Question

A typed answer is prefered.

Some of the reactions for coal gasification are as follows: Reaction 1: C(s) + H.O(g) « CO(g) + H.(g) AH = 131.3 kJ/mol Reaction 2: CO(g) + H2O(g) «CO(g) + H.(g) AH = -41.2 kJ/mol Reaction 3: CO(g) + 3H,(g) O(g) « H.O(g) + CH,(g) AH' = -205.9 kJ/mol 1. Write the equilibrium constant expression for reaction 2. 2. For all three reactions, predict the way the equilibrium will shift in response to an increase in pressure. EXPLAIN. 3. For all three reactions, predict the direction in which each equilibrium shifts in response to an increase in reaction volume. EXPLAIN.

Answer 1

1. Reaction 2 CO(g) + H2O (2) Comet H2 (9) DH = - 41.2 kJ k = [co] [2] [co] [H₂o Where ka equilibrium Constant 2. Due to an towards a moles. increase in pressure the equilibrium shifts direction which has less number of gaseous 1. C(s) + H2O (8) CO(g) + H2 (8) No.of gasesus moles - so the reaction proceeds towards back due to increase in pressure 2. CO(g) + H₂O (8) CO2(g) + H2 (5) No. « Задел са 1+1 = 1+1 moles change in so there will be no effect of pressure on the reaction 3. Co(s) + 3H2O(g) – H2O(g) + CH4 (9) No. or a 1 + 3 gaseous moles ltl - 4 So the reaction will more towards right due to increase in pressure as there are less no. of moles on right side.