1. Tardigrade
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  3. Chemistry
  4. At 298 K N 2( g )+3 H 2( g ) leftharpoons 2 NH 3( g ), K 1=4 × 105 N 2( g )+ O 2( g ) leftharpoons 2 NO ( g ), K 2=1.6 × 1012 H 2( g )+(1/2) O 2( g ) leftharpoons H 2 O ( g ), K 3=1.0 × 10-13 Based on above equilibria, the equilibrium constant of the reaction, 2 NH 3( g )+(5/2) O 2( g ) leftharpoons 2 NO ( g )+3 H 2 O ( g ) is × 10-33 (Nearest integer)

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Q. At $298 K$
$ N _2( g )+3 H _2( g ) \rightleftharpoons 2 NH _3( g ), K _1=4 \times 10^5 $
$ N _2( g )+ O _2( g ) \rightleftharpoons 2 NO ( g ), K _2=1.6 \times 10^{12} $
$ H _2( g )+\frac{1}{2} O _2( g ) \rightleftharpoons H _2 O ( g ), K _3=1.0 \times 10^{-13}$
Based on above equilibria, the equilibrium constant of the reaction,
$2 NH _3( g )+\frac{5}{2} O _2( g ) \rightleftharpoons 2 NO ( g )+3 H _2 O ( g )$
is ___$\times 10^{-33}$
(Nearest integer)

JEE MainJEE Main 2023Equilibrium

Solution:

$ N _2( g )+3 H _2( g ) \rightleftharpoons 2 NH _3( g ), K _1=4 \times 10^5 \ldots( i )$
$ N _2( g )+ O _2( g ) \rightleftharpoons 2 NO ( g ), K _2=1.6 \times 10^{12} \ldots( ii )$
$ H _2( g )+\frac{1}{2} O _2( g ) \rightleftharpoons H _2 O ( g ), K _3=1.0 \times 10^{-13} \ldots (iii)$
$(ii)+3 \times( iii)-( i)$
$ 2 NH _3( g )+\frac{5}{2} O _2( g ) \rightleftharpoons 2 NO ( g )+3 H _2 O ( g )$
$k _{ eq }=\frac{ k _2 \times k _3^3}{ k _1}=\frac{1.6 \times 10^{12} \times\left(10^{-13}\right)^3}{4 \times 10^5} $
$ =\frac{1.6}{4} \times 10^{-32}=4 \times 10^{-33}$