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 are 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%`

Answer

To solve the problem, we need to analyze the equilibrium of the reaction between hydrogen (\(H_2\)) and iodine (\(I_2\)) to form hydrogen iodide (\(HI\)). The reaction can be represented as:

\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step-by-step Solution:

1. **Initial Concentration**:
- Given that we have equimolar concentrations of \(H_2\) and \(I_2\), we can assume that initially, we have 1 mole of each in a 2 L flask. Therefore, the initial concentrations are:
\[ [H_2]_{initial} = 0.5 \, \text{M}, \quad [I_2]_{initial} = 0.5 \, \text{M}, \quad [HI]_{initial} = 0 \, \text{M} \]

2. **Change in Concentration**:
- Let \(x\) be the amount of \(H_2\) and \(I_2\) that reacts at equilibrium. Thus, at equilibrium, the concentrations will be:
\[ [H_2]_{equilibrium} = 0.5 - x \, \text{M}, \quad [I_2]_{equilibrium} = 0.5 - x \, \text{M}, \quad [HI]_{equilibrium} = 2x \, \text{M} \]

3. **Equilibrium Constant**:
- The equilibrium constant \(K_c\) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
- Since it is given that the forward and backward rate constants are equal, we have:
\[ K_c = 1 \]

4. **Setting Up the Equation**:
- Substituting the equilibrium concentrations into the expression for \(K_c\):
\[ 1 = \frac{(2x)^2}{(0.5 - x)(0.5 - x)} \]
- This simplifies to:
\[ 1 = \frac{4x^2}{(0.5 - x)^2} \]

5. **Cross Multiplying**:
- Cross-multiplying gives:
\[ (0.5 - x)^2 = 4x^2 \]

6. **Expanding and Rearranging**:
- Expanding the left side:
\[ 0.25 - x + x^2 = 4x^2 \]
- Rearranging gives:
\[ 3x^2 + x - 0.25 = 0 \]

7. **Solving the Quadratic Equation**:
- Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-0.25)}}{2 \cdot 3} \]
- This simplifies to:
\[ x = \frac{-1 \pm \sqrt{1 + 3}}{6} = \frac{-1 \pm 2}{6} \]
- Thus, we have two possible solutions for \(x\):
\[ x = \frac{1}{6} \quad \text{(valid)} \quad \text{or} \quad x = -\frac{3}{6} \quad \text{(not valid)} \]

8. **Calculating the Percentage**:
- The amount of \(H_2\) that has reacted is \(x = \frac{1}{6}\) moles. The initial concentration was 1 mole, so the percentage that has reacted is:
\[ \text{Percentage} = \left(\frac{x}{0.5} \times 100\right) = \left(\frac{\frac{1}{6}}{0.5} \times 100\right) = \left(\frac{1}{6} \times 200\right) = \frac{200}{6} \approx 33.33\% \]

### Final Answer:
Thus, the percentage of the initial concentration of \(H_2\) that has reached equilibrium is approximately **33.33%**.