Question

You are a member of a research team of chemists discussing plans for a plant to produce ammonia: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ (a) The plant will operate at close to \(700 \mathrm{~K},\) at which \(K_{\mathrm{p}}\) is \(1.00 \times 10^{-4},\) and employs the stoichiometric \(1 / 3\) ratio of \(\mathrm{N}_{2} / \mathrm{H}_{2}\). At equilibrium, the partial pressure of \(\mathrm{NH}_{3}\) is 50 . atm. Calculate the partial pressures of each reactant and \(P_{\text {total }}\) (b) One member of the team suggests the following: since the partial pressure of \(\mathrm{H}_{2}\) is cubed in the reaction quotient, the plant could produce the same amount of \(\mathrm{NH}_{3}\) if the reactants were in a \(1 / 6\) ratio of \(\mathrm{N}_{2} / \mathrm{H}_{2}\) and could do so at a lower pressure, which would cut operating costs. Calculate the partial pressure of each reactant and \(P_{\text {total }}\) under these conditions, assuming an unchanged partial pressure of \(50 .\) atm for \(\mathrm{NH}_{3}\). Is the suggestion valid?

Short answer

The partial pressures for case (a) are \(P_{N_2} = 2.77 \text{atm}\), \(P_{H_2} = 8.31 \text{atm}\) and \(P_{total} = 61.08 \text{atm}\). For case (b), they are \(P_{N_2} = 1.90 \text{atm}\), \(P_{H_2} = 11.40 \text{atm}\), and \(P_{total} = 63.30 \text{atm}\). The suggestion is not valid.

Step-by-step solution

Write the Equilibrium Constant Expression

Write the expression for the equilibrium constant, \(K_p\), for the reaction. For the given reaction, \[\mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\] the equilibrium constant expression in terms of partial pressures is \[K_p = \frac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}\]

Substitute Known Values (Case a)

Substitute the known values into the equilibrium expression. Given \(K_p = 1.00 \times 10^{-4}\) and \(P_{NH_3} = 50\text{ atm}\), we set up the equation: \[1.00 \times 10^{-4} = \frac{(50)^2}{P_{N_2}(P_{H_2})^3}\]

Solve for the Partial Pressures (Case a)

Assume an initial pressure ratio \(P_{N_2} = x\) and \(P_{H_2} = 3x\). Plug these into the equilibrium expression: \[1.00 \times 10^{-4} = \frac{2500}{x(3x)^3} = \frac{2500}{27x^4}\] Solving for x, we get: \[x = \left(\frac{2500}{27 \times 10^{-4}}\right)^{1/4}\] Calculating, \[x = 2.77 \text{ atm}\] Therefore, \(P_{N_2} = 2.77\text{ atm}\) and \(P_{H_2} = 3 \times 2.77 = 8.31 \text{ atm}\). Consequently, \(P_{total} = P_{N_2} + P_{H_2} + P_{NH_3} = 2.77 + 8.31 + 50 = 61.08 \text{ atm}\)

Reevaluate with New Ratio (Case b)

Consider the proposed ratio \(\frac{1}{6}\) for \(\mathrm{N}_2\) to \(\mathrm{H}_2\). Let \(P_{N_2} = y\) and \(P_{H_2} = 6y\). Substitute into equilibrium expression: \[1.00 \times 10^{-4} = \frac{(50)^2}{y(6y)^3} = \frac{2500}{216y^4}\] Solve for y: \[y = \left(\frac{2500}{216 \times 10^{-4}}\right)^{1/4} = 1.90 \text{ atm}\] Therefore, \(P_{N_2} = 1.90\text{ atm}\) and \(P_{H_2} = 6 \times 1.90 = 11.40 \text{ atm}\). Hence, \(P_{total} = P_{N_2} + P_{H_2} + P_{NH_3} = 1.90 + 11.40 + 50 = 63.30 \text{ atm}\)

Validate the Suggestion

Compare the \(P_{total}\) calculated for the new ratio to that of the stoichiometric ratio. Since \(P_{total} \text{at the proposed lower ratio}\) (63.30 atm) is higher than the \(P_{total} \text{at the stoichiometric ratio}\) (61.08 atm), the proposed suggestion to use a \(\frac{1}{6}\) ratio is not valid for reducing operating costs.

Key concepts

Partial Pressure

Partial pressure is a crucial concept in understanding gas mixtures and equilibrium. When dealing with gases, each gas in a mixture exerts its own pressure, known as its partial pressure. For example, in the reaction to produce ammonia, \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\), the gases involved are nitrogen (\text{N}_2\text{)} and hydrogen (\text{H}_2\text{)} as reactants, and ammonia (\text{NH}_3\text{)} as the product. The partial pressure of a gas is directly proportional to its mole fraction in the mixture and the total pressure. This means that if you know the total pressure and the concentration of each gas, you can calculate the partial pressures. These partial pressures are essential when using the equilibrium constant expression. For instance, with the equation for equilibrium constant \(K_p = \frac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}\), understanding partial pressures helps you plug in the correct values to find unknown pressures of reactants or products at equilibrium. It's important to grasp that partial pressures contribute to the equilibrium constant in this way.

Le Chatelier's Principle

Le Chatelier's Principle explains how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing conditions such as pressure, temperature, or concentration of components, the position of equilibrium will shift to counteract the change. Applying Le Chatelier's Principle to the ammonia reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\), you can predict how changing conditions will affect the production of ammonia.

• For example, increasing the pressure shifts the equilibrium towards the side with fewer gas molecules, which in this reaction is the product side (ammonia).


• Conversely, decreasing pressure would favor the reactants, reducing ammonia production.

One of the team members suggested using a different \(\text{N}_2 / \text{H}_2\) ratio, thinking it could produce the same amount of ammonia at lower pressure by exploiting Le Chatelier’s Principle. However, this idea needs to account for how partial pressures and total pressure interact and often won't work as intended. As demonstrated in the given exercise, changing the reactants' ratio resulted in a higher total pressure, disproving the hypothesis.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using balanced chemical equations to calculate the amounts of reactants and products. The ammonia synthesis reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\) serves as an excellent example of stoichiometric principles in action.

• In this reaction, one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. This 1:3:2 ratio is critical when calculating the partial pressures and understanding how much of each substance you'd get at equilibrium.


• To solve for unknown pressures, we substituted the expression \(P_{N_2} = x\) and \(P_{H_2} = 3x\) (following the 1:3 ratio) into the equilibrium constant equation. This approach enables the determination of these partial pressures accurately.

When the suggestion to alter the ratio to 1:6 was tested, even though the ammonia partial pressure was kept constant, we observed that it altered the total pressure in an unfavorable way, thereby demonstrating how stoichiometry deeply influences equilibrium.