For the reaction `H_2 +I_2 hArr 2HI`, the equilibrium concentration of `H_2,I_2` and HI are 8.0, 3.0 and 28.0 mol per litre respectively, the equilibrium constant of the reaction

Options

A. `30.66 `
B. `32.66`
C. `34.66`
D. `36.66`

Explanation

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

\[ H_2 + I_2 \rightleftharpoons 2 HI \]

given the equilibrium concentrations of \( H_2 \), \( I_2 \), and \( HI \) as 8.0 mol/L, 3.0 mol/L, and 28.0 mol/L respectively, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by the formula:

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

### Step 2: Substitute the equilibrium concentrations into the expression

Now, we can substitute the given equilibrium concentrations into the expression:

- \([HI] = 28.0 \, \text{mol/L}\)
- \([H_2] = 8.0 \, \text{mol/L}\)
- \([I_2] = 3.0 \, \text{mol/L}\)

So, we have:

\[ K_c = \frac{(28.0)^2}{(8.0)(3.0)} \]

### Step 3: Calculate the numerator and denominator

First, calculate the numerator:

\[ (28.0)^2 = 784.0 \]

Next, calculate the denominator:

\[ (8.0)(3.0) = 24.0 \]

### Step 4: Divide the numerator by the denominator

Now, divide the numerator by the denominator to find \( K_c \):

\[ K_c = \frac{784.0}{24.0} \approx 32.67 \]

### Step 5: Final result

Thus, the equilibrium constant \( K_c \) for the reaction is approximately:

\[ K_c \approx 32.67 \]