Coal, which is primarily carbon, can be converted to natural gas, primarily ch4, by the following exothermic reaction: c(s)+2h2(g)⇌ch4(g) a. what is the effect of adding more c to the reaction mixture? b. what is the effect of adding more h2 to the reaction mixture? c. what is the effect of raising the temperature of the reaction mixture? d. what is the effect of lowering the volume of the reaction mixture? e. what is the effect of adding a catalyst to the reaction mixture?

A. No effect on equilibrium.

B. Favors forward reaction.

C. Favors reverse reaction.

D. Favors forward reaction.

E. No effect on equilibrium.

Explanation:

To answer this question we need to bear in mind Le Chatelier's Principle: if a system at equilibrium is disturbed by changing some condition, it will react to counteract the perturbation.

Let us consider the exothermic reaction:

C(s) + 2H₂(g) ⇄ CH₄(g)

A. What is the effect of adding more C to the reaction mixture?

Carbon is a solid and solids are not part of the equilibrium (because their concentration remains almost constant) so this will have no effect.

B. What is the effect of adding more H₂ to the reaction mixture?

If the concentration of H₂ is increased, the system will try to decrease it by favoring the forward reaction and consuming this reactant.

C. What is the effect of raising the temperature of the reaction mixture?

If the temperature is raised, the system will try to decrease it. Since the reaction is exothermic (releases heat), the way of decreasing temperature will be by favoring the reverse reaction.

D. What is the effect of lowering the volume of the reaction mixture?

Lowering the volume will increase the pressure of the gases (Boyle's Law). The system will try to decrease pressure by shifting to the side where there are fewer moles of gases, in this case, to the right, favoring forward reaction.

E. What is the effect of adding a catalyst to the reaction mixture?

Catalysts increase both direct and inverse rates of reaction, so they have no effect on the equilibrium.