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Chemical Equilibrium question

2023 · 29 Jan · Shift 2 · Q18
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Chemical Equilibrium question

2023 · 29 Jan · Shift 2 · Q18

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
At 298 K N2 (g)+3H2 (g)⇌ 2NH3 (g), K1=4×105N2 (g)+O2 (g)⇌ 2NO (g), K2=1.6×1012H2 (g)+12O2 (g)⇌ H2O (g), K3=1.0×10−13\mathrm{N_2~(g)+3H_2~(g)\rightleftharpoons~2NH_3~(g),~K_1=4\times10^5}\mathrm{N_2~(g)+O_2~(g)\rightleftharpoons~2NO~(g),~K_2=1.6\times10^{12}}\mathrm{H_2~(g)+\frac{1}{2}O_2~(g)\rightleftharpoons~H_2O~(g),~K_3=1.0\times10^{-13}}N2​ (g)+3H2​ (g)⇌ 2NH3​ (g), K1​=4×105N2​ (g)+O2​ (g)⇌ 2NO (g), K2​=1.6×1012H2​ (g)+21​O2​ (g)⇌ H2​O (g), K3​=1.0×10−13 Based on above equilibria, then equilibrium constant of the reaction, 2NH3(g)+52O2 (g)⇌ 2NO (g)+3H2O (g)\mathrm{2NH_3(g)+\frac{5}{2}O_2~(g)\rightleftharpoons~2NO~(g)+3H_2O~(g)}2NH3​(g)+25​O2​ (g)⇌ 2NO (g)+3H2​O (g) is ‾\underline{\hspace{2cm}}​×10−33\times10^{-33}×10−33 (Nearest integer).
Numerical answer
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