Question

Consider the following reversible reaction.

Answer 1

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

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. It is expressed as $K_{eq}$

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

The equilibrium constant in terms of concentration is written as :

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

Thus the correct answer choice is B.

Answer 2

Answer:

Explanation:

For any system in equilibrium, the molar concentration of all the species on the reactant side are related to those on the product side by a constant known as the equilibrium constant $K_{eq}$.

For a given reaction:

              aA + bB ⇄ cC

 $K_{eq}$ =   $\frac{[C]^{c} }{[A]^{a} [B]^{b} }$

The reaction equation is given as:

     $CO_{g}$ + 2H₂$_{g}$ ⇆CH₃OH $_{g}$

Note: All the species are in gaseous phase.

            $K_{eq}$ = $\frac{[CH_{3}OH ]}{[CO] [H_{2}] }$