At 298 K \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \mathrm{K}_{1}=4 \times 10^{5}\) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g}), \mathrm{K}_{2}=1.6 \times 10^{12}\) \(\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}), \mathrm{K}_{3}=1.0 \times 10^{-13}\) Based on above equilibria, the equilibrium constant of the reaction, \(2 \mathrm{NH}_{3}(\mathrm{~g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})\) is \(\_\_\_\_\) \(\times 10^{-33}\) (Nearest integer)

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Equilibrium

At 298 K

N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g), K_{1} = 4 \times 1 0^{5}

N_{2} (g) + O_{2} (g) ⇌ 2 NO (g), K_{2} = 1.6 \times 1 0^{12}

H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g), K_{3} = 1.0 \times 1 0^{- 13}

Based on above equilibria, the equilibrium constant of the reaction,

2 NH_{3} (g) + \frac{5}{2} O_{2} (g) ⇌ 2 NO (g) + 3 H_{2} O (g)

____

\times 1 0^{- 33}

(Nearest integer)

Answer: 4

Solution

The target change is assembled from the given equilibria: the ammonia equilibrium is reversed, so its constant is inverted; the nitric oxide equilibrium is kept as it stands; and the water-forming equilibrium is taken three times, so its constant is cubed. Multiplying these together gives

K = \frac{K _{2} K _{3}^{3}}{K _{1}} = \frac{( 1.6 \times 1 0 ^{12} ) ( 1.0 \times 1 0 ^{- 13} ) ^{3}}{4 \times 1 0 ^{5}} = \frac{1.6 \times 1 0 ^{- 27}}{4 \times 1 0 ^{5}} = 4 \times 1 0^{- 33}

. The required figure is four.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Equilibrium

Solution

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