For the equilibrium system, CO(g) + 3H2(g) <----> H2O(g) + CH4(g) delta H= -206 kJ/mol....thanks?

CO(g) + 3H₂(g) ⇌ H₂O(g) + CH₄(g) ...... ΔH < 0

(a)
The yield of CH₄ will increase.
When some H₂ gas is added, according Le Chatelier's principle, the equilibrium position will shift to the right in order to remove some H₂.

(b)
The yield of CH₄ will decrease.
When some CO gas is removed, according Le Chatelier's principle, the equilibrium position will shift to the left in order to produce some H₂.

(c)
The yield of CH₄ will decrease.
As ΔH is negative, the forward reaction is exothermic and the backward reaction is endothermic. According to Le Chatelier's principle, increase in temperature favors the endothermic reaction, such that the backward reaction. Hence, the equilibrium position will shift to the left.

(d)
The yield of CH₄ will remain the same.
The catalyst increases both the rate of forward reaction and that of backward reaction to the same extent. Therefore, the addition of catalyst will just increase the reaction rate but has no effect on equilibrium position.