Equimolar concentrations of `H_(2)` and `I_(2)` are heated to equilibrium in a 2 L flask. At equilibrium, the forward and backward rate constants arer found to be equal. What percentage of initial concentration of `H_(2)` has reached at equilibrium ?

Options

A. `33%`
B. `66%`
C. `50%`
D. `40%`

Explanation

To solve the problem, we need to analyze the equilibrium reaction between hydrogen (H₂) and iodine (I₂) to form hydrogen iodide (HI). The reaction can be represented as:

\[ \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \]

### Step-by-step Solution:

1. **Initial Concentrations**:
Since we have equimolar concentrations of H₂ and I₂, let's assume the initial concentration of both is 1 M in a 2 L flask. Therefore, the initial moles of H₂ and I₂ are:
\[ \text{Initial moles of H}_2 = 1 \text{ mol} \quad \text{and} \quad \text{Initial moles of I}_2 = 1 \text{ mol} \]

2. **Change in Concentration**:
Let \( x \) be the amount of H₂ and I₂ that reacts at equilibrium. Therefore, at equilibrium:
- Moles of H₂ left = \( 1 - x \)
- Moles of I₂ left = \( 1 - x \)
- Moles of HI formed = \( 2x \)

3. **Equilibrium Expression**:
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(1 - x)(1 - x)} = \frac{4x^2}{(1 - x)^2} \]

4. **Given Condition**:
We are told that the forward and backward rate constants are equal, which implies:
\[ K_c = 1 \]

5. **Setting Up the Equation**:
Setting the equilibrium expression equal to 1:
\[ \frac{4x^2}{(1 - x)^2} = 1 \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 4x^2 = (1 - x)^2 \]

7. **Expanding the Equation**:
Expanding the right side:
\[ 4x^2 = 1 - 2x + x^2 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 4x^2 - x^2 + 2x - 1 = 0 \implies 3x^2 + 2x - 1 = 0 \]

9. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 3, \quad b = 2, \quad c = -1 \]
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \]

10. **Finding the Roots**:
This gives us two possible solutions:
\[ x = \frac{2}{6} = \frac{1}{3} \quad \text{and} \quad x = \frac{-6}{6} = -1 \quad (\text{not valid}) \]

11. **Calculating the Percentage**:
The valid solution is \( x = \frac{1}{3} \). This means that \( \frac{1}{3} \) moles of H₂ have reacted. The percentage of H₂ that has reacted is:
\[ \text{Percentage reacted} = \left(\frac{x}{\text{Initial moles}} \times 100\right) = \left(\frac{1/3}{1} \times 100\right) = 33.33\% \]

12. **Final Answer**:
Therefore, the percentage of initial concentration of H₂ that has reached equilibrium is:
\[ \text{Percentage of H}_2 \text{ remaining} = 100\% - 33.33\% = 66.67\% \]