NOCl\text{NOCl}NOCl. (If you were unable to answer Part 1, assume that the mole fraction of NOCl\text{NOCl}NOCl is 0.3500.3500.350. This is not the correct answer.)
The table below shows the mole fractions of the three components in a different equilibrium mixture of the same system:
| Gas | Mole fraction |
|---|---|
| NOCl\text{NOCl}NOCl | 0.4000.4000.400 |
| NO\text{NO}NO | 0.4000.4000.400 |
| Cl2\text{Cl}_2Cl2 | 0.2000.2000.200 |
For this specific equilibrium mixture, Kp=44.0 kPaK_p = 44.0\text{ kPa}Kp=44.0 kPa.
The equilibrium mixture in Part 3 is compressed into a smaller volume at constant temperature. Deduce the effect, if any, of this volume reduction on the equilibrium yield of chlorine and on the numerical value of KpK_pKp.
The equilibrium mixture in Part 3 is allowed to reach a new equilibrium at a lower temperature. Explain why the equilibrium yield of chlorine decreases.