Equilibrium concentration of `HI, I_(2) and H_(2)` is 0.7, 0.1 and 0.1 M respectively. The equilibrium constant for the reaction, `I_(2)+H_(2)hArr 2HI` is :

Options

A. 0.36
B. 36
C. 49
D. 0.49

Answer

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

\[ I_2 + H_2 \rightleftharpoons 2 HI \]

given the equilibrium concentrations of \( HI \), \( I_2 \), and \( H_2 \) as 0.7 M, 0.1 M, and 0.1 M respectively, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by the formula:

\[ K_c = \frac{[HI]^2}{[I_2][H_2]} \]

where:
- \([HI]\) is the equilibrium concentration of hydrogen iodide,
- \([I_2]\) is the equilibrium concentration of iodine,
- \([H_2]\) is the equilibrium concentration of hydrogen.

### Step 2: Substitute the equilibrium concentrations into the expression

From the question, we know:
- \([HI] = 0.7 \, \text{M}\)
- \([I_2] = 0.1 \, \text{M}\)
- \([H_2] = 0.1 \, \text{M}\)

Now, substituting these values into the expression for \( K_c \):

\[ K_c = \frac{(0.7)^2}{(0.1)(0.1)} \]

### Step 3: Calculate \( K_c \)

Calculating the numerator and denominator:

- Numerator: \((0.7)^2 = 0.49\)
- Denominator: \((0.1)(0.1) = 0.01\)

Now, substituting these values into the equation:

\[ K_c = \frac{0.49}{0.01} = 49 \]

### Step 4: State the final answer

Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 49 \]