For the reaction, `H_(2) + I_(2)hArr 2HI, K = 47.6` . If the initial number of moles of each reactant and product is 1 mole then at equilibrium

Options

A. `[1_(2)]=[H_(2)],[l_(2)]gt[Hl]`
B. `[1_(2)]lt[H_(2)],[l_(2)]=[Hl]`
C. `[1_(2)]=[H_(2)],[l_(2)]lt[Hl]`
D. `[1_(2)]gt[H_(2)],[l_(2)]=[Hl]`

Answer

To solve the problem, we need to analyze the equilibrium of the reaction:

\[ H_2 + I_2 \rightleftharpoons 2HI \]

Given that the equilibrium constant \( K_c = 47.6 \) and the initial number of moles of each reactant and product is 1 mole, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant

The equilibrium constant expression for the reaction is:

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

### Step 2: Set up the initial concentrations

Since we have 1 mole of each substance and assuming the volume of the reaction vessel is 1 L (for simplicity), the initial concentrations are:

\[ [H_2] = 1 \, \text{M}, \quad [I_2] = 1 \, \text{M}, \quad [HI] = 1 \, \text{M} \]

### Step 3: Determine the change in concentrations at equilibrium

Let \( x \) be the change in concentration of \( H_2 \) and \( I_2 \) that reacts to form \( HI \). At equilibrium, the concentrations will be:

\[ [H_2] = 1 - x, \quad [I_2] = 1 - x, \quad [HI] = 1 + 2x \]

### Step 4: Substitute the equilibrium concentrations into the \( K_c \) expression

Substituting these equilibrium concentrations into the \( K_c \) expression gives:

\[ K_c = \frac{(1 + 2x)^2}{(1 - x)(1 - x)} \]

### Step 5: Set the equation equal to the given \( K_c \)

Now we set this equal to the given \( K_c \):

\[ 47.6 = \frac{(1 + 2x)^2}{(1 - x)^2} \]

### Step 6: Solve for \( x \)

Cross-multiplying gives:

\[ 47.6(1 - x)^2 = (1 + 2x)^2 \]

Expanding both sides:

\[ 47.6(1 - 2x + x^2) = 1 + 4x + 4x^2 \]

This simplifies to:

\[ 47.6 - 95.2x + 47.6x^2 = 1 + 4x + 4x^2 \]

Rearranging gives:

\[ (47.6 - 4)x^2 - (95.2 + 4)x + (47.6 - 1) = 0 \]

\[ 43.6x^2 - 99.2x + 46.6 = 0 \]

### Step 7: Use the quadratic formula to find \( x \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 43.6 \), \( b = -99.2 \), and \( c = 46.6 \).

Calculating the discriminant:

\[ D = (-99.2)^2 - 4 \cdot 43.6 \cdot 46.6 \]

Calculating \( D \):

\[ D = 9840.64 - 8124.64 = 1716 \]

Now, substituting back into the quadratic formula:

\[ x = \frac{99.2 \pm \sqrt{1716}}{2 \cdot 43.6} \]

Calculating \( \sqrt{1716} \approx 41.4 \):

\[ x = \frac{99.2 \pm 41.4}{87.2} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{140.6}{87.2} \approx 1.61 \) (not possible since it exceeds initial concentration)
2. \( x = \frac{57.8}{87.2} \approx 0.66 \)

### Step 8: Calculate the equilibrium concentrations

Now substituting \( x \) back to find equilibrium concentrations:

\[ [H_2] = 1 - 0.66 = 0.34 \, \text{M} \]
\[ [I_2] = 1 - 0.66 = 0.34 \, \text{M} \]
\[ [HI] = 1 + 2(0.66) = 2.32 \, \text{M} \]

### Conclusion

At equilibrium, we have:

- \( [H_2] = 0.34 \, \text{M} \)
- \( [I_2] = 0.34 \, \text{M} \)
- \( [HI] = 2.32 \, \text{M} \)

Thus, the correct relationships are:

- \( [H_2] = [I_2] < [HI] \)