Question

Cobalt metal can be prepared by reducing cobalt (II) oxide with carbon monoxide.

\(CoO(s) + CO(g) \rightleftharpoons Co(s) + C{O_2}(g)\)

\({K_c} = 4.90 \times 1{0^2}at55{0^o}C\)

What concentration of \(CO\) remains in an equilibrium mixture with \(\left[ {C{O_2}} \right] = 0.100M\)

Short answer

The concentration of CO at equilibrium is \(2.04 \times 1{0^{ - 4}}M\)

Step-by-step solution

Given information

\(CoO(s) + CO(g) \rightleftharpoons Co(s) + C{O_2}(g)\)

1. Value of equilibrium constant \({K_C} = 4.90 \times 1{0^2}\)
2. The concentration of \(C{O_2}\) is \(0.100\,M\)

The value of equilibrium molar concentration needs to be calculated using the equation relating equilibrium constant and molar concentrations.

Calculate concentration of CO at equilibrium

\(\begin{array}{*{20}{c}}{{K_c}}&{ = \frac{{\left[ {{\rm{C}}{{\rm{O}}_2}} \right]}}{{[{\rm{CO}}]}}}\\{[CO]}&{ = \frac{{\left[ {C{O_2}} \right]}}{{{K_c}}}}\\{}&{ = \frac{{0.100}}{{4.90 \times {{10}^2}}}}\\{}&{ = 2.04 \times {{10}^{ - 4}}{\rm{M}}}\end{array}\)