Questions: Calculating an equilibrium constant from a heterogeneous equilibrium Ni(CO)4(g) ⇌ Ni(s) + 4CO(g) At a certain temperature, a 3.76 L reaction vessel containing a mixture of nickel carbonyl, nickel, and carbon monoxide at equilibrium has the following composition: Ni(CO)4 0.277 g Ni 10.0 g CO 1.96 g Calculate the value of the equilibrium constant, Kc, for this reaction. Round your answer to 3 significant digits.

Solution Steps

The balanced chemical equation for the reaction is: \[ \text{Ni(CO)}_4(g) \rightleftharpoons \text{Ni}(s) + 4\text{CO}(g) \]

Since \( K_c \) is based on concentrations of gases and aqueous species, we only consider the gaseous species: \[ K_c = \frac{[\text{CO}]^4}{[\text{Ni(CO)}_4]} \]

First, we need to convert the given masses to moles using their molar masses:

• Molar mass of Ni(CO)\(_4\): \( 58.69 + 4 \times 12.01 + 4 \times 16.00 = 170.69 \, \text{g/mol} \)
• Molar mass of CO: \( 12.01 + 16.00 = 28.01 \, \text{g/mol} \)

For Ni(CO)\(_4\): \[ \text{Moles of Ni(CO)}_4 = \frac{0.277 \, \text{g}}{170.69 \, \text{g/mol}} = 0.001622 \, \text{mol} \]

For CO: \[ \text{Moles of CO} = \frac{1.96 \, \text{g}}{28.01 \, \text{g/mol}} = 0.06999 \, \text{mol} \]

Next, we calculate the concentrations by dividing the moles by the volume of the reaction vessel (3.76 L): \[ [\text{Ni(CO)}_4] = \frac{0.001622 \, \text{mol}}{3.76 \, \text{L}} = 0.0004315 \, \text{M} \] \[ [\text{CO}] = \frac{0.06999 \, \text{mol}}{3.76 \, \text{L}} = 0.01861 \, \text{M} \]

Using the concentrations in the equilibrium expression: \[ K_c = \frac{[\text{CO}]^4}{[\text{Ni(CO)}_4]} = \frac{(0.01861)^4}{0.0004315} \]

\[ K_c = \frac{(0.00001198)}{0.0004315} = 0.02777 \]

Final Answer