One mole each of hydrogen and iodine are allowed to react at certain temperature till the equilibrium is reached. Calculate the composition of equilibrium mixture if K= 66.5.

Explanation

To solve the problem, 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 written as:
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g) \]

### Step 2: Set up the initial concentrations
Initially, we have 1 mole of hydrogen and 1 mole of iodine. Therefore, the initial concentrations are:
- \([H_2] = 1 \, \text{mol}\)
- \([I_2] = 1 \, \text{mol}\)
- \([HI] = 0 \, \text{mol}\)

### Step 3: Define the change in concentration at equilibrium
Let \(x\) be the amount of hydrogen and iodine that reacts at equilibrium. The changes in concentration will be:
- \([H_2] = 1 - x\)
- \([I_2] = 1 - x\)
- \([HI] = 2x\) (since 2 moles of HI are produced for every mole of H₂ and I₂ that reacts)

### Step 4: Write the expression for the equilibrium constant \(K\)
The equilibrium constant expression for the reaction is given by:
\[ K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]
Substituting the equilibrium concentrations into the expression gives:
\[ K = \frac{(2x)^2}{(1 - x)(1 - x)} = \frac{4x^2}{(1 - x)^2} \]

### Step 5: Substitute the value of \(K\) and solve for \(x\)
Given \(K = 66.5\), we can set up the equation:
\[ 66.5 = \frac{4x^2}{(1 - x)^2} \]

Cross-multiplying gives:
\[ 66.5(1 - x)^2 = 4x^2 \]

Expanding the left side:
\[ 66.5(1 - 2x + x^2) = 4x^2 \]
\[ 66.5 - 133x + 66.5x^2 = 4x^2 \]

Rearranging the equation:
\[ 66.5x^2 - 4x^2 - 133x + 66.5 = 0 \]
\[ 62.5x^2 - 133x + 66.5 = 0 \]

### Step 6: Use the quadratic formula to find \(x\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- \(a = 62.5\)
- \(b = -133\)
- \(c = 66.5\)

Calculating the discriminant:
\[ b^2 - 4ac = (-133)^2 - 4(62.5)(66.5) \]
\[ = 17689 - 16650 = 1039 \]

Now substituting into the quadratic formula:
\[ x = \frac{133 \pm \sqrt{1039}}{2 \times 62.5} \]
Calculating \(\sqrt{1039} \approx 32.2\):
\[ x = \frac{133 \pm 32.2}{125} \]

Calculating the two possible values for \(x\):
1. \(x = \frac{165.2}{125} \approx 1.3216\) (not possible since it exceeds initial moles)
2. \(x = \frac{100.8}{125} \approx 0.8064\)

### Step 7: Calculate the equilibrium concentrations
Now that we have \(x \approx 0.8064\):
- \([H_2] = 1 - x = 1 - 0.8064 = 0.1936\)
- \([I_2] = 1 - x = 1 - 0.8064 = 0.1936\)
- \([HI] = 2x = 2 \times 0.8064 = 1.6128\)

### Final Answer
The equilibrium composition of the mixture is:
- \([H_2] \approx 0.194 \, \text{mol}\)
- \([I_2] \approx 0.194 \, \text{mol}\)
- \([HI] \approx 1.613 \, \text{mol}\)