Problem 100 - Chemistry: Principles and Reactions

Question

Consider the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ At \(500^{\circ} \mathrm{C}\), a flask initially has all three gases, each at a partial pressure of \(0.200\) atm. When equilibrium is established, the partial pressure of HI is determined to be \(0.48\) atm. What is \(\Delta G^{\circ}\) for the reaction at \(500^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

Solution: The standard Gibbs free energy change for the reaction at 500°C is 3.73 kJ/mol.

Step 1: Calculate the equilibrium pressures of H2 and I2

We know the initial pressures of the three gases, and the equilibrium pressure of HI. Using the balanced equation, we can find the equilibrium pressures of H2 and I2. For every 2 moles of HI, 1 mole of H2 and 1 mole of I2 are produced. So, we can see the stoichiometry as: $$ 2HI → H_2 + I_2 $$ Initial pressures: $$ [HI]_0 = 0.2\,\mathrm{atm}, [H_2]_0 = [I_2]_0 = 0.2\,\mathrm{atm} $$ Equilibrium pressures: $$ [HI]_{eq} = 0.48\,\mathrm{atm}, [H_2]_{eq} = [I_2]_{eq} = 0.2\,\mathrm{atm} - x $$ As 2 moles of HI are consumed to produce 1 mole of H2 and 1 mole of I2, therefore: $$ 2x = (0.2\,\mathrm{atm} - [HI]_{eq}) = (0.2\,\mathrm{atm} - 0.48\,\mathrm{atm}) $$ Solving for x: $$ x = \frac{-0.28\,\mathrm{atm}}{2} = -0.14\,\mathrm{atm} $$ Thus, equilibrium pressures are: $$ [H_2]_{eq} = [I_2]_{eq} = 0.2\,\mathrm{atm} - (-0.14\,\mathrm{atm}) = 0.34\,\mathrm{atm} $$

Step 2: Calculate the equilibrium constant (K)

Using the equilibrium concentrations, we can find the equilibrium constant (K). The reaction quotient, Q, is defined as: $$ Q = \frac{[H_2]_{eq}[I_2]_{eq}}{[HI]_{eq}^2} $$ Since, at equilibrium, Q = K: $$ K = \frac{0.34 \times 0.34}{0.48^2} = 0.4993 $$

Step 3: Use the van't Hoff equation to find \(\Delta G^{\circ}\)

The van't Hoff equation relates the equilibrium constant, temperature, and standard Gibbs free energy change and is given by: $$ \Delta G^{\circ} = -RT\ln K $$ Where \(\Delta G^{\circ}\) is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and ln K is the natural logarithm of the equilibrium constant. First, convert the temperature in Celsius to Kelvin: $$ T = 500°C + 273.15 = 773.15\,\mathrm{K} $$ Now, calculate the standard Gibbs free energy change: $$ \Delta G^{\circ} = - (8.314\,\mathrm{\frac{J}{mol\cdot K}}) (773.15\,\mathrm{K}) \ln(0.4993) = 3731.85\,\mathrm{J/mol} $$

Step 4: Convert the solution to appropriate units

Since the standard Gibbs free energy change is usually reported in kJ/mol, we need to convert the value obtained: $$ \Delta G^{\circ} = 3731.85\,\mathrm{J/mol} \times \frac{1\,\mathrm{kJ}}{1000\,\mathrm{J}} = 3.73\,\mathrm{kJ/mol} $$ The standard Gibbs free energy change for the reaction at 500°C is 3.73 kJ/mol.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a crucial concept in chemical reactions. It helps us understand the ratio of the concentrations of products to reactants at equilibrium. This constant is only dependent on temperature, meaning it remains the same as long as the temperature does not change. For a given reaction like \( 2 \, ext{HI} ightarrow ext{H}_2 + ext{I}_2 \), the equilibrium constant expression is derived from the balanced chemical equation.

For this reaction, \( K \) is expressed as:

• \( K = \frac{[ ext{H}_2][ ext{I}_2]}{[ ext{HI}]^2} \)

To find the equilibrium constant, you need the equilibrium concentrations of all reactants and products, which can often be determined experimentally or through given data in a problem. Remember, the equilibrium constant is dimensionless and provides insights into the extent of the reaction. A larger \( K \) indicates a product-favored reaction, while a smaller \( K \) suggests a reactant-favored process.

Reaction Quotient

The reaction quotient, represented as \( Q \), offers a snapshot of the relative concentration or pressure of reactants and products at any point in time other than equilibrium. It has the same form as the expression for the equilibrium constant \( K \) for a reaction. So, for our example, the expression for \( Q \) is:

• \( Q = \frac{[ ext{H}_2][ ext{I}_2]}{[ ext{HI}]^2} \)

The value of \( Q \) is used to determine the direction in which a reaction is headed:

• If \( Q < K \), the reaction will proceed forward, forming more products.
• If \( Q > K \), the reaction will proceed in reverse, forming more reactants.
• If \( Q = K \), the reaction is at equilibrium.

Comparing \( Q \) to \( K \) helps chemists predict how long it will take for a chemical system to reach equilibrium. This makes \( Q \) extremely useful for understanding and manipulating reaction conditions.

Van't Hoff Equation

The van't Hoff equation is a powerful tool that relates the equilibrium constant \( K \) to the standard Gibbs free energy change \( \Delta G^{\circ} \) at a specific temperature \( T \). The equation is:

• \( \Delta G^{\circ} = -RT\ln K \)

Here, \( R \) is the gas constant (8.314 J/mol·K) and \( T \) is the absolute temperature in Kelvin. The equation allows us to predict how \( K \) might change with temperature, given the enthalpy and entropy changes of the reaction. It is particularly useful in calculating \( \Delta G^{\circ} \), informing us whether a process is spontaneous:

• If \( \Delta G^{\circ} < 0 \), the process is spontaneous, favoring products.
• If \( \Delta G^{\circ} > 0 \), the process is non-spontaneous, favoring reactants.
• If \( \Delta G^{\circ} = 0 \), the system is at equilibrium.

This equation highlights the intrinsic link between thermodynamics and chemical equilibria.

Partial Pressure

In gaseous reactions, the partial pressure of a gas is its individual pressure in a mixture. It's a vital concept when dealing with gaseous equilibria, affecting the equilibrium constant expression and calculations. Partial pressure is measured in units like atmospheres (atm) or torr, and is given by the formula:

• \( P_i = \frac{n_i}{n_{ ext{total}}} \times P_{ ext{total}} \)

Here, \( P_i \) is the partial pressure, \( n_i \) is the amount (in moles) of the gas, and \( P_{ ext{total}} \) is the total pressure of the gas mixture. In our example, knowing each gas's partial pressure helps us determine how it affects the equilibrium state:

• Make changes in partial pressures to drive a reaction toward equilibrium.
• Use partial pressure values to calculate \( K \) and hence, \( \Delta G^{\circ} \).

Understanding partial pressures provides insights into how reactant and product concentrations shift during the course of a chemical reaction. This is critical in industrial chemical processes where control over reaction conditions leads to optimum yield.