Methane can be steam-reformed to produce synthesis gas, and sulfur dioxide can be oxidized to form sulfur trioxide in the Contact Process over a heated catalyst.
The equations for the two reactions are: Reaction 1: CH4(g)+H2O(g)⇌CO(g)+3H2(g)ΔH=+206 kJ/mol\text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO(g)} + 3\text{H}_2\text{(g)} \quad \Delta H = +206\text{ kJ/mol}CH4(g)+H2O(g)⇌CO(g)+3H2(g)ΔH=+206 kJ/mol Reaction 2: 2SO2(g)+O2(g)⇌2SO3(g)ΔH=−197 kJ/mol2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)} \quad \Delta H = -197\text{ kJ/mol}2SO2(g)+O2(g)⇌2SO3(g)ΔH=−197 kJ/mol
Assume that both reactions reach a position of equilibrium.
For Reaction 1, predict whether using a high or a low temperature would produce the higher yield of carbon monoxide, CO\text{CO}CO. Give a reason for your choice.
For Reaction 2, predict whether using a high or a low pressure would produce the higher yield of sulfur trioxide, SO3\text{SO}_3SO3. Give a reason for your choice.
The catalyst increases the rate of both the forward reaction and the backward reaction. Suggest why the catalyst has no effect on the position of equilibrium.
Reaction 1 can be represented by a reaction profile diagram.
For the diagram above, describe how to complete the profile to show the products of the reaction and the enthalpy change, ΔH\Delta HΔH.
Describe how to represent the activation energy, EEE, for the forward reaction on this profile.
State the effect, if any, of the catalyst on the enthalpy change for the reaction.
