Water decomposes at 2300 K $$\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$$ The percent of water decomposing at 2300 K and 1 bar is ______ (Nearest integer). Equilibrium constant for the reaction is $2 \times 10^{-3}$ at 2300 K

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Equilibrium

Water decomposes at 2300 K

H_{2} O (g) \to H_{2} (g) + \frac{1}{2} O_{2} (g)

The percent of water decomposing at 2300 K and 1 bar is ______ (Nearest integer).

Equilibrium constant for the reaction is

2 \times 1 0^{- 3}

at 2300 K

Answer: 2

Starting from one mole of steam that dissociates to a degree

α

, the equilibrium amounts are

1 - α

of water vapour,

α

of hydrogen and

\frac{α}{2}

of oxygen. With

α

small the total pressure stays near one bar and

1 - α \approx 1

, so the expression

K_{p} = \frac{p _{H_{2}} p _{O_{2}}^{1/2}}{p _{H_{2} O}}

reduces to

\frac{α ( α /2 ) ^{1/2}}{1} = \frac{α ^{3/2}}{2} = 2 \times 1 0^{- 3}

. Then

α^{3/2} = 22 \times 1 0^{- 3}

, giving

α = 2 \times 1 0^{- 2}

, so the extent of decomposition at 2300 K is two percent.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Integer answer