If for `H_(2(g)) + (1)/(2)S_(2(g))` and `H_(2(g)) + Br_(2(g))hArr2HBr_(g)`
The equilibrium constants are `K_(1)` and `K_(2)` respectively, the reaction `Br_(2(g)) + H_(2)S_(g) hArr2HBr_(g) + (1)/(2)S_(2(g))` would have equilibrium constant

Options

A. `K_(1)xxK_(2)`
B. `K_(1)//K_(2)`
C. `K_(2)//K_(1)`
D. `K_(2)^(2)//K_(1)`

Explanation

To determine the equilibrium constant for the reaction:

\[ \text{Br}_2(g) + \text{H}_2S(g) \rightleftharpoons 2 \text{HBr}(g) + \frac{1}{2} \text{S}_2(g) \]

we will use the given reactions and their equilibrium constants \( K_1 \) and \( K_2 \).

### Step 1: Write down the given reactions and their equilibrium constants.

1. **First Reaction:**
\[ \text{H}_2(g) + \frac{1}{2} \text{S}_2(g) \rightleftharpoons \text{H}_2S(g) \quad \text{with } K_1 \]

2. **Second Reaction:**
\[ \text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) \quad \text{with } K_2 \]

### Step 2: Reverse the first reaction.

To obtain the desired reaction, we need to reverse the first reaction. When we reverse a reaction, the equilibrium constant becomes the reciprocal.

\[ \text{H}_2S(g) \rightleftharpoons \text{H}_2(g) + \frac{1}{2} \text{S}_2(g) \quad \text{with } K = \frac{1}{K_1} \]

### Step 3: Keep the second reaction as it is.

The second reaction remains unchanged:

\[ \text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) \quad \text{with } K_2 \]

### Step 4: Add the two reactions together.

Now, we add the reversed first reaction and the second reaction:

1. From the reversed first reaction:
\[ \text{H}_2S(g) \rightleftharpoons \text{H}_2(g) + \frac{1}{2} \text{S}_2(g) \]

2. From the second reaction:
\[ \text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) \]

Adding these two reactions gives:

\[ \text{H}_2S(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) + \frac{1}{2} \text{S}_2(g) \]

### Step 5: Determine the equilibrium constant for the overall reaction.

When adding reactions, the equilibrium constants multiply:

\[ K = K_2 \times \frac{1}{K_1} = \frac{K_2}{K_1} \]

### Final Answer:

The equilibrium constant for the reaction

\[ \text{Br}_2(g) + \text{H}_2S(g) \rightleftharpoons 2 \text{HBr}(g) + \frac{1}{2} \text{S}_2(g) \]

is

\[ K = \frac{K_2}{K_1} \]

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