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

\[ H_2 + I_2 \rightleftharpoons 2 HI \]

given that 24 mL of HI are produced from the reaction of 15 mL of \( H_2 \) and 17.1 mL of \( I_2 \) at \( 444^\circ C \), we can follow these steps:

### Step 1: Write down the initial volumes
We start with the initial volumes of the reactants:
- Volume of \( H_2 \) = 15 mL
- Volume of \( I_2 \) = 17.1 mL
- Volume of \( HI \) = 0 mL (since it hasn't formed yet)

### Step 2: Set up the change in volumes
Let \( x \) be the volume of \( H_2 \) and \( I_2 \) that reacts to form \( HI \). According to the stoichiometry of the reaction:
- For every 1 mL of \( H_2 \) and \( I_2 \) that reacts, 2 mL of \( HI \) is produced.

At equilibrium, the volumes will be:
- Volume of \( H_2 \) = \( 15 - x \)
- Volume of \( I_2 \) = \( 17.1 - x \)
- Volume of \( HI \) = \( 2x \)

### Step 3: Use the information given to find \( x \)
We know that the total volume of \( HI \) produced is 24 mL:
\[ 2x = 24 \implies x = 12 \text{ mL} \]

### Step 4: Calculate the equilibrium volumes
Now we can calculate the volumes of \( H_2 \) and \( I_2 \) at equilibrium:
- Volume of \( H_2 \) at equilibrium = \( 15 - 12 = 3 \) mL
- Volume of \( I_2 \) at equilibrium = \( 17.1 - 12 = 5.1 \) mL

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 6: Calculate the concentrations
To find the concentrations, we need the total volume at equilibrium:
\[ \text{Total volume} = 3 + 5.1 + 24 = 32.1 \text{ mL} \]

Now we can calculate the concentrations:
- Concentration of \( HI \) = \( \frac{24}{32.1} \)
- Concentration of \( H_2 \) = \( \frac{3}{32.1} \)
- Concentration of \( I_2 \) = \( \frac{5.1}{32.1} \)

### Step 7: Substitute into the \( K_c \) expression
Substituting the concentrations into the equilibrium expression:
\[ K_c = \frac{\left(\frac{24}{32.1}\right)^2}{\left(\frac{3}{32.1}\right)\left(\frac{5.1}{32.1}\right)} \]

### Step 8: Simplify the expression
This simplifies to:
\[ K_c = \frac{(24)^2}{3 \times 5.1} = \frac{576}{15.3} \approx 37.647 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) at \( 444^\circ C \) is approximately **37.647**.

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