`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

Answer

To find the equilibrium constant \( K_c \) for the reaction

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

we will follow these steps:

### Step 1: Write the balanced equation
The balanced equation for the reaction is:

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

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

Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that react at equilibrium. Therefore, at equilibrium, we have:
- Moles of \( H_2 = 4.5 - x \)
- Moles of \( I_2 = 4.5 - x \)
- Moles of \( HI = 2x \)

### Step 3: Use the information given
We know from the problem that at equilibrium, there are 3 moles of \( HI \):
\[ 2x = 3 \implies x = \frac{3}{2} = 1.5 \]

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

### Step 5: Calculate concentrations
The volume of the vessel is given as 10 liters. The concentrations at equilibrium can be calculated as follows:
- Concentration of \( H_2 = \frac{3 \text{ moles}}{10 \text{ L}} = 0.3 \, \text{M} \)
- Concentration of \( I_2 = \frac{3 \text{ moles}}{10 \text{ L}} = 0.3 \, \text{M} \)
- Concentration of \( HI = \frac{3 \text{ moles}}{10 \text{ L}} = 0.3 \, \text{M} \)

### Step 6: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:

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

### Step 7: Substitute the concentrations into the expression
Substituting the concentrations we found:

\[ K_c = \frac{(0.3)^2}{(0.3)(0.3)} = \frac{0.09}{0.09} = 1 \]

### Conclusion
The equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 1 \]