Equilibrium concentration of HI,I2 and H2 is 0.7,0.1 and 0.1M respectively , Find Kc

Explanation

To find the equilibrium constant \( K_c \) for the reaction involving hydrogen (H₂), iodine (I₂), and hydrogen iodide (HI), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction can be represented as:
\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

2. **Identify the Equilibrium Concentrations:**
From the problem, we have the following equilibrium concentrations:
- \([HI] = 0.7 \, M\)
- \([I_2] = 0.1 \, M\)
- \([H_2] = 0.1 \, M\)

3. **Write the Expression for \( K_c \):**
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]
For our reaction, this becomes:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

4. **Substitute the Equilibrium Concentrations into the Expression:**
Plugging in the values we have:
\[ K_c = \frac{(0.7)^2}{(0.1)(0.1)} \]

5. **Calculate \( K_c \):**
First, calculate \( (0.7)^2 \):
\[ (0.7)^2 = 0.49 \]
Then, calculate the denominator:
\[ (0.1)(0.1) = 0.01 \]
Now, substitute these values into the equation:
\[ K_c = \frac{0.49}{0.01} = 49 \]

6. **Final Result:**
Therefore, the equilibrium constant \( K_c \) is:
\[ K_c = 49 \]