The equilibrium constant for the reaction
$H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)$
is 0.44 at 1260 K. The equilibrium constant for the reaction
$2 H_{2} (g) + 2 C O_{2} (g) ⇌ 2 C O (g) + 2 H_{2} O (g)$
at 1260 K is equal to

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TestHub · Accepted Answer

2 H 2 ( g ) + 2 C O ( g ) ⇌ 2 C O ( g ) + 2 H 2 O ( g ) is obtained by reversing the reaction H 2 O ( g ) + C O ( g ) ⇌ H 2 ( g ) + C O 2 ( g ) and then multiplying it by the factor '2'. Hence the equilibrium constant ( K' ) for this reaction is given by K'= ( 1 K ) 2 = ( 1 0.44 ) 2 = 5.16

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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