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(a) Methanol is manufactured from carbon (IV) oxide and hydrogen gas according to the equation: #CO_2# (g) + #3H_2#(g)#rightarrow# #CH_3OH#(g) + #H_2O#(g) The reaction is carried out in the presence of a chromium catalyst at 700K and 30kPa.
Under these conditions, equilibrium is reached when 2% of the carbon (IV) oxide is converted to methanol
(i)How does the rate of the forward reaction compare with that of the reverse reaction when 2% of the carbon (IV) oxide is converted to methanol?
(ii)Explain how each of the following would affect the yield of methanol:
I Reduction in pressure
II Using a more efficient catalyst
(iii) If the reaction is carried out at 500K and 30Pa, the percentage of carbon (IV) oxide converted to methanol is
higher than 2%
I what is the sign of ,H for the reaction? Give a reason
II Explain why in practice the reaction is carried out at 700K but NOT at 500K