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).

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).

Accepted Answer

# Brief Explanations:
To solve this, we use Le Chatelier's principle. The forward reaction is $2H_2S(g)
ightarrow2H_2(g)+S_2(g)$. The reverse reaction is the formation of $H_2S$ from $H_2$ and $S_2$.

- For the effect of product concentration ($H_2$ or $S_2$) on forward reaction rate:
  - An increase in product concentration ($H_2$ or $S_2$) shifts the equilibrium to the left (favors reverse reaction), so forward reaction rate decreases.
  - A decrease in product concentration ($H_2$ or $S_2$) shifts the equilibrium to the right (favors forward reaction), so forward reaction rate increases.

Now analyze each option:
- Option 1: Increase in $S_2$ (product) should decrease forward rate, not increase. Eliminate.
- Option 2: Decrease in $H_2$ (product) should increase forward rate, not decrease. Eliminate.
- Option 3: Decrease in $H_2$ (product) shifts equilibrium right, so forward rate increases. Correct.
- Option 4: Increase in $S_2$ (product) shifts equilibrium left, so forward rate decreases. Correct.
- Option 5: Increase in $H_2$ (product) shifts equilibrium left, so forward rate decreases. Correct.

Wait, let's re - check. The question is about how forward reaction rate varies with product concentration. The forward reaction rate is affected by product concentration via Le Chatelier:

- When product ($H_2$ or $S_2$) concentration increases, the reverse reaction is favored, so forward reaction rate decreases (because the system tries to consume the excess products, so less forward reaction).
- When product concentration decreases, the forward reaction is favored (to produce more products), so forward reaction rate increases.

Now re - evaluate each option:

1. "It increases with an increase in the concentration of $S_2(g)$." → Wrong. Increase in product ($S_2$) should decrease forward rate.
2. "It decreases with a decrease in the concentration of $H_2(g)$." → Wrong. Decrease in product ($H_2$) should increase forward rate.
3. "It increases with a decrease in the concentration of $H_2(g)$." → Correct. Decrease in $H_2$ (product) means the system will try to make more products, so forward reaction rate increases.
4. "It decreases with an increase in the concentration of $S_2(g)$." → Correct. Increase in $S_2$ (product) means the system will try to consume $S_2$ (reverse reaction), so forward reaction rate decreases.
5. "It decreases with increase in the concentration of $H_2(g)$." → Correct. Increase in $H_2$ (product) means the system will try to consume $H_2$ (reverse reaction), so forward reaction rate decreases.

But maybe the question is about the relationship between forward reaction rate and product concentration (not equilibrium shift in terms of rate change direction). Let's think in terms of reaction rates. The forward reaction rate depends on reactant concentration ($H_2S$) and the reverse reaction rate depends on product concentrations ($H_2$ and $S_2$). At equilibrium, forward and reverse rates are equal. When product concentration increases, reverse reaction rate increases (since rate of reverse reaction $r_{reverse}=k_{reverse}[H_2]^2[S_2]$). For the forward reaction, its rate $r_{forward}=k_{forward}[H_2S]^2$. But when we are not at equilibrium, if product concentration increases, the system will move towards equilibrium, so forward reaction rate will decrease (because reverse rate is higher, so net forward rate decreases). If product concentration decreases, forward reaction rate will increase (because the system needs to make more products, so forward rate becomes higher than reverse rate).

So:

- Option 3: Decrease in $H_2$ (product) → forward rate increases. Correct.
- Option 4: Increase in $S_2$ (product) → forward rate decreases. Correct.
- Option 5: Increase in $H_2$ (product) → forward rate decreases. Correct.

But maybe the question has multiple correct answers. Let's check the options again.

# Answer:
C. It increases with a decrease in the concentration of $H_2(g)$, D. It decreases with an increase in the concentration of $S_2(g)$, E. It decreases with increase in the concentration of $H_2(g)$

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