Problem 76 - Chemistry

Question

A mixture of 0.47 mole of \(\mathrm{H}_{2}\) and 3.59 moles of \(\mathrm{HCl}\) is heated to \(2800^{\circ} \mathrm{C}\). Calculate the equilibrium partial pressures of \(\mathrm{H}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{HCl}\) if the total pressure is 2.00 atm. For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$$ \(K_{P}\) is 193 at \(2800^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The solution to the problem requires the application of the Le Chatelier's principle, the use the ideal gas law to convert between moles and pressure and the equilibrium constant. At the end of the calculations, values are obtained for the equilibrium partial pressures of \(\mathrm{H}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{HCl}\).

Step 1 - Defining Initial and Change in Moles

From the reaction's stoichiometry, we know the initial amount of HCl is 3.59 moles and we have 0.47 moles of \(H_{2}\). We do not know the initial amount of \(Cl_{2}\), but we know the total pressure is 2 atm. We define \(x\) as the change in moles of H2 and Cl2, and 2x as the change in molar quantity of HCl. At equilibrium, the moles of \(H_{2}\) and \(Cl_{2}\) will both be \(0.47 - x\) while for HCl it'll be \(3.59 - 2x\). Putting these into an ICE table, we get: \nH2, \tInitial \(0.47 - x\) \tChange \(x\) \tEquilibrium \(0.47 - x\) \nCl2, \tInitial \(0.47 - x\) \tChange \(x\) \tEquilibrium \(0.47 - x\) \nHCl, \tInitial \(3.59 - 2x\) \tChange -\(2x\) \tEquilibrium \(3.59 - 2x\)

Step 2 - Calculating the Partial Pressures

Next, convert the molar quantities at equilibrium into partial pressures using the ideal gas law: \[P = \frac{n \cdot R \cdot T}{V}\]\nSince the total pressure is given, and assuming the volume of the container and the temperature are constant, the partial pressures for each component will be equal to their mole fraction times the total pressure. Therefore, at equilibrium the partial pressures are:\n \(P_{H_{2}} = \frac{(0.47 - x)}{(0.47 -x) + (0.47 -x ) + (3.59-2x)} \cdot 2\)\n \[P_{Cl_{2}} = \frac{(0.47 - x)}{(0.47 -x) + (0.47 -x ) + (3.59-2x)} \cdot 2\]\n \[P_{HCl} = \frac{(3.59 -2x)}{(0.47 -x) + (0.47 -x ) + (3.59-2x)} \cdot 2\]

Step 3 - Solving for x using the Equilibrium Constant

Now we can substitute the expressions for the partial pressures into the expression for \(K_P\):\n\(193 = K_P = \frac{P_{HCl}^{2}}{P_{H_{2}} \cdot P_{Cl_{2}}}\)\nSubstituting the expression for the partial pressures from step 2 and solving for \(x\) will give the changes in molar quantities at equilibrium. From which we calculate the equilibrium partial pressures.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_P\) in gaseous reactions, quantifies the ratio of product concentrations to reactant concentrations at equilibrium under constant pressure. For a given reaction, let's consider \(\text{\(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\)}\)the expression for the equilibrium constant \(K_P\) is:\[K_P = \frac{(P_{\mathrm{HCl}})^2}{P_{\mathrm{H}_2} \cdot P_{\mathrm{Cl}_2}}\]The value of \(K_P\) provides insight into the position of equilibrium. A \(K_P\) substantially greater than 1 indicates that the reaction heavily favors product formation at equilibrium, meaning the products' pressures are significantly higher compared to those of the reactants. Conversely, a \(K_P\) value less than 1 implies the reactants dominate the equilibrium state. In our specific case, a \(K_P\) of 193 suggests that at 2800°C, the production of \(\mathrm{HCl}\) is strongly favored over the reactants. Thus, at high temperature, \(\mathrm{HCl}\) will predominantly be at play within the reaction atmosphere.

Partial Pressure

In a mixtures of gases, each gas exerts a pressure proportional to the number of moles it contributes to the mixture, known as its partial pressure. Total pressure of the system is the sum of all individual partial pressures due to Dalton's Law of Partial Pressures. Mathematically represented as:\[P_{\text{total}} = P_{\mathrm{H}_2} + P_{\mathrm{Cl}_2} + P_{\mathrm{HCl}}\]Each component's partial pressure can be determined by its mole fraction, \(X\), multiplied by the total pressure. The mole fraction is calculated by dividing the moles of one gas by the total amount of moles in the gas mixture. Applying this to our problem:

• For \(\mathrm{H}_2\): \(P_{\mathrm{H}_2} = X_{\mathrm{H}_2} \cdot P_{\text{total}}\)
• For \(\mathrm{Cl}_2\): \(P_{\mathrm{Cl}_2} = X_{\mathrm{Cl}_2} \cdot P_{\text{total}}\)
• For \(\mathrm{HCl}\): \(P_{\mathrm{HCl}} = X_{\mathrm{HCl}} \cdot P_{\text{total}}\)

Given our reaction constraints, the partial pressures must adjust themselves so that the equilibrium condition and total pressure criteria are met. Hence, the derivation of partial pressures is crucial for calculating the equilibrium state in problems involving gases.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle relating the pressure, volume, temperature, and amount of gas quantitatively. It's written as \(PV = nRT\) where:

• P: Pressure of the gas
• V: Volume of the gas
• n: Amount of substance in moles
• R: Universal gas constant (0.0821 atm L/mol K)
• T: Temperature in Kelvin

This law elegantly ties temperature and the physical properties of gases in confined spaces. In the context of chemical equilibrium, while calculating partial pressures, the ideal gas law implies that under constant temperature and volume, changes in partial pressures directly correspond to changes in mole fractions. By deploying the ideal gas law alongside Dalton's Law, we can reshape the equation to express pressure as a function of mole fraction. Thus, it becomes an essential tool for calculations involved with gaseous chemical equilibria problems.

ICE Table

When dealing with chemical equilibria, ICE tables help in organizing and calculating the changes in concentrations or pressures of reactants and products. ICE stands for:

• I: Initial Pressure or Concentration
• C: Change in Pressure or Concentration
• E: Equilibrium Pressure or Concentration

For the reaction \(\mathrm{H}_2 + \mathrm{Cl}_2 \rightleftharpoons 2 \mathrm{HCl}\), the ICE table lets us systematically determine unknowns like \(x\), the change in moles that occurs as the system approaches equilibrium. Here’s the basic setup for that reaction, given the known amounts:

• Initial: \(\mathrm{H_2}\) and \(\mathrm{Cl_2}\) are each \(0.47 - x\); \(\mathrm{HCl}\) starts at \(3.59 - 2x\).
• Change: The decrease or increase in moles, such as \(x\) for \(\mathrm{H_2}\) and \(\mathrm{Cl_2}\) leads to conclusion on equilibrium shifts.
• Equilibrium: The resulting expression \((0.47 - x)\) for \(\mathrm{H_2} \) or \((3.59 - 2x)\) for \(\mathrm{HCl}\) depict pressures at equilibrium.

The ICE table dramatically simplifies the problem-solving process involving chemical reactions by providing a clear framework to insert values and solve for unknowns like \( x\). It thereby assists in determining equilibrium conditions precisely.