Problem 82 - Chemistry and Chemical Reactivity

Question

Ammonia gas is synthesized by combining hydrogen and nitrogen: $$ 3 \mathrm{H}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) $$ (a) If you want to produce 562 g of \(\mathrm{NH}_{3}\), what volume of \(\mathrm{H}_{2}\) gas, at \(56^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg, is required? (b) Nitrogen for this reaction will be obtained from air. What volume of air, measured at \(29^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg pressure, will be required to provide the nitrogen needed to produce \(562 \mathrm{g}\) of \(\mathrm{NH}_{3} ?\) Assume the sample of air contains 78.1 mole \(\%\) N \(_{2}\)

Step-by-Step Solution

Answer

562 g of \(\mathrm{NH}_3\) requires approximately 1379.9 L of \(\mathrm{H}_2\) and 539.5 L of air.

Step 1: Determine Moles of Ammonia

First, find the molar mass of ammonia (\(\mathrm{NH}_3\)): \(14.01 + 3 \times 1.008 = 17.034 \, \mathrm{g/mol}\). Next, calculate the moles of \(\mathrm{NH}_3\) needed. Using the formula: \[ \text{moles of } \mathrm{NH}_3 = \frac{\text{mass of } \mathrm{NH}_3}{\text{molar mass of } \mathrm{NH}_3} = \frac{562 \, \mathrm{g}}{17.034 \, \mathrm{g/mol}} \approx 32.99 \, \mathrm{mol} \]

Step 2: Calculate Moles of Hydrogen Gas Required

For every 2 moles of \(\mathrm{NH}_3\), 3 moles of \(\mathrm{H}_2\) are required. So, calculate moles of \(\mathrm{H}_2\) using stoichiometry: \[ \text{moles of } \mathrm{H}_2 = \frac{32.99 \, \mathrm{mol} \, \mathrm{NH}_3 \times 3 \, \mathrm{mol} \, \mathrm{H}_2}{2 \, \mathrm{mol} \, \mathrm{NH}_3} = 49.485 \, \mathrm{mol} \]

Step 3: Convert Moles of Hydrogen to Volume

Use the ideal gas law \(PV = nRT\) to find volume. Convert \(56^{\circ} \mathrm{C}\) to Kelvin: \(56 + 273.15 = 329.15 \, \mathrm{K}\). Convert pressure from mm Hg to atm: \(745 \, \mathrm{mmHg} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{mmHg}} \approx 0.9803 \, \mathrm{atm}\). Use \(R = 0.0821 \, \mathrm{L\cdot atm/K\cdot mol}\). Solve for \(V\): \[ V = \frac{nRT}{P} = \frac{49.485 \, \mathrm{mol} \times 0.0821 \, \mathrm{L\cdot atm/K\cdot mol} \times 329.15 \, \mathrm{K}}{0.9803 \, \mathrm{atm}} \approx 1379.9 \, \mathrm{L} \]

Step 4: Calculate Moles of Nitrogen Required

For every 2 moles of \(\mathrm{NH}_3\), 1 mole of \(\mathrm{N}_2\) is needed. Calculate moles of \(\mathrm{N}_2\) using stoichiometry: \[ \text{moles of } \mathrm{N}_2 = \frac{32.99 \, \mathrm{mol} \times 1}{2} = 16.495 \, \mathrm{mol} \]

Step 5: Convert Moles of Nitrogen to Volume of Air

Since air is 78.1% \(\mathrm{N}_2\), calculate volume of air. Convert \(29^{\circ} \mathrm{C}\) to Kelvin: \(29 + 273.15 = 302.15 \, \mathrm{K}\). Pressure is \(745 \, \mathrm{mmHg} = 0.9803 \, \mathrm{atm}\). Solve for volume of pure \(\mathrm{N}_2\) using the ideal gas law: \[ V_{\mathrm{N}_2} = \frac{nRT}{P} = \frac{16.495 \, \mathrm{mol} \times 0.0821 \times 302.15}{0.9803} \approx 421.38 \, \mathrm{L} \] Since air is only 78.1% \(\mathrm{N}_2\), calculate total volume of air: \[ V_{\text{air}} = \frac{421.38}{0.781} \approx 539.48 \, \mathrm{L} \]

Key Concepts

Ideal Gas Law

The ideal gas law is a cornerstone of understanding gas behavior in chemistry. It connects pressure, volume, temperature, and the number of moles of gas. The formula is given as \[ PV = nRT \] where:

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

It’s essential to use consistent units, typically atm for pressure, liters for volume, and Kelvin for temperature. The ideal gas law is very handy in stoichiometry when you want to determine the volume of a gas involved in a reaction, provided you know the number of moles and conditions like temperature and pressure where the reaction takes place.

Ammonia Synthesis

Ammonia synthesis is a critical chemical process, often discussed in chemistry for its industrial significance. The Haber process is the most common method used, where nitrogen \( \mathrm{N}_2 \) and hydrogen \( \mathrm{H}_2 \) are combined under specific conditions to produce ammonia \( \mathrm{NH}_3 \). In chemical notation, the reaction is: \[ 3\, \mathrm{H}_2(\mathrm{g}) + \mathrm{N}_2(\mathrm{g}) \rightarrow 2\, \mathrm{NH}_3(\mathrm{g}) \] This reaction is exothermic, releasing energy, which means it gives off heat. It's vital to control temperature and pressure well, as they affect the yield and efficiency. Typically, high pressure and temperature and the use of a catalyst facilitate this reaction, ensuring optimal ammonia production.

Mole Calculations

Mole calculations are pivotal in stoichiometry, allowing chemists to bridge the gap between the macro scale we observe and the micro atomic level. A mole is defined as containing exactly 6.022 x \( 10^{23} \) particles, be it atoms, molecules, or ions. This concept helps convert between mass and number of particles. For example, to find the number of moles of ammonia \( \mathrm{NH}_3 \)needed in a reaction, we first determine its molar mass:

• \( \mathrm{NH}_3 \): 14.01 of N + (3 x 1.008 of H) = 17.034 g/mol

Then, using the formula—moles = mass/molar mass—we calculate how much is needed for a reaction.This forms the bedrock of stoichiometry, as you can predict how much reactants are needed to form a desired amount of product.

Chemical Reactions

Chemical reactions are transformations where reactants convert into products, involving the breaking and forming of bonds. In a balanced chemical reaction like the one for ammonia synthesis, it's important to obey the conservation of mass, meaning the same number of each type of atom should be present on both sides of the equation.We also focus on stoichiometry to understand the ratios of reactants needed. For the reaction \( 3\, \mathrm{H}_2 (\mathrm{g}) + \mathrm{N}_2 (\mathrm{g}) \rightarrow 2\, \mathrm{NH}_3 (\mathrm{g}) \),stoichiometry tells us 3 moles of hydrogen are required for every 1 mole of nitrogen to produce 2 moles of ammonia.Understanding these proportions allows us to calculate amounts of reactants or products given some initial mass or volume data, making reactions predictable and manageable in practical situations.