Question

10. At a certain temperature, Kc equals 1.4 × 102 for the reaction:2 CO(g) + O2(g) ⇌ 2 CO2(g). If a 2.50-L flask contains 0.400 mol of CO2 and 0.100 mol of O2 at equilibrium, how many moles of CO are also present in the flask?

Answer 1

Answer: Moles of CO present in the flask are 0.175

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

$2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g)$

Equilibrium concentration of $CO_2$ = $\frac{moles}{volume}=\frac{0.400}{2.50L}=0.16M$

Equilibrium concentration of $O_2$ = $\frac{moles}{volume}=\frac{0.100}{2.50L}=0.04M$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]^2}{[CO]^2\times [O_2]}$

$1.4\times 10^2=\frac{(0.16)^2}{[CO]^2\times 0.04}$

$[CO]=0.07M$

Equilibrium concentration of $CO$ = $\frac{moles}{volume}=\frac{moles}{2.50L}$

$0.07=\frac{moles}{2.50L}$

$moles=0.175$

Thus moles of CO present in the flask are 0.175