`4.5` moles each of hydrogen and iodine heated in a sealed 10 litrevesel. At equilibrium, 3 moles of HI was foun. The equilibrium constant for `H_(2)(g) + I_(2) (g)hArr2HI(g)` is

Options

A. 1
B. 10
C. 5
D. `0.33`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 2: Determine initial moles and changes at equilibrium
Initially, we have:
- Moles of \( H_2 \) = 4.5 moles
- Moles of \( I_2 \) = 4.5 moles
- Moles of \( HI \) = 0 moles

Let \( x \) be the moles of \( H_2 \) and \( I_2 \) that react. According to the stoichiometry of the reaction:
- Moles of \( H_2 \) at equilibrium = \( 4.5 - x \)
- Moles of \( I_2 \) at equilibrium = \( 4.5 - x \)
- Moles of \( HI \) at equilibrium = \( 2x \)

### Step 3: Use the information given about equilibrium
We are given that at equilibrium, there are 3 moles of \( HI \):
\[ 2x = 3 \]
From this, we can solve for \( x \):
\[ x = \frac{3}{2} = 1.5 \]

### Step 4: Calculate the moles of \( H_2 \) and \( I_2 \) at equilibrium
Now substituting \( x \) back into the expressions for \( H_2 \) and \( I_2 \):
- Moles of \( H_2 \) at equilibrium = \( 4.5 - 1.5 = 3 \) moles
- Moles of \( I_2 \) at equilibrium = \( 4.5 - 1.5 = 3 \) moles

### Step 5: Calculate concentrations
The volume of the vessel is 10 liters. The concentration of each species at equilibrium can be calculated using the formula:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]

- Concentration of \( HI \):
\[ [HI] = \frac{3 \text{ moles}}{10 \text{ L}} = 0.3 \text{ M} \]

- Concentration of \( H_2 \):
\[ [H_2] = \frac{3 \text{ moles}}{10 \text{ L}} = 0.3 \text{ M} \]

- Concentration of \( I_2 \):
\[ [I_2] = \frac{3 \text{ moles}}{10 \text{ L}} = 0.3 \text{ M} \]

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

Substituting the concentrations we calculated:
\[ K_c = \frac{(0.3)^2}{(0.3)(0.3)} \]

### Step 7: Calculate \( K_c \)
Calculating the above expression:
\[ K_c = \frac{0.09}{0.09} = 1 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 1 \]

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