If `0.5 `mole `H_(2)` is reacted with 0.5 mole `I_(2)` in a ten `-` litre container at `444^(@)C` and at same temperature value of equilibrium constant `K_(C)` is 49, the ratio of `[Hl]` and `[l_(2)]` will be `:`

Options

A. 7
B. `(1)/(7)`
C. `sqrt((1)/(7))`
D. 49

Explanation

To solve the problem, we need to determine the ratio of the concentrations of HI (hydrogen iodide) to I2 (iodine) at equilibrium given the initial amounts of H2 and I2, and the equilibrium constant Kc.

### Step-by-Step Solution:

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

2. **Initial Concentrations**:
We start with 0.5 moles of H2 and 0.5 moles of I2 in a 10-liter container.
The initial concentrations are:
\[ [H_2] = \frac{0.5 \text{ moles}}{10 \text{ L}} = 0.05 \text{ M} \]
\[ [I_2] = \frac{0.5 \text{ moles}}{10 \text{ L}} = 0.05 \text{ M} \]
\[ [HI] = 0 \text{ M} \] (initially)

3. **Change in Concentrations at Equilibrium**:
Let \( x \) be the amount of H2 and I2 that reacts at equilibrium.
The changes in concentrations will be:
\[ [H_2] = 0.05 - x \]
\[ [I_2] = 0.05 - x \]
\[ [HI] = 2x \] (since 2 moles of HI are produced for every mole of H2 and I2 that reacts)

4. **Equilibrium Constant Expression**:
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
Substituting the equilibrium concentrations, we have:
\[ K_c = \frac{(2x)^2}{(0.05 - x)(0.05 - x)} \]

5. **Substituting the Given Value of Kc**:
We know that \( K_c = 49 \). Therefore, we can set up the equation:
\[ 49 = \frac{(2x)^2}{(0.05 - x)^2} \]
Simplifying this gives:
\[ 49(0.05 - x)^2 = 4x^2 \]

6. **Expanding and Rearranging**:
Expanding the left side:
\[ 49(0.0025 - 0.1x + x^2) = 4x^2 \]
\[ 0.1225 - 4.9x + 49x^2 = 4x^2 \]
Rearranging gives:
\[ 45x^2 - 4.9x + 0.1225 = 0 \]

7. **Solving the Quadratic Equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 45, b = -4.9, c = 0.1225 \).
Calculate the discriminant:
\[ b^2 - 4ac = (-4.9)^2 - 4(45)(0.1225) \]
\[ = 24.01 - 22.05 = 1.96 \]
Now calculate \( x \):
\[ x = \frac{4.9 \pm \sqrt{1.96}}{90} \]
\[ = \frac{4.9 \pm 1.4}{90} \]
This gives two possible values for \( x \). We take the positive root.

8. **Finding the Ratio of Concentrations**:
Once we have \( x \), we can find the concentrations at equilibrium:
\[ [HI] = 2x \]
\[ [I_2] = 0.05 - x \]
The ratio \( \frac{[HI]}{[I_2]} = \frac{2x}{0.05 - x} \).
From the earlier calculation, we know that \( K_c = 49 \) implies \( \frac{[HI]}{[I_2]} = 7 \).

### Final Answer:
The ratio of \([HI]\) to \([I_2]\) is \( 7:1 \).

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