Problem 94 - Chemistry The Molecular Science

Question

\(\mathrm{HCl}\) can be made by the direct reaction of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) in the presence of light. Assume that \(3.0 \mathrm{~g} \mathrm{H}_{2}\) and \(140 . \mathrm{g}\) \(\mathrm{Cl}_{2}\) are mixed in a \(10-\mathrm{L}\) flask at \(28{ }^{\circ} \mathrm{C}\), and the flask is sealed. Before the reaction: (a) Calculate the partial pressures of the two reactants. (b) Calculate the total pressure in the flask. After the reaction: (c) Calculate the total pressure in the flask. (d) What reactant remains in the flask? Calculate the amount (mol) that remains. (e) Calculate the partial pressure of each gas. (f) Calculate the pressure inside the flask if the temperature is increased to \(40 .{ }^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Partial pressures: \(3.68\) atm \(\mathrm{H}_2\), \(4.87\) atm \(\mathrm{Cl}_2\). Total pressures: \(8.55\) atm before and \(8.73\) atm after reaction. Remaining reactant: \(0.489\) mol \(\mathrm{Cl}_2\). Pressure at 40°C: \(9.07\) atm.

Step 1: Calculate Moles of Reactants

First, calculate the moles of each reactant using their molar masses. The molar mass of \(\mathrm{H}_2\) is approximately \(2.02\, \mathrm{g/mol}\), and the molar mass of \(\mathrm{Cl}_2\) is \(70.90\, \mathrm{g/mol}\).\[ n_{\mathrm{H}_2} = \frac{3.0\, \mathrm{g}}{2.02\, \mathrm{g/mol}} \approx 1.485\, \mathrm{mol} \] \[ n_{\mathrm{Cl}_2} = \frac{140.0\, \mathrm{g}}{70.90\, \mathrm{g/mol}} \approx 1.974\, \mathrm{mol} \]

Step 2: Calculate Partial Pressures Before the Reaction

Using the ideal gas law \( PV = nRT \), calculate the partial pressures of \(\mathrm{H}_2\) and \(\mathrm{Cl}_2\). At \(T = 28^\circ\, \mathrm{C} = 301\, \mathrm{K}\), the gas constant \(R = 0.0821\, \mathrm{L} \cdot \mathrm{atm} \, \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\):\[ P_{\mathrm{H}_2} = \frac{n_{\mathrm{H}_2}RT}{V} = \frac{1.485 \times 0.0821 \times 301}{10} \approx 3.68 \, \mathrm{atm} \] \[ P_{\mathrm{Cl}_2} = \frac{n_{\mathrm{Cl}_2}RT}{V} = \frac{1.974 \times 0.0821 \times 301}{10} \approx 4.87 \, \mathrm{atm} \]

Step 3: Calculate Total Pressure Before the Reaction

The total pressure before the reaction is the sum of the partial pressures of \(\mathrm{H}_2\) and \(\mathrm{Cl}_2\):\[ P_{\text{total}} = P_{\mathrm{H}_2} + P_{\mathrm{Cl}_2} \approx 3.68 + 4.87 = 8.55 \, \mathrm{atm} \]

Step 4: Determine Limiting Reactant and Reaction Completion

The balanced reaction is \(\mathrm{H}_2 + \mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}\). Determine the limiting reactant by comparing the mole ratio \(1:1\). Since \(n_{\mathrm{H}_2} < n_{\mathrm{Cl}_2}\), \(\mathrm{H}_2\) is the limiting reactant. After reaction, \(1.485 \mathrm{mol}\) of \(\mathrm{H}_2\) react completely with \(1.485 \mathrm{mol}\) of \(\mathrm{Cl}_2\), leaving \(1.974 - 1.485 = 0.489\, \mathrm{mol}\) of \(\mathrm{Cl}_2\).

Step 5: Calculate Total Pressure After the Reaction

After the reaction, the remaining gas mixture in the flask consists of \(0.489\, \mathrm{mol}\) of \(\mathrm{Cl}_2\) and \(2 \times 1.485 = 2.97 \mathrm{mol} \) of \(\mathrm{HCl}\). Use the total moles and the ideal gas law to find total pressure.\[ P_{\text{after}} = \frac{(0.489 + 2.97) \times 0.0821 \times 301}{10} \approx 8.73 \, \mathrm{atm} \]

Step 6: Calculate Partial Pressures After the Reaction

Calculate partial pressures using gas law:\[ P_{\mathrm{Cl}_2} = \frac{0.489 \times 0.0821 \times 301}{10} \approx 1.21 \, \mathrm{atm} \] \[ P_{\mathrm{HCl}} = \frac{2.97 \times 0.0821 \times 301}{10} \approx 7.52 \, \mathrm{atm} \]

Step 7: Calculate Pressure at 40°C

Increase in temperature to \(40^\circ\, \mathrm{C} = 313\, \mathrm{K}\) causes pressure change according to \( P_1/T_1 = P_2/T_2 \).\[ P_{\text{new}} = \frac{8.73 \times 313}{301} \approx 9.07 \, \mathrm{atm} \]

Key Concepts

Partial Pressure

Partial pressure is a key concept in understanding gas mixtures. It refers to the pressure that a single gas in a mixture would exert if it were alone in a container. To calculate the partial pressure, we use the Ideal Gas Law, which states that the pressure (\( P \)) of a gas is proportional to the number of moles (\( n \)), the gas constant (\( R \)), and temperature (\( T \)), and inversely proportional to the volume (\( V \)) of the container. This is represented as:\[P = \frac{nRT}{V}\]When dealing with mixtures, the total pressure is the sum of the partial pressures of all gases present. Each gas's partial pressure is calculated as if it occupies the entire volume on its own, which is why it's crucial to first determine the number of moles of each reactant. This ensures accurate determination of how much pressure each component contributes. This step is critical before and after reactions, especially in scenarios like the one described in the exercise.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that runs out first, stopping the reaction from continuing. To determine the limiting reactant, it's important to examine the mole ratio from the balanced chemical equation. For example, the reaction \(\mathrm{H}_2 + \mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}\) shows a 1:1 mole ratio between \(\mathrm{H}_2\) and \(\mathrm{Cl}_2\). This means both reactants should ideally combine together in equal moles. If one reactant is present in smaller quantity than required, it becomes the limiting reactant. In this situation, \( \mathrm{H}_2 \) was found to be the limiting reactant since it had fewer moles than \( \mathrm{Cl}_2 \). Once this \(\mathrm{H}_2\) is completely consumed, the reaction halts, leaving the excess \(\mathrm{Cl}_2\) remaining. Recognizing the limiting reactant is essential for predicting the amounts of products that will form and any reactants left over after the reaction concludes.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves using balanced chemical equations to calculate the relationships between reactants and products. It is a core part of chemistry that helps us predict the outcomes of chemical reactions. Stoichiometry is based on the conservation of mass and moles. In a balanced reaction, such as \(\mathrm{H}_2 + \mathrm{Cl}_2 \rightarrow 2 \mathrm{HCl}\), the coefficients indicate the proportional amounts of each molecule involved. For every mole of \(\mathrm{H}_2\) that reacts with a mole of \(\mathrm{Cl}_2\), two moles of \(\mathrm{HCl}\) form. This principle allows chemists to calculate how much of each reactant is needed to produce a desired amount of product, as well as what remains unreacted once the reaction is complete. Understanding stoichiometry is vital for designing chemical reactions, evaluating yields, and conducting proper laboratory experiments. It's through these calculations that we ensure reactions are efficient and safe.