When 1 mole of `H_(2(g))` is heated with one mole of `I_(2(g)),` it was found that 1.48 moles of `HI_((g))` is formed at equilibrium. Its `K_c` is

Options

A. 16
B. 32
C. 8
D. 24

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction between hydrogen gas and iodine gas to form hydrogen iodide is:
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g) \]

### Step 2: Set up the initial concentrations
Initially, we have:
- 1 mole of \( \text{H}_2 \)
- 1 mole of \( \text{I}_2 \)
- 0 moles of \( \text{HI} \)

### Step 3: Define the change in moles at equilibrium
Let \( x \) be the amount of \( \text{H}_2 \) and \( \text{I}_2 \) that reacts. According to the stoichiometry of the reaction:
- \( \text{H}_2 \) decreases by \( x \)
- \( \text{I}_2 \) decreases by \( x \)
- \( \text{HI} \) increases by \( 2x \)

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \( [\text{H}_2] = 1 - x \)
- \( [\text{I}_2] = 1 - x \)
- \( [\text{HI}] = 2x \)

### Step 5: Use the given information to find \( x \)
We know that at equilibrium, 1.48 moles of \( \text{HI} \) are formed:
\[ 2x = 1.48 \]
\[ x = \frac{1.48}{2} = 0.74 \]

### Step 6: Calculate the equilibrium concentrations
Now substituting \( x \) back into the equilibrium expressions:
- \( [\text{H}_2] = 1 - 0.74 = 0.26 \)
- \( [\text{I}_2] = 1 - 0.74 = 0.26 \)
- \( [\text{HI}] = 1.48 \)

### Step 7: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]

### Step 8: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{(1.48)^2}{(0.26)(0.26)} \]

### Step 9: Calculate \( K_c \)
Calculating the values:
- \( (1.48)^2 = 2.1904 \)
- \( (0.26)(0.26) = 0.0676 \)

Now substituting these values into the equation:
\[ K_c = \frac{2.1904}{0.0676} \]

### Step 10: Final calculation
Calculating \( K_c \):
\[ K_c \approx 32.4 \]

Thus, the value of \( K_c \) is **32.4**.

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