In a reaction between hydrogen and iodine `6.84 mol` of hydrogen and `4.02` mol of iodine are found to be in equilibrium with `42.85` mol of hydrogen iodide at `350^(@)C.` Calculate the equilibrium constant.

Explanation

To calculate the equilibrium constant for the reaction between hydrogen and iodine, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen (H₂) and iodine (I₂) to form hydrogen iodide (HI) can be represented as:
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]

### Step 2: Identify the equilibrium concentrations
From the problem, we have:
- Moles of H₂ at equilibrium = 6.84 mol
- Moles of I₂ at equilibrium = 4.02 mol
- Moles of HI at equilibrium = 42.85 mol

### Step 3: Write the expression for the equilibrium constant (K)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]

### Step 4: Convert moles to concentrations
Assuming the volume of the reaction vessel is V (in liters), the equilibrium concentrations can be expressed as:
- \([\text{H}_2] = \frac{6.84}{V}\)
- \([\text{I}_2] = \frac{4.02}{V}\)
- \([\text{HI}] = \frac{42.85}{V}\)

### Step 5: Substitute the concentrations into the equilibrium expression
Substituting the equilibrium concentrations into the K expression:
\[ K_c = \frac{\left(\frac{42.85}{V}\right)^2}{\left(\frac{6.84}{V}\right)\left(\frac{4.02}{V}\right)} \]

### Step 6: Simplify the expression
This simplifies to:
\[ K_c = \frac{(42.85)^2}{(6.84)(4.02)} \]
The \(V\) terms cancel out.

### Step 7: Calculate the values
Now we will calculate:
1. \( (42.85)^2 = 1836.1225 \)
2. \( (6.84)(4.02) = 27.4968 \)

### Step 8: Divide the results to find K
Now, we can find \(K_c\):
\[ K_c = \frac{1836.1225}{27.4968} \approx 66.7758 \]

### Final Answer
Thus, the equilibrium constant \(K_c\) is approximately:
\[ K_c \approx 66.78 \]

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