1. Tardigrade
  2. Question
  3. Chemistry
  4. Equilibrium constants for the following reactions at 1200 K are given: 2H2O(g) leftharpoons 2H2(g) + O2(g); K1 = 6.4 × 10-8 2CO2(g) leftharpoons 2CO(g) + O2(g); K2 = 1.6 × 10-6 The equilibrium constant for the reaction H2(g) + CO2 (g) leftharpoons CO(g) + H2O(g) at 1200 K will be

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Q. Equilibrium constants for the following reactions at $1200\, K$ are given :
$2H_{2}O_{(g)} {\rightleftharpoons} 2H_{2}{(g)} + O_{2}$$_{(g)}; K_1 = 6.4 × 10^{-8}$
$2CO_{2}$$_{(g)} {\rightleftharpoons} 2CO_{(g)} + O_{2}$$_{(g)}; K_{2} = 1.6 × 10^{-6}$
The equilibrium constant for the reaction
$H_{2}$$_{(g)}$ $ + CO_{2}$ $_{(g)} {\rightleftharpoons} CO_{(g)} + H_{2}O_{(g)}$ at $1200\, K$ will be

WBJEEWBJEE 2017Equilibrium

Solution:

For the given reactions,
$2 H _{2} O \rightleftharpoons 2 H _{2}+ O _{2}, K_{1}=6.4 \times 10^{-8}$
or, $H _{2} O \rightleftharpoons H _{2}+\frac{1}{2} O _{2}, K_{1}'=\sqrt{K_{1}}$
or $H _{2}+\frac{1}{2} O _{2} \rightleftharpoons H _{2} O , K_{1}''=1 / \sqrt{K_{1}}\,\,\,\, \ldots$ (i)
$2 CO _{2} \rightleftharpoons 2 CO + O _{2}, K_{2}=1.6 \times 10^{-6}$
or $CO _{2} \rightleftharpoons CO +\frac{1}{2} O _{2}, K_{2}'=\sqrt{K_{2}} \,\,\,\,\ldots$ (ii)
Form (i) and (ii) [on adding]
$K=\frac{\sqrt{K_{2}}}{\sqrt{K_{1}}}=\frac{\sqrt{1.6 \times 10^{-6}}}{\sqrt{6.4 \times 10^{-8}}}$
$K=\sqrt{\frac{10^{2}}{4}}=\sqrt{25}=5$