1. IC3.27.4 Increasing the yield © Oxford University Press 2011 Increasing the yield

2. IC3.27.4 Increasing the yield © Oxford University Press 2011 Making sulfuric acid involves several stages, including this equilibrium reaction: sulfur dioxide + oxygen sulfur trioxide 2SO 2 (g) + O 2 (g) 2SO 3 (g) ∆H = –197 kJ/mol

3. IC3.27.4 Increasing the yield © Oxford University Press 2011  The reaction is exothermic. So lowering the temperature increases the yield of sulfur trioxide, SO 3.  The reaction is very slow at low temperatures, so a temperature of around 450 °C is chosen as a compromise.

4. IC3.27.4 Increasing the yield © Oxford University Press 2011 Increasing the pressure also increases the yield of sulfur trioxide. This is because there are fewer molecules shown on the right of the symbol equation than on the left: 2SO 2 (g) + O 2 (g) 2SO 3 (g) Number of molecules on left = (2 + 1) = 3 Number of molecules on right = 2

5. IC3.27.4 Increasing the yield © Oxford University Press 2011 For all equilibrium gas reactions, an increase in pressure favours the reaction that produces the fewest molecules, as shown by the symbol equation for the reaction.

6. IC3.27.4 Increasing the yield © Oxford University Press 2011 The Haber process reaction equation shows a total of 4 molecules on the left and 2 molecules on the right: N 2 (g) + 3H 2 (g) 2NH 3 (g) So an increase in pressure shifts the equilibrium to the right, and increases the yield of ammonia. This explains why a high pressure is chosen for the Haber process.

7. IC3.27.4 Increasing the yield © Oxford University Press 2011 1.Predict whether a high or low pressure would maximise the yield of nitrogen monoxide (NO) in this reaction. Explain. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) There are more molecules on the right of the equation, so a low pressure is needed to maximise the yield of nitrogen monoxide.

8. IC3.27.4 Increasing the yield © Oxford University Press 2011 2.Predict the effect of increasing pressure on this equilibrium reaction. Explain. H 2 (g) + I 2 (g) 2HI(g) There are equal numbers of molecules on both sides of the reaction, so increasing the pressure would have no effect on this reaction.