The equilibrium constant for the reaction: H 2 ​ (g)+I 2 ​ (g)⇔2HI(g) is 64, at certain temperature.The equilibrium concentrations of H2 and HI are 2 and 16 mollit −1 respectively. The equilibrium concentration (in molLit −1 ) of I2 ​ is ?

Explanation

To find the equilibrium concentration of I2 in the reaction:

\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]

we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (K)
The equilibrium constant (K) for the reaction can be expressed as:

\[ K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]

### Step 2: Substitute the known values into the equation
From the problem, we know:
- \( K = 64 \)
- \( [\text{H}_2] = 2 \, \text{mol/L} \)
- \( [\text{HI}] = 16 \, \text{mol/L} \)

Now, substituting these values into the equilibrium expression:

\[ 64 = \frac{(16)^2}{(2)([\text{I}_2])} \]

### Step 3: Simplify and solve for [I2]
Calculating \( (16)^2 \):

\[ 64 = \frac{256}{(2)([\text{I}_2])} \]

Now, multiply both sides by \( (2)([\text{I}_2]) \):

\[ 64 \cdot (2)([\text{I}_2]) = 256 \]

This simplifies to:

\[ 128[\text{I}_2] = 256 \]

Now, divide both sides by 128:

\[ [\text{I}_2] = \frac{256}{128} = 2 \, \text{mol/L} \]

### Final Answer
The equilibrium concentration of I2 is:

\[ [\text{I}_2] = 2 \, \text{mol/L} \]

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