equilibrum constant PMT

Question 1

Q3a For this reaction at 1050 K the equilibrium constant, Kp = 7.62 x 105 Pa For this reaction at 500 K the equilibrium constant, Kp = 3.94 x 104 Pa
Explain how this information can be used to deduce that the forward reaction is endothermic.

Answer 1

Kp is higher at higher temperature or converse
At higher temperature more dissociation occurs / more products are formed

Question 2

Q7b A data book value for the H–H bond enthalpy is 436 kJ mol−1.
Suggest one reason why this value is different from your answer to part (a).

Answer 2

Mean bond enthalpies are not the same as the actual bond enthalpies in CO2 (and / or methanol and / or water)

Question 3

Q7c Suggest one environmental advantage of manufacturing methanol fuel by this reaction.
CO2(g) + 3H2(g) CH3OH(g) + H2O(g)

Answer 3

The carbon dioxide (produced on burning methanol) is used up in this reaction

Question 4

Q7d Use Le Chatelier’s principle to justify why the reaction is carried out at a high pressure rather than at atmospheric pressure.
CO2(g) + 3H2(g) CH3OH(g) + H2O(g)

Answer 4

4 mol of gas form 2 mol
At high pressure the position of equilibrium moves to the right to lower the pressure / oppose the high pressure
This increases the yield of methanol

Question 5

Q7e Suggest why the catalyst used in this process may become less efficient if the carbon dioxide and hydrogen contain impurities.

Answer 5

mpurities (or sulfur compounds) block the active sites