Question

Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentrations of CO and H2 were 0.15 M and0.36 M,respectively. What is the equilibrium concentration of CH3OH? The value of Keq for this reaction is 14.5 at the temperature of the experiment. What is the equilibrium concentration of CH3OH?
A) 14.5B) 7.61 x 10 ^ -3C) 2.82 x 10 ^ -1D) 3.72 x 10 ^ -3E) 1.34 x 10 ^ -3

Answer 1

Answer : The equilibrium concentration of $CH_3OH$ will be, (C) $2.82\times 10^{-1}$

Explanation :  Given,

Equilibrium constant = 14.5

Concentration of $CO$ at equilibrium = 0.15 M

Concentration of $H_2$ at equilibrium = 0.36 M

The balanced equilibrium reaction is,

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

The expression of equilibrium constant for the reaction will be:

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

Now put all the values in this expression, we get:

$14.5=\frac{[CH_3OH]}{(0.15)\times (0.36)^2}$

$[CH_3OH]=2.82\times 10^{-1}M$

Therefore, the equilibrium concentration of $CH_3OH$ will be, (C) $2.82\times 10^{-1}$