Answer
a) No effect
b) Shift right
c) Shift left
d) Shift right
e) No effect
Work Step by Step
Write the equilibrium expression of the reaction
$C(s) + 2H_{2}(g) \leftrightharpoons CH_{4}(g)$
Equilibrium constant$ k_{eq} = \frac{[CH_{4}]}{[H_{2}]}$
Now, if a chemical system is disturbed at equilibrium, the system minimizes the disturbance by shifting right or left direction (Le Chatelier's principle).
a) Adding more C to the reaction mixture will have no effect on the system. This is because: here the state of C is Solid. So, increasing or decreasing the concentration of C in the system will have no effect in the equilibrium system.
b) If we add more $H_{2}$ to the reaction mixture the concentration of $H_{2}$ increases. So, to minimize the disturbance of the system at equilibrium, the concentration of product should also be increased. This can happen if the forward reaction rate increases i.e. the reaction shifts towards right.
c) If we raised the temp. of the reaction mixture, the reaction will shift towards left. This is because: this is an exothermic reaction. That means heat is emitted in forward reaction. So, if we increase temp. of the reaction, the extra heat must be absorbed by the system to minimize the disturbance. Now, if the forward reaction is exothermic then the backward reaction will be endothermic i.e. heat is absorbed in backward reaction. So, the extra heat will be absorbed if the reaction shifts toward left.
d) If we lowered the volume of the reaction mixture, the pressure of the system is increased. To minimize this disturbance, the reaction shifts toward the direction where there are fewer moles of gas particles. Now, no. of moles of gas particles at left is equal to 2 and at right is equal to 1. So, the reaction will shift towards right to minimize the disturbance.
e) Adding a catalyst to the reaction mixture will have no effect on the equilibrium system. Catalysts do not disturb the equilibrium of the system.
