Question

In this reaction, how does the rate of forward reaction vary with the concentration of the product? 2H2S(g) ⇌ 2H2(g) + S2(g) It increases with an increase in the concentration of S2(g). It decreases with a decrease in the concentration of H2(g). It increases with a decrease in the concentration of H2(g). It decreases with an increase in the concentration of S2(g). It decreases with increase in the concentration of H2(g).
MULTIPLE ANSWERS

Answer 1

Answer: C) It increases with a decrease in the concentration of $H_2(g)$. D) It decreases with increase in the concentration of $H_2(g)$.

Explanation: According to Le Chatelier's principle, if an equilibrium reaction is disturbed, the reaction would try to undo the change imposed.

Thus if the concentration of the products in increased, the reaction would shift in a direction where the concentration of products is decreasing i.e in backward direction.

If the concentration of the products in decreased, the reaction would shift in a direction where the concentration of products is increasing i.e in forward direction.

When the concentration of either of the products, i.e $H_2$ or $S_2$ is increased, the reaction will shift in backward direction and the rate of forward direction decreases.

Answer 2

The forward reaction rate depends mostly on the concentration of reactants rather than products. Changes in product concentrations thus do not immediately influence the forward reaction rate. Such changes, however, would alter the concentration of reactants. Their impacts on the forward reaction rate are thus indirect. Two of the options are likely.

• Rate of the forward reaction increases with an increase in $[\text{S}_2 \; (g)]$;
• Rate of the forward reaction decreases with a decrease in $[\text{H}_2 \; (g)]$;

Increasing $[\text{S}_2]$, the concentration of sulfur, will increase the rate of the backward reaction that converts sulfur and hydrogen back to hydrogen sulfide $\text{H}_2\text{S}$. An increase in the hydrogen sulfide concentration would spur the forward reaction. Overall, it appears as if increasing the concentration of sulfur $\text{S}_2 \; (g)$ increases the rate of the forward reaction.

Similarly, reducing the concentration of hydrogen reduces the rate of the backward reaction. The rate of hydrogen sulfide $\text{H}_2\text{S}$ production declines, whereas it is being consumed at nearly the same speed. As a result, there shall be a decrease in the concentration of the reactant $\text{H}_2\text{S}$. The rate of the forward reaction decreases accordingly.

Alternatively, consider the forward and backward reactions as two counteracting forces behind the chemical equilibrium. Rates of the reaction in the two directions are equal in cases that the system has achieved equilibrium. Increasing the concentration of a product would speeds up the backward reaction. The rate of the reaction in the backward direction now exceeds that of the forward reaction. The system would speed up the forward reaction to catch up with the backward reaction. It would hence establish a new equilibrium.