Problem 71 - Chemistry: Principles and Reactions

Question

An intermediate reaction used in the production of nitrogencontaining fertilizers is that between ammonia and oxygen: $$ 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ A \(150.0\) - \(\mathrm{L}\) reaction chamber is charged with reactants to the following partial pressures at \(500^{\circ} \mathrm{C}: P_{\mathrm{NH}_{3}}=1.3 \mathrm{~atm}, P_{\mathrm{O}_{2}}=1.5 \mathrm{~atm} .\) What is the limiting reactant?

Step-by-Step Solution

Answer

Answer: To find the limiting reactant, calculate the ratio of the moles of NH3 and O2 using the Ideal Gas Law equation and their stoichiometric coefficients. Compare this ratio to the expected stoichiometric ratio. If the calculated ratio is greater than 4/5, NH3 is the limiting reactant; if the ratio is less than 4/5, O2 is the limiting reactant; and if the ratio equals 4/5, both reactants are limiting.

Step 1: Convert partial pressures to moles using Ideal Gas Law

The Ideal Gas Law equation is given by: \(PV=nRT\) We'll need to rearrange the equation to solve for n (moles), where n = \(\frac{PV}{RT}\). R is the ideal gas constant, 0.0821 L atm/mol K, and T is the temperature in Kelvins. Given the temperature in Celsius, we need to convert it to Kelvin. T = \(500 + 273.15 = 773.15 K\) Now we can calculate the moles of ammonia (NH3) and oxygen (O2). n(NH3) = \(\frac{P_{NH_{3}} \cdot V}{R \cdot T}\) = \(\frac{1.3 \cdot 150.0}{0.0821 \cdot 773.15}\) n(O2) = \(\frac{P_{O_{2}} \cdot V}{R \cdot T}\) = \(\frac{1.5 \cdot 150.0}{0.0821 \cdot 773.15}\)

Step 2: Calculate mole ratios

The reaction equation shows the stoichiometric coefficients (the numbers in front of the molecules) for each reactant. This allows us to compare how many moles of each reactant are needed for the reaction. Mole ratio of NH3/O2 = \(\frac{4}{5} \Rightarrow \frac{\text{moles of NH3}}{\text{moles of O2}}=\frac{4}{5}\) Divide the moles of each reactant by their respective stoichiometric coefficients: NH3: \(\frac{n(NH3)}{4}\) O2: \(\frac{n(O2)}{5}\)

Step 3: Determine the limiting reactant

Compare the ratios of moles divided by stoichiometric coefficients: \(\frac{\frac{n(NH3)}{4}}{\frac{n(O2)}{5}}\) If the ratio calculated is greater than \(\frac{4}{5}\), NH3 is the limiting reactant. If the ratio calculated is less than \(\frac{4}{5}\), O2 is the limiting reactant. If the ratio calculated is equal to \(\frac{4}{5}\), both reactants are limiting and the reaction will proceed to completion. Find out which reactant is limiting by calculating the ratio as mentioned above and comparing it to the expected stoichiometric ratio (\(\frac{4}{5}\)).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that helps to relate the properties of gases. It is especially useful when you need to convert between different states or calculate unknown values. The formula of the Ideal Gas Law is \(PV = nRT\), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume of the gas
• \(n\) is the number of moles
• \(R\) is the ideal gas constant, which is \(0.0821\, \text{L atm/mol K}\)
• \(T\) is the temperature in Kelvin

To use the equation, ensure the units of pressure, volume, and temperature align with the constant \(R\). In the given exercise, we rearrange the Ideal Gas Law to solve for \(n\), the number of moles. Start by calculating the temperature in Kelvin by adding 273.15 to the Celsius value.By applying this law, you can find out how much of each reactant is present in terms of moles, which is crucial for understanding how chemical reactions proceed.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It allows chemists to predict the quantities of substances consumed and produced. In the given reaction, stoichiometry comes into play when examining the balanced chemical equation:\[4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}\]The coefficients in the equation represent the molar relationships between reactants and products. For our exercise, it shows that four moles of \(\text{NH}_3\) react with five moles of \(\text{O}_2\). This ratio, \(\frac{4}{5}\), is vital when determining the limiting reactant.Understanding these stoichiometric ratios will help you see what amounts of each reactant are necessary for the complete reaction. It allows you to balance the numbers carefully, ensuring that you know how much of each reactant you'll require or have left over.

Chemical Reaction Calculation

Calculating chemical reactions involves determining which reactant will run out first, halting the reaction. Known as the limiting reactant, this part of chemistry ensures reactions are fully understood.First, you find the moles of each reactant using the Ideal Gas Law, which you've already done. Then, use stoichiometry to divide the moles of each reactant by their respective coefficients from the balanced equation.The formula becomes:\[\frac{\text{moles of NH}_3}{4}, \quad \frac{\text{moles of O}_2}{5}\]After dividing, compare these results. The smallest value determines the limiting reactant because it restricts how much the reaction can proceed. This calculation highlights the importance of using stoichiometric ratios to understand these chemical relationships comprehensively.

Partial Pressures

Partial pressures are essential to understanding gases and their behavior in a mixture. In a system where multiple gases are present, each gas exerts its pressure as if it were alone. This pressure is known as the partial pressure.In the original exercise, you are given the partial pressures of \(\text{NH}_3\) and \(\text{O}_2\). These values help calculate the moles of each gas using the Ideal Gas Law. Use:\[P_{\text{NH}_3} = 1.3 \text{ atm}, \quad P_{\text{O}_2} = 1.5 \text{ atm}\]Understanding partial pressures enables you to grasp how each gas contributes to the total pressure in the chamber. This concept is vital in identifying the amount of reactants available to participate in the reaction and is integral in not only determining concentrations but also in establishing the conditions for ideal gas calculations.