Sulfur trioxide decomposes at a high temperature:
2SO3(g)⇌2SO2(g)+O2(g)ΔH=+198 kJ mol−12\text{SO}_3(\text{g}) \rightleftharpoons 2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \quad \Delta H = +198\text{ kJ mol}^{-1}2SO3(g)⇌2SO2(g)+O2(g)ΔH=+198 kJ mol−1
A 0.500 mol0.500\text{ mol}0.500 mol sample of sulfur trioxide is placed in a sealed flask and heated at a constant temperature until equilibrium is reached. At equilibrium, the flask contains 0.170 mol0.170\text{ mol}0.170 mol of oxygen. Calculate the mole fraction of each substance at equilibrium. Give your answers to 3 decimal places.
The total pressure in the flask in Part 1 is 150 kPa150\text{ kPa}150 kPa at equilibrium. Calculate the partial pressure, in kPa\text{kPa}kPa, of SO3\text{SO}_3SO3. (If you were unable to answer Part 1, assume that the mole fraction of SO3\text{SO}_3SO3 is 0.3000.3000.300. This is not the correct answer.)
The table below shows the mole fractions of the three gases in a different equilibrium mixture:
| Gas | Mole fraction |
|---|---|
| SO3\text{SO}_3SO3 | 0.3000.3000.300 |
| SO2\text{SO}_2SO2 | 0.5000.5000.500 |
| O2\text{O}_2O2 | 0.2000.2000.200 |
For this equilibrium mixture, Kp=65.0 kPaK_p = 65.0\text{ kPa}Kp=65.0 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 oxygen 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 oxygen decreases.