Video: Using the Equilibrium Constant and CO₂ Concentration to Calculate the Concentration of CO in the Reduction of Cobalt Oxide into Cobalt Metal

Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide, as shown in the equation: CoO (s) + CO (g), ⇌ Co (s) + CO₂ (g), K_(c) = 4.90 × 10² at 550°C. What concentration of CO remains in an equilibrium mixture with [CO₂] = 0.100 M?

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Video Transcript

Cobalt metal can be prepared by reducing cobalt(II)oxide with carbon monoxide, as shown in the equation: CoO solid plus CO gas in equilibrium with CO solid and CO two gas, with K c equal to 4.90 times 10 to the power of two at 550 degrees Celsius. What concentration of CO remains in the equilibrium mixture with the concentration of CO two equal to 0.100 molars?

The question describes the reduction of cobalt(II)oxide. In this case, both definitions of reducing apply. We are removing oxygen or adding electrons to cobalt oxide, going from a plus two oxidation state to a zero oxidation state. In this equilibrium, the reducing agent is carbon monoxide which removes an oxygen in order to turn itself into carbon dioxide. We’re being asked to work out the concentration of this reducing agent carbon monoxide when the concentration of carbon dioxide CO₂ is 0.100 molars. For that, we’ll need K c, the equilibrium constant.

The equilibrium constant is a fixed value for a given temperature. We can use the equilibrium constant expression to work out the concentrations of one component from the concentration of the others. For an equilibrium where chemicals A and B are in equilibrium with C and D, this is the equilibrium constant expression. Any pure solids or liquids are not included in the equilibrium expression. Cobalt oxide and cobalt are both solids under these conditions. So we can begin our equilibrium constant expression with the concentration of carbon dioxide at the top and finish with our concentration of carbon monoxide at the bottom.

To find the concentration of carbon monoxide as per the question, we need to rearrange the equation in terms of the concentration of carbon monoxide, which is equal to the concentration of carbon dioxide divided by the equilibrium constant. Before we substitute in our values, we should verify the units for K c. K c is usually given without units. But concentrations for K c are always expressed in molars. We only have one concentration term in our expression in the numerator and one in the denominator. These units cancel out, meaning that K c in this case is unitless.

Now we can substitute the concentration of carbon dioxide and the value for the equilibrium constant into our expression. This gives a value for the carbon monoxide concentration of 2.04082 times 10 to the minus four molars. All the values in our calculation are given to three significant figures. So we should give our answer to three significant figures also. So our answer for the concentration of carbon monoxide in an equilibrium mixture where the carbon dioxide concentration is 0.100 molar is 2.04 times 10 to the minus four molars.

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