The equilbrium constant of the reaction `H_(2)(g)+I_(2)(g)hArr 2HI(g)` 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 question regarding the equilibrium constant of the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) when the volume of the container is reduced to half, we will follow these steps:

### Step 1: Understand the Equilibrium Constant
The equilibrium constant \( K_c \) for the reaction is given by the formula:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
where \( [HI] \), \( [H_2] \), and \( [I_2] \) are the molar concentrations of hydrogen iodide, hydrogen, and iodine, respectively.

### Step 2: Analyze the Change in Volume
When the volume of the container is reduced to half, the concentrations of all gases will change. The concentration of a gas is given by the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
If the volume is halved, the concentration of each gas will double.

### Step 3: Calculate the Change in Concentration
Let’s denote the initial concentrations as:
- \( [H_2] = a \)
- \( [I_2] = a \)
- \( [HI] = 0 \)

After the reaction reaches equilibrium, if we assume that \( x \) moles of \( H_2 \) and \( I_2 \) react to form \( 2x \) moles of \( HI \), the concentrations at equilibrium will be:
- \( [H_2] = a - x \)
- \( [I_2] = a - x \)
- \( [HI] = 2x \)

When the volume is halved, the new concentrations will be:
- \( [H_2] = 2(a - x) \)
- \( [I_2] = 2(a - x) \)
- \( [HI] = 2(2x) = 4x \)

### Step 4: Substitute into the Equilibrium Expression
Now substituting these new concentrations into the equilibrium expression:
\[ K_c' = \frac{[HI]^2}{[H_2][I_2]} = \frac{(4x)^2}{(2(a-x))(2(a-x))} = \frac{16x^2}{4(a-x)^2} = \frac{4x^2}{(a-x)^2} \]

### Step 5: Compare with Original Equilibrium Constant
Since the original equilibrium constant \( K_c \) is given as:
\[ K_c = \frac{x^2}{(a-x)^2} \]
We can see that \( K_c' \) after halving the volume does not change the ratio because the factor of 4 in the numerator and denominator cancels out.

### Conclusion
Thus, the value of the equilibrium constant \( K_c \) remains unchanged when the volume is halved, and it will still be:
\[ K_c = 50 \]

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

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