24 mL of HI are produced from the reaction of 15mL of `H_2` and 17.1 mL of `I_2` vapour at `444 ^@C` . The equilibrium constant for the reaction.
`H_2+I_2 hArr 2HI "at" 444^@C` is

Options

A. 27.5
B. 37.647
C. 73.647
D. 57.25

Answer

To find the equilibrium constant \( K_c \) for the reaction

\[ H_2 + I_2 \rightleftharpoons 2HI \]

given the volumes of the reactants and products, we can follow these steps:

### Step 1: Identify the initial volumes of reactants and products.
- Initial volume of \( H_2 = 15 \, \text{mL} \)
- Initial volume of \( I_2 = 17.1 \, \text{mL} \)
- Volume of \( HI \) produced at equilibrium = \( 24 \, \text{mL} \)

### Step 2: Set up the change in volume for the reaction.
Let \( x \) be the volume of \( H_2 \) and \( I_2 \) that reacts. According to the stoichiometry of the reaction:
- 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \).
- Therefore, if \( x \) mL of \( H_2 \) and \( I_2 \) react, \( 2x \) mL of \( HI \) will be produced.

### Step 3: Relate the produced volume of \( HI \) to \( x \).
From the problem, we know that \( 2x = 24 \, \text{mL} \) (since 24 mL of \( HI \) is produced). Thus:
\[ x = \frac{24}{2} = 12 \, \text{mL} \]

### Step 4: Calculate the equilibrium volumes of \( H_2 \) and \( I_2 \).
- Volume of \( H_2 \) at equilibrium:
\[ \text{Volume of } H_2 = 15 \, \text{mL} - x = 15 - 12 = 3 \, \text{mL} \]
- Volume of \( I_2 \) at equilibrium:
\[ \text{Volume of } I_2 = 17.1 \, \text{mL} - x = 17.1 - 12 = 5.1 \, \text{mL} \]

### Step 5: Write the expression for the equilibrium constant \( K_c \).
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]
In terms of volume, this can be expressed as:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(24)^2}{(3)(5.1)} \]

### Step 6: Calculate \( K_c \).
Substituting the values:
\[ K_c = \frac{24^2}{3 \times 5.1} = \frac{576}{15.3} \approx 37.64 \]

### Final Result:
The equilibrium constant \( K_c \) at \( 444^\circ C \) is approximately \( 37.64 \).

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