This question is about some gas mixtures at equilibrium.
This reaction can be used to produce nitrogen monoxide:
N2(g)+O2(g)⇌2NO(g)ΔH=+180.6 kJ mol−1\text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{NO}(\text{g}) \quad \Delta H = +180.6\text{ kJ mol}^{-1}N2(g)+O2(g)⇌2NO(g)ΔH=+180.6 kJ mol−1
A mixture of 4.00 mol4.00\text{ mol}4.00 mol of N2(g)\text{N}_2(\text{g})N2(g) and 4.00 mol4.00\text{ mol}4.00 mol of O2(g)\text{O}_2(\text{g})O2(g) is allowed to reach equilibrium at a constant temperature in a 15.0 dm315.0\text{ dm}^315.0 dm3 container.
At equilibrium, there are 2.40 mol2.40\text{ mol}2.40 mol of NO(g)\text{NO}(\text{g})NO(g).
Calculate the mole fraction of NO(g)\text{NO}(\text{g})NO(g) in the equilibrium mixture.
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:
Ethanol can be synthesized industrially by the direct hydration of ethene:
C2H4(g)+H2O(g)⇌C2H5OH(g)ΔH=−46 kJ mol−1\text{C}_2\text{H}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{C}_2\text{H}_5\text{OH}(\text{g}) \quad \Delta H = -46\text{ kJ mol}^{-1}C2H4(g)+H2O(g)⇌C2H5OH(g)ΔH=−46 kJ mol−1
The table below shows the mole fractions of each gas in an equilibrium mixture at a total pressure of 5000 kPa5000\text{ kPa}5000 kPa.
| Gas | Mole fraction |
|---|---|
| Ethene, C2H4\text{C}_2\text{H}_4C2H4 | 0.4200.4200.420 |
| Steam, H2O\text{H}_2\text{O}H2O | 0.4800.4800.480 |
| Ethanol, C2H5\text{C}_2\text{H}_5C2H5OH | 0.1000.1000.100 |
Give an expression for KpK_{\text{p}}Kp for this reaction.
Calculate the value of KpK_{\text{p}}Kp at 5000 kPa5000\text{ kPa}5000 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.