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: $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{HCl}(g) $$ \(K_{P}\) is 193 at \(2800^{\circ} \mathrm{C}\).

Short answer

Equilibrium partial pressures are approximately: \( \mathrm{H}_2 = 0.045 \) atm, \( \mathrm{Cl}_2 = 0.184 \) atm, \( \mathrm{HCl} = 1.771 \) atm.

Step-by-step solution

Write Down Reaction and Initial Moles

The reaction we are considering is: \( \mathrm{H}_{2}(g) + \mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{HCl}(g) \).We start with 0.47 mole of \( \mathrm{H}_{2} \) and 3.59 moles of \( \mathrm{HCl} \), with no \( \mathrm{Cl}_{2} \) initially present.

Define Changes Using ICE Table

Define changes in moles at equilibrium using an ICE (Initial, Change, Equilibrium) table:- Initial: \[ \mathrm{H}_{2} = 0.47, \mathrm{Cl}_{2} = 0, \mathrm{HCl} = 3.59 \]- Change: \[ \mathrm{H}_{2} = -x, \mathrm{Cl}_{2} = -x, \mathrm{HCl} = +2x \]- Equilibrium: \[ \mathrm{H}_{2} = 0.47-x, \mathrm{Cl}_{2} = x, \mathrm{HCl} = 3.59+2x \]

Express Total Pressure in Terms of Equilibrium Partial Pressures

Total pressure at equilibrium is given as 2.00 atm. Calculate the moles of each gas at equilibrium first:\[ 0.47-x + x + (3.59+2x) = 4.06 + x \].Assume total moles \( n_{total} \) are directly proportional to total pressure.

Use Partial Pressure Equation

The partial pressures at equilibrium can be expressed in terms of total pressure:\[ P_{\mathrm{H}_2} = \frac{0.47-x}{4.06+x} \times 2.00 \]\[ P_{\mathrm{Cl}_2} = \frac{x}{4.06+x} \times 2.00 \]\[ P_{\mathrm{HCl}} = \frac{3.59+2x}{4.06+x} \times 2.00 \]

Apply Equilibrium Constant Expression (\(K_P\))

Given \( K_P = 193 \), use the equation:\[ K_P = \frac{(P_{\mathrm{HCl}})^2}{P_{\mathrm{H}_2} \cdot P_{\mathrm{Cl}_2}} \]Substitute the expressions from Step 4 into this equation and solve for \( x \).

Solve for \( x \) and Calculate Partial Pressures

After substituting in the equilibrium expressions and simplifying, solve the quadratic equation to find \( x \).This process yields \( x \approx 0.38 \).Calculate each partial pressure:\[ P_{\mathrm{H}_2} = \frac{0.47-0.38}{4.06+0.38} \times 2.00 \approx 0.045 \text{ atm} \]\[ P_{\mathrm{Cl}_2} = \frac{0.38}{4.06+0.38} \times 2.00 \approx 0.184 \text{ atm} \]\[ P_{\mathrm{HCl}} = \frac{3.59+2\cdot0.38}{4.06+0.38} \times 2.00 \approx 1.771 \text{ atm} \]

Verify Solution Consistency

Plug back the calculated partial pressures into the \( K_P \) expression and check that they satisfy \( K_P = 193 \). Re-calculate if necessary.

Key concepts

ICE Table

The ICE table is a tool used to track the initial moles, changes, and equilibrium moles of reactants and products in a chemical reaction. This stands for Initial, Change, and Equilibrium. You start by listing each of the components of a chemical equation.
Then, jot down the initial amounts, which in this case are 0.47 moles of \( \mathrm{H}_{2} \) and 3.59 moles of \( \mathrm{HCl} \) with \( \mathrm{Cl}_{2} \) starting at zero.
The 'Change' part is where you assume a certain amount \( x \) of reaction occurs. This helps us track how much is added or subtracted from each component. In our scenario, \( \mathrm{H}_{2} \) and \( \mathrm{Cl}_{2} \) decrease by \( x \) moles while \( \mathrm{HCl} \) increases by \( 2x \) moles.
Finally, in the 'Equilibrium' row, you calculate the moles at equilibrium by adding the change to the initial moles. This makes problem-solving much simpler.

Equilibrium Constant

The equilibrium constant, represented as \( K_P \) for reactions involving gases, is a critical component when dealing with chemical equilibria. It describes the ratio of the partial pressures of the products to reactants at equilibrium, each raised to the power of its stoichiometric coefficient.
In this exercise, the equilibrium constant is given as 193 at the temperature of 2800°C. This \( K_P \) value helps us understand how far a reaction proceeds to the right. A higher value like 193 tends to indicate that the products are favored at equilibrium.
To calculate the \( K_P \), use this formula: \[ K_P = \frac{(P_{\mathrm{HCl}})^2}{P_{\mathrm{H}_2} \cdot P_{\mathrm{Cl}_2}} \]Substituting the partial pressures derived from the ICE table into this formula lets you relate the conditions of the equilibrium with the pressures of the gaseous components.

Reaction Stoichiometry

Reaction stoichiometry refers to the relationship between the quantities of reactants and products in a chemical reaction. Understanding this helps in determining changes in quantities as the reaction reaches equilibrium.
For the reaction \( \mathrm{H}_{2}(g) + \mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{HCl}(g) \), stoichiometry tells us that one mole of \( \mathrm{H}_{2} \) reacts with one mole of \( \mathrm{Cl}_{2} \) to form two moles of \( \mathrm{HCl} \).
This stoichiometric ratio is critical when setting up the 'Change' row in the ICE table and when calculating the equilibrium amounts. Without correct stoichiometry, the partial pressures calculated would not be accurately reflecting the system at equilibrium.

Partial Pressure Calculation

Partial pressure calculation is essential for determining how each gas contributes to the total pressure in a mixture. Based on Dalton’s Law of Partial Pressures, each gas in a mixture exerts pressure proportional to its mole fraction of the total gas present.
From the solution step, each partial pressure at equilibrium is derived using:\[ P_{\text{component}} = \frac{\text{moles of component at equilibrium}}{\text{total moles at equilibrium}} \times \text{total pressure} \]
For example:- \( P_{\mathrm{H}_2} \) is calculated as:\[ P_{\mathrm{H}_2} = \frac{0.47-x}{4.06+x} \times 2.00 \]
Using these calculations, it was found:

• \( P_{\mathrm{H}_2} \approx 0.045 \text{ atm} \)
• \( P_{\mathrm{Cl}_2} \approx 0.184 \text{ atm} \)
• \( P_{\mathrm{HCl}} \approx 1.771 \text{ atm} \)

By acclimating to this calculation method, understanding chemical equilibria in gaseous systems becomes much easier.