Question

\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$ \mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g) $$ At \(1600 \mathrm{~K}\) the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2}\). If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

Short answer

The reduction of NiO to Ni is unlikely to occur under the given conditions, as the calculated equilibrium pressure of CO₂ is \(9.0 \times 10^4\,\text{torr}\), which is much higher than the maximum allowed total pressure of 760 torr in the furnace.

Step-by-step solution

Write the expression for the equilibrium constant

We need to write the equilibrium constant expression for the reaction. In this case, the equilibrium constant is given in terms of pressure (Kp), so the expression will include the partial pressures of the gases involved. As the reaction is in equilibrium: \[K_p = \frac{P_{Ni} \cdot P_{CO_2}}{P_{NiO} \cdot P_{CO}}\]

Identify the given information

We have: - \(K_p = 6.0 \times 10^2\) - Pressure of CO: \(P_{CO} = 150 \,\text{torr}\) - Since Ni and NiO are solids, their partial pressures do not affect the equilibrium constant. So, \(P_{NiO} = P_{Ni} = 1\)

Calculate the equilibrium pressure of carbon dioxide

Now, let's plug in the given values into the equilibrium constant expression: \[K_p = \frac{P_{CO_2}}{P_{CO}}\] Rearrange for \(P_{CO_2}\): \[6.0 \times 10^2 = \frac{P_{CO_2}}{150}\] Now, simply solve for \(P_{CO_2}\): \[P_{CO_2} = 6.0 \times 10^2 \times 150\] \[P_{CO_2} = 9.0 \times 10^4 \,\text{torr}\]

Analyze the pressure values

In order for the reaction to occur, the total pressure in the furnace should not exceed 760 torr. The calculated \(P_{CO_2}\) is much higher than this limit, and adding the given pressure of CO (150 torr) would further increase the total pressure. Therefore, the reduction of NiO to Ni is unlikely to occur under these given conditions.