Problem 57 - Chemistry and Chemical Reactivity

Question

Most metal oxides can be reduced with hydrogen to the pure metal. (Although such reactions work well, this expensive method is not used often for large- scale preparations.) The equilibrium constant for the reduction of iron(II) oxide is 0.422 at \(700^{\circ} \mathrm{C}\). Fstimate \(\Delta G_{\text {ren }}^{\circ}\) $$ \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$

Step-by-Step Solution

Answer

\(\Delta G^{\circ}_{\text{rxn}} \approx 6.987 \, \text{kJ/mol}\).

Step 1: Identify the Relationship Between Equilibrium Constant and Gibbs Free Energy

The relationship between the standard Gibbs free energy change \(\Delta G^{\circ}_{ ext{rxn}}\) and the equilibrium constant \(K\) can be expressed using the equation:\[\Delta G^{\circ}_{ ext{rxn}} = -RT \ln K\]where \(R\) is the universal gas constant \(8.314 \, \text{J/mol}\cdot\text{K}\), \(T\) is the absolute temperature in Kelvin, and \(K\) is the equilibrium constant.

Step 2: Convert the Temperature to Kelvin

The given temperature is \(700^{\circ} \text{C}\). To convert this to Kelvin, add 273.15:\[T = 700 + 273.15 = 973.15 \, \text{K}\]

Step 3: Insert Values Into the Gibbs Free Energy Equation

Use the equation from Step 1, with \(R = 8.314 \, \text{J/mol}\cdot\text{K}\), \(T = 973.15 \, \text{K}\), and \(K = 0.422\).\[\Delta G^{\circ}_{\text{rxn}} = -8.314 \, \text{J/mol}\cdot\text{K} \times 973.15 \, \text{K} \times \ln(0.422)\]

Step 4: Calculate the Natural Logarithm and Solve

First, calculate \(\ln(0.422)\):\[\ln(0.422) \approx -0.862\]Now substitute in:\[\Delta G^{\circ}_{\text{rxn}} = -8.314 \, \text{J/mol}\cdot\text{K} \times 973.15 \, \text{K} \times (-0.862)\]\[\Delta G^{\circ}_{\text{rxn}} \approx 6987 \, \text{J/mol}\]

Step 5: Convert Joules to Kilojoules

The final step is converting Joules to Kilojoules, since \(1 \, \text{kJ} = 1000 \, \text{J}\). \[\Delta G^{\circ}_{\text{rxn}} \approx 6.987 \, \text{kJ/mol}\]

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, denoted as \( K \), represents the ratio of the concentrations of products to reactants at equilibrium. It provides a measure of the extent a reaction will proceed. For the specific reaction of reducing iron(II) oxide with hydrogen, the equilibrium constant is 0.422 at \( 700^{\circ} \text{C} \). This tells us how favorably the reaction goes to completion under standard conditions.
Understanding the equilibrium constant is critical as it helps predict the direction a reaction will shift to reach equilibrium. A high \( K \) value means the reaction favors product formation, whereas a low \( K \) signifies stronger favor towards reactants. In this problem, the \( K \) value of 0.422 indicates that at \( 700^{\circ} \text{C} \), more iron(II) oxide and hydrogen remain unreacted at equilibrium.

Metal Oxides Reduction

The reduction of metal oxides, like iron(II) oxide, typically involves removing oxygen from the oxide to yield the pure metal. For instance, using hydrogen gas can reduce iron(II) oxide to yield pure iron and water vapor. Although this method can be effective, it's less commonly used in industrial settings due to cost factors, given hydrogen's expense.
Metal oxide reduction reactions are crucial to understand the principles of thermodynamics and kinetic feasibility. Such reactions are often measured by their equilibrium constant and Gibbs free energy changes, showing how these parameters can influence the likelihood of achieving pure metal.

Temperature Conversion

Temperature plays a vital role in chemical reactions and is a key factor when calculating Gibbs free energy. In thermodynamics, temperatures must be expressed in Kelvin to use them in equations like \( \Delta G^{\circ}_{\text{rxn}} = -RT \ln K \).
To convert Celsius to Kelvin, simply add 273.15 to the Celsius value. In this example, the temperature of \( 700^{\circ} \text{C} \) is converted to \( 973.15 \text{ K} \). Kelvin provides an absolute scale helpful for scientific computation, as it begins at absolute zero.

Natural Logarithm

The natural logarithm, represented as \( \ln \), is a mathematical tool used in thermodynamics to relate Gibbs free energy with the equilibrium constant. Understanding \( \ln K \) is essential because it shows how changes in equilibrium constants affect the free energy of a system.
For instance, in the Gibbs free energy equation, \( \ln(0.422) \approx -0.862 \), a crucial part of solving for \( \Delta G^{\circ}_{\text{rxn}} \). The negative value indicates the direction shift needed to reach equilibrium. The natural logarithm simplifies power and exponential relationships, making it indispensable in calculating the flow and progress of non-spontaneous reactions.