The equilibrium constant for the reaction
${H}_{2}O \left(g\right) + CO \left(g\right) ⇌$ {H}_{2}\left(g\right) $+$ C{O}_{2}\left(g\right)
is 0.44 at 1260 K. $The equilibrium constant for the reaction<br />2{H}_{2} \left(g\right)$ + 2C{O}_{2} \left(g\right) $⇌$ 2CO\left(g\right)+ $2{H}_{2}O\left(g\right)$
at 1260 K is equal to

Question

The equilibrium constant for the reaction

{H}_{2}O \left(g\right) + CO \left(g\right) ⇌

{H}_{2}\left(g\right)

+

C{O}_{2}\left(g\right)
is 0.44 at 1260 K.

The equilibrium constant for the reaction<br />2{H}_{2} \left(g\right)

+ 2C{O}_{2} \left(g\right)

⇌

2CO\left(g\right)+

2{H}_{2}O\left(g\right)

at 1260 K is equal to

Quantrex Academy · Accepted Answer

Correct answer: C. 5.16

2 H _ 2 ( g ) + 2CO ( g ) ⇌ 2CO ( g ) + 2 H _ 2 O ( g ) is obtained by reversing the reaction H _ 2 O ( g ) + CO ( 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

The equilibrium constant for the reaction H _ 2 O (g) + CO (g) ⇌ H...

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