At 298 K N₂~(g)+3H₂~(g) ~2NH₃~(g),~K₁=4×10⁵ N₂~(g)+O₂~(g) ~2NO~(g),~K₂=1.6×10¹² H₂~(g)+1/2O₂~(g) ~H₂O~(g),~K₃=1.0×10⁻¹³ Based on above equilibria, then equilibrium constant of the reaction, 2NH₃(g)+5/2O₂~(g) ~2NO~(g)+3H₂O~(g) is _ _ _ _ _ _ ×10⁻³³ (Nearest integer).

Equilibrium: JEE Main 2023 Chemistry question

JEE Main 2023 · 29 Jan, Evening shift · Q4

JEE Main 202329 Jan, Evening shiftChemistryHardEquilibrium123× since 2002

At 298 K

\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}}

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

\mathrm{2NH_3(g)+\frac{5}{2}O_2~(g)\rightleftharpoons~2NO~(g)+3H_2O~(g)}

is ____________

\times10^{-33}

(Nearest integer).

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2023 Equilibrium Question with Answer (29 Jan Evening shift, Q4) | JEE Arc