JEE Chemistry 2023 Question with Solution

Given: Initial moles of $\text{H}_2 = 4.5$, initial moles of $\text{I}_2 = 4.5$, equilibrium moles of $\text{HI} = 3$, vessel volume $= 10 \, \text{L}$.

Find: The equilibrium constant $K_c$ for

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

Using the ICE table:

• $\text{H}_2: 4.5,\ -x,\ 4.5-x$
• $\text{I}_2: 4.5,\ -x,\ 4.5-x$
• $\text{HI}: 0,\ +2x,\ 2x$

Since equilibrium moles of $\text{HI}$ are $3$,

$$2x = 3$$

So,

$$x = \frac{3}{2} = 1.5$$

Therefore, equilibrium moles are:

• $\text{H}_2 = 4.5 - 1.5 = 3$
• $\text{I}_2 = 4.5 - 1.5 = 3$
• $\text{HI} = 3$

Now divide by volume $10 \, \text{L}$ to get equilibrium concentrations:

$$[\text{H}_2] = \frac{3}{10} = 0.3$$ $$[\text{I}_2] = \frac{3}{10} = 0.3$$ $$[\text{HI}] = \frac{3}{10} = 0.3$$

The equilibrium constant is

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

Substituting the values,

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

Therefore, the equilibrium constant is $1$.

Given: Equal initial moles of $\text{H}_2$ and $\text{I}_2$ are taken, and $3$ moles of $\text{HI}$ are present at equilibrium.

Find: The value of $K_c$.

For the reaction

$$\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$$

if $x$ moles each of $\text{H}_2$ and $\text{I}_2$ react, then $2x$ moles of $\text{HI}$ are formed.

Given,

$$2x = 3$$

Hence,

$$x = 1.5$$

So at equilibrium:

$$\text{H}_2 = 4.5 - 1.5 = 3$$ $$\text{I}_2 = 4.5 - 1.5 = 3$$ $$\text{HI} = 3$$

Since the vessel volume is $10 \, \text{L}$,

$$[\text{H}_2] = [\text{I}_2] = [\text{HI}] = \frac{3}{10} = 0.3$$

Now,

$$K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.3)^2}{(0.3)(0.3)} = 1$$

Therefore, the correct numerical value is $1$.