Problem 115 - Chemistry The Science in Context

Question

Ammonia is produced industrially from the reaction of hydrogen with nitrogen under pressure in a sealed reactor. What is the percent decrease in pressure of a sealed reaction vessel during the reaction between \(3.60 \times 10^{3} \mathrm{mol}\) of \(\mathrm{H}_{2}\) and \(1.20 \times 10^{3} \mathrm{mol}\) of \(\mathrm{N}_{2}\) if half of the \(\mathrm{N}_{2}\) is consumed?

Step-by-Step Solution

Answer

Answer: The percent decrease in pressure of the sealed reaction vessel during the reaction is 25%.

Step 1: Determine the stoichiometric equation for the reaction

Write down the balanced chemical equation for the reaction of hydrogen and nitrogen to produce ammonia. $$ \mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3 $$

Step 2: Calculate the initial moles of hydrogen, nitrogen, and ammonia

According to the exercise, we have: - \(3.60 \times 10^3 \text{ mol}\) of \(\mathrm{H}_{2}\) - \(1.20 \times 10^3 \text{ mol}\) of \(\mathrm{N}_{2}\) Initially, there are 0 moles of ammonia.

Step 3: Calculate the final moles of hydrogen, nitrogen, and ammonia

Since half of the nitrogen is consumed, the final moles of nitrogen are: $$ \frac{1.20 \times 10^3}{2} = 600 \text{ mol} $$ Since the stoichiometric ratio for the reaction is 1 mol of \(\mathrm{N}_2\) to 3 mols of \(\mathrm{H}_2\), the moles of \(\mathrm{H}_{2}\) consumed are: $$ 3 \times 600 \text{ mol} = 1800 \text{ mol} $$ Subtracting the moles of \(\mathrm{H}_{2}\) consumed from the initial moles, we get the final moles of \(\mathrm{H}_{2}\) as: $$ 3.60 \times 10^3 \text{ mol} - 1.8 \times 10^3 \text{ mol} = 1.80 \times 10^3 \text{ mol} $$ The final moles of ammonia are: $$ 2 \times 600 \text{ mol} = 1200 \text{ mol} $$

Step 4: Calculate the initial and final pressure using the ideal gas law

Since the volume and temperature remain constant, and assuming the pressure is the same for each gas, we can use the formula for the ideal gas law: $$ PV = nRT $$ Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. Since the initial pressure is the same for all gases, we can write: $$ P_iV = n_iRT $$ where \(n_i = n_{H_2} + n_{N_2}\) Similarly, for the final pressure: $$ P_fV = n_fRT $$ where \(n_f = n_{H_2} + n_{N_2} + n_{NH_3}\) Taking the ratio of the initial to final pressure, we get: $$ \frac{P_i}{P_f} = \frac{n_f}{n_i} $$ Substituting the values of initial and final moles, we get: $$ \frac{P_i}{P_f} = \frac{1.80 \times 10^3 + 600 + 1200}{3.60 \times 10^3 + 1.20 \times 10^3} = \frac{3600}{4800} = \frac{3}{4} $$

Step 5: Calculate the percent decrease in pressure

To calculate the percent decrease in pressure, we can use the formula: $$ \text{Percent Decrease} = \frac{\text{Initial Pressure - Final Pressure}}{\text{Initial Pressure}} \times 100 $$ Substituting the ratio calculated in Step 4, we get: $$ \text{Percent Decrease} = \frac{P_i - \frac{3}{4} P_i}{P_i} \times 100 = \frac{1}{4} \times 100 = 25\% $$ The percent decrease in pressure of the sealed reaction vessel during the reaction is 25%.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a key principle in chemistry that describes the behavior of gases under ideal conditions. This law is represented by the formula \(PV = nRT\), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume the gas occupies
• \(n\) is the number of moles of gas
• \(R\) is the ideal gas constant, which is approximately 8.314 J/(mol\cdot K)
• \(T\) is the temperature in Kelvin

To solve problems using the ideal gas law, it's crucial to keep these variables in check and ensure that the units are consistent—usually using L for volume, atm for pressure, mol for the amount of substance, and Kelvin for temperature. In chemical reactions, particularly those in a closed system, this law can help calculate changes in pressure when connected to changes in the amount of substances involved. For example, in the reaction producing ammonia from nitrogen and hydrogen, the ratio of moles before and after the reaction helps determine the changes in pressure, assuming constant volume and temperature.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the relationships between the quantities of reactants and products in a chemical reaction. It's based on the principle of conservation of mass, stating that the mass of the reactants must equal the mass of the products.

In the chemical equation \(\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3\), stoichiometry helps us understand the proportions of nitrogen, hydrogen, and ammonia. For example, one mole of \(\mathrm{N}_2\) reacts with three moles of \(\mathrm{H}_2\) to produce two moles of \(\mathrm{NH}_3\).

Understanding these proportions allows us to calculate how much of each reactant is consumed during the reaction and how much product is formed. If half of the nitrogen is used, as in our earlier exercise, stoichiometry helps us work out the resulting amounts of hydrogen and ammonia formed through simple mole-based calculations.

Industrial Chemical Processes

Industrial chemical processes involve large-scale chemical reactions often used to synthesize necessary products like fuels, plastics, and other materials. One famous industrial chemical process is the Haber Process, which synthesizes ammonia from nitrogen and hydrogen under high pressure.

In industrial settings, understanding how gases behave and relate to one another is crucial for efficiency and safety. The key aspects include maintaining optimal pressure and temperature conditions to maximize yield and reduce costs. Using a catalyst to speed up the reaction can also improve efficiency.

In the exercise scenario, pressure changes in the reaction vessel were calculated to determine how much the pressure decreases as nitrogen and hydrogen convert to ammonia. This information is vital to ensure that the reaction conditions are controlled, preventing any potential hazards associated with significant pressure changes in sealed reactors.