Chemistry Q74 – JEE-MAIN 2025

The equilibrium constant for decomposition of

H_{2} O (g)

H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g) (Δ G^{\circ} = 92.34 {kJ mol}^{- 1})

8.0 \times 10^{- 3}

2300 K

and total pressure at equilibrium is

1 bar

. Under this condition, the degree of dissociation (

α

) of water is _______

\times 10^{- 2}

(nearest integer value). [Assume

α

is negligible with respect to 1]

Enter Calculated Numerical Value:

Clue 1 of 3: Set up the ICE table

Determine the equilibrium moles and total moles in terms of the degree of dissociation, $α$ .

🧠 Full Solution Path

Step 1: Set up Equilibrium and Partial Pressures✦ Active

The given reaction is: $H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g)$ . Let the initial moles of $H_{2} O$ be 1. If $α$ is the degree of dissociation, then at equilibrium:

\begin{array}{lccc} Species & H_{2} O & H_{2} & O_{2} \\ Initial moles & 1 & 0 & 0 \\ Change & - α & + α & + α / 2 \\ Equilibrium moles & 1 - α & α & α / 2 \end{array}

Total moles at equilibrium, $n_{t o t a l} = (1 - α) + α + \frac{α}{2} = 1 + \frac{α}{2}$ . Given that $α ≪ 1$ , we can approximate $n_{t o t a l} \approx 1$ . Partial pressures at equilibrium ( $P_{t o t a l} = 1 bar$ ):

P_{H_{2} O} = \frac{1 - α}{1 + α / 2} P_{t o t a l} \approx (1 - α) P_{t o t a l}

P_{H_{2}} = \frac{α}{1 + α / 2} P_{t o t a l} \approx α P_{t o t a l}

P_{O_{2}} = \frac{α / 2}{1 + α / 2} P_{t o t a l} \approx \frac{α}{2} P_{t o t a l}

💡 Teacher's Secret Hint

Remember to use the total moles at equilibrium to calculate partial pressures, and then apply the approximation.

Step 2: Write and Simplify the

K_{p}

Expression○ Expand

The equilibrium constant $K_{p}$ is given by:

K_{p} = \frac{P_{H_{2}} \cdot (P_{O_{2}})^{1 / 2}}{P_{H_{2} O}}

Substitute the partial pressure expressions:

K_{p} = \frac{(\frac{α}{1 + α / 2} P_{t o t a l}) \cdot {(\frac{α / 2}{1 + α / 2} P_{t o t a l})}^{1 / 2}}{(\frac{1 - α}{1 + α / 2} P_{t o t a l})}

K_{p} = \frac{α \cdot (α / 2)^{1 / 2}}{(1 - α) \cdot (1 + α / 2)^{1 / 2}} \cdot (P_{t o t a l})^{1 / 2}

Applying the approximation $α ≪ 1$ (so $1 - α \approx 1$ and $1 + α / 2 \approx 1$ ):

K_{p} \approx \frac{α \cdot (α / 2)^{1 / 2}}{1} \cdot (P_{t o t a l})^{1 / 2}

K_{p} \approx \frac{α^{3 / 2}}{\sqrt{2}} \cdot (P_{t o t a l})^{1 / 2}

💡 Teacher's Secret Hint

Ensure the exponents for partial pressures in the $K_{p}$ expression match the stoichiometric coefficients.

Step 3: Solve for

α

○ Expand

Given $K_{p} = 8.0 \times 10^{- 3}$ and $P_{t o t a l} = 1 bar$ :

8.0 \times 10^{- 3} = \frac{α^{3 / 2}}{\sqrt{2}} \cdot (1)^{1 / 2}

α^{3 / 2} = 8.0 \times 10^{- 3} \times \sqrt{2}

α^{3 / 2} = 8.0 \times 10^{- 3} \times 1.414

α^{3 / 2} = 0.011312

α = (0.011312)^{2 / 3}

α \approx 0.05035

We need to express $α$ in the format _______ $\times 10^{- 2}$ (nearest integer value). $α = 0.05035 = 5.035 \times 10^{- 2}$ . The nearest integer value for the blank is 5.

💡 Teacher's Secret Hint

Double-check your calculation for the fractional exponent and the final rounding to the nearest integer.

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Chemistry Question 74 – JEE-MAIN 2025

Clue 1 of 3: Set up the ICE table

🧠 Full Solution Path