Question

Consider the following reversible reaction. Upper C upper O (g) plus 2 upper H subscript 2 (g) double-headed arrow upper C upper H subscript 3 upper O upper H (g). What is the equilibrium constant expression for the given system? K e q equals StartFraction StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket superscript 2 over StartBracket upper C upper H subscript 3 upper O upper H EndBracket EndFraction. K e q equals StartFraction StartBracket upper C upper H subscript 3 upper H upper O EndBracket over StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket superscript 2 EndFraction. K e q equals StartFraction StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket over StartBracket upper C upper H subscript 3 upper O upper H EndBracket EndFraction. K e q equals sart fraction StartBracket upper C upper H subscript 3 upper O StartBracket upper C upper O EndBracket StartBracket upper H 2 EndBracket EndFraction.

Answer 1

Answer:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

Explanation:

Hello,

In this case, for the described reaction at equilibrium:

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The analysis of the law of mass action allows us to write the equilibrium expression as shown below:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

Which is written considering that carbon monoxide, hydrogen and methanol are all in gaseous phase, for that reason all of them are included in the expression due to homogeneous equilibrium. Moreover, since hydrogen has a stoichiometric coefficient of 2, it is squared in the law of mass action.

Best regards.