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

\[ I_2 + H_2 \rightleftharpoons 2 HI, \]

we can follow these steps:

### Step 1: Write the equilibrium constant expression
The equilibrium constant expression for the reaction is given by:

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

### Step 2: Substitute the equilibrium concentrations
From the problem, we know the equilibrium concentrations:
- \([HI] = 0.7 \, M\)
- \([I_2] = 0.1 \, M\)
- \([H_2] = 0.1 \, M\)

Substituting these values into the equilibrium constant expression:

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

### Step 3: Calculate the value
Now, we calculate the numerator and the denominator:

1. Calculate the numerator:
\[ (0.7)^2 = 0.49 \]

2. Calculate the denominator:
\[ (0.1)(0.1) = 0.01 \]

Now substitute these values back into the equation:

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

### Step 4: Final calculation
Now, perform the division:

\[ K_c = 49 \]

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

\[ \boxed{49} \]