The equilibrium constant of the reaction, `H_(2)+I_(2)hArr 2HI` is 50. If the volume of the container is reduced to half of its original value, the value of equilibrium constant will be

Options

A. 25
B. 50
C. 75
D. 100

Answer

To solve the problem, we need to understand how the equilibrium constant (K) is affected by changes in volume. Here's a step-by-step breakdown of the solution:

### Step 1: Write the equilibrium expression
For the reaction \( H_2 + I_2 \rightleftharpoons 2HI \), the equilibrium constant \( K \) is defined as:
\[ K = \frac{[HI]^2}{[H_2][I_2]} \]
where \( [HI] \), \( [H_2] \), and \( [I_2] \) are the molar concentrations of the respective substances at equilibrium.

### Step 2: Understand the effect of volume on concentration
When the volume of the container is reduced to half, the concentrations of all gases will change. The concentration \( C \) of a gas is given by:
\[ C = \frac{n}{V} \]
where \( n \) is the number of moles and \( V \) is the volume. If the volume is halved (i.e., \( V \) becomes \( \frac{V}{2} \)), the new concentrations will be:
\[ [H_2] = \frac{n_{H_2}}{\frac{V}{2}} = \frac{2n_{H_2}}{V} \]
\[ [I_2] = \frac{n_{I_2}}{\frac{V}{2}} = \frac{2n_{I_2}}{V} \]
\[ [HI] = \frac{n_{HI}}{\frac{V}{2}} = \frac{2n_{HI}}{V} \]

### Step 3: Substitute the new concentrations into the equilibrium expression
Now substituting these new concentrations into the equilibrium expression:
\[ K' = \frac{(2[HI])^2}{(2[H_2])(2[I_2])} \]
This simplifies to:
\[ K' = \frac{4[HI]^2}{4[H_2][I_2]} = \frac{[HI]^2}{[H_2][I_2]} = K \]

### Step 4: Conclusion about the equilibrium constant
Since \( K' = K \), we conclude that the equilibrium constant does not change with the change in volume. Therefore, the value of the equilibrium constant remains the same at 50.

### Final Answer:
The value of the equilibrium constant will be **50**.

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