Question

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 will occur under the given furnace conditions, which operate at a CO pressure of 150 torr and a total pressure not exceeding 760 torr. This is because when comparing the reaction quotient and equilibrium constant, we find that more products will be formed and the maximum CO2 pressure at equilibrium is less than the limit imposed by the total pressure.

Step-by-step solution

Write down the given information

Equation: $\mathrm{NiO}(s)+\mathrm{CO}(g)\rightleftharpoons \mathrm{Ni}(s)+{\mathrm{CO}}_2(g)$ $K_p=6.0\times {10}^2$ at 1600 K Pressure of CO: 150 torr Total pressure: 760 torr (we know that pressure will not exceed this value)

Identify the partial pressures of the gases at equilibrium

We will assume that the initial pressure of CO2 is zero since the reduction process has not yet started. Let the change in pressure of CO at equilibrium be -x (losing x amount of pressure due to reaction) So, the pressure of CO2 at equilibrium will be equal to the pressure change, x. Pressure of CO = 150 - x Pressure of CO2 = x

Write the expression for reaction quotient (Qp) and equilibrium constant (Kp)

Reaction quotient, $Q_p=\frac{P_{{\mathrm{CO}}_2}}{P_{\mathrm{CO}}}$, as we can ignore the contribution of solids Ni and NiO in the expression. Therefore, substituting the expressions we found in step 2, $Q_p=\frac{x}{150-x}$ We are given that the equilibrium constant, $K_p=6.0\times {10}^2$.

Compare Qp and Kp to determine if reduction occurs

To determine if the reduction will occur, we must compare $Q_p$ and $K_p$. If $Q_p<K_p$, then more products will be formed, which means the reduction will occur. So, we have to check whether $\frac{x}{150-x}<6.0\times {10}^2$.

Solve for x to find the maximum CO2 pressure at equilibrium

Let's solve the inequality to find the maximum pressure of CO2 possible at equilibrium: $\frac{x}{150-x}<6.0\times {10}^2$ Rearranging and cross multiplying, we get: $x<150\times 6.0\times {10}^2-x\times 6.0\times {10}^2$ $x<\frac{150\times 6.0\times {10}^2}{1+6.0\times {10}^2}$ Using a calculator, we find that: $x<1.4925\text{ torr}$

Check for reduction using the obtained result

Since total pressure never exceeds 760 torr and we found that the maximum pressure of CO2 at equilibrium is 1.4925 torr, this indicates that the reaction will proceed in the forward direction, resulting in the reduction of NiO to Ni. Hence, reduction will occur under the given furnace conditions.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a foundational concept in chemical equilibrium that helps predict how a system at equilibrium reacts to changes in concentration, pressure, volume, or temperature.

According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the system responds by offsetting the disturbance. In the context of our exercise, where $\mathrm{NiO}(s)+\mathrm{CO}(g)\rightleftharpoons \mathrm{Ni}(s)+{\mathrm{CO}}_2(g)$, an increase in the pressure of \mathrm{CO}(g) will result in the equilibrium shifting to the right to produce more \mathrm{Ni}(s) and \mathrm{CO}_2(g), thus promoting the reduction of \mathrm{NiO}(s).

Practical applications: In industrial processes, Le Chatelier's principle is used to manipulate conditions to maximize desired product formation, such as in the reduction of NiO by increasing the pressure of CO.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) represents the ratio of partial pressures of products to reactants at equilibrium for a gas-phase reaction and is a measure of the extent to which a reaction proceeds. It is expressed in terms of partial pressures (P) and is temperature-dependent. $K_p=\frac{P_{\text{products}}}{P_{\text{reactants}}}$ where each pressure is raised to the power of its stoichiometric coefficient.

In the given exercise, the $K_p$ of $6.0\times {10}^2$ suggests that at 1600 K, the reaction favors the formation of products. This numerical value indicates a position of equilibrium far to the right, therefore suggesting that under ideal conditions, NiO reduction is likely. However, to make further predictions about the reaction's behavior with the given conditions, we must also consider the reaction quotient (Qp).

Reaction Quotient (Qp)

The reaction quotient (Qp) is used to calculate the direction in which a reaction will proceed to reach equilibrium.

For our gas-phase reaction, $Q_p=\frac{P_{{\mathrm{CO}}_2}}{P_{\mathrm{CO}}}$, similar to $K_p$ but using the initial or current pressures instead of the equilibrium pressures. At the beginning of a reaction or at any given point before equilibrium is reached, the value of $Q_p$ offers insight into the system's state.

If $Q_p<K_p$, the reaction will proceed towards the products, as there are fewer products than at equilibrium. Conversely, if $Q_p>K_p$, the reaction shifts towards reactants to achieve equilibrium. In the example, finding that the maximum pressure for ${\mathrm{CO}}_2$ is 1.4925 torr implies a low initial $Q_p$, validating that the system will shift towards producing more ${\mathrm{CO}}_2$ and $\mathrm{Ni}$, favoring the reduction process.