Question

The equilibrium constant under standard conditions \((1 \mathrm{~atm}, 298 \mathrm{~K}, \mathrm{pH} 7,0)\) for the reaction catalyzed by fumarase, Fumarate \(+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{L}\)-malate is \(4.00\). At \(310 \mathrm{~K}\), the equilibrium constant is \(8.00\). A. What is the standard free energy change, \(\Delta G^{\circ}\), for this reaction? B. What is the standard enthalpy change, \(\Delta H^{\circ}\), for this reaction? Assume the standard enthalpy change is independent of temperature. C. What is the standard entropy change, \(\Delta 5^{\circ}\), at \(298 \mathrm{~K}\) for this reaction? D. If the concentration of both reactants is equal to \(0.01 \mathrm{M}\), what is the free energy change at \(298 \mathrm{~K}\) ? As the reaction proceeds to equilibrium, will more fumarate or L-malate form? E. What is the free energy change for this reaction when equilibrium is reached?

Short answer

A. -3.44 kJ/mol, B. -11.1 kJ/mol, C. -25.8 J/mol⋅K, D. -3.44 kJ/mol, L-malate will form, E. 0 J/mol.

Step-by-step solution

Calculate Standard Free Energy Change (Delta G°)

The standard free energy change, \( \Delta G^{\circ} \), for a reaction can be calculated using the formula: \[ \Delta G^{\circ} = -RT \ln K \]where \( R \) is the universal gas constant \( (8.314 \text{ J mol}^{-1} \text{K}^{-1}) \), \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant. **A.** At 298 K, since \( K = 4.00 \):\[ \Delta G^{\circ}_{298} = -8.314 \times 298 \times \ln(4.00) \approx -RT \times 1.386 \approx -3437.6 \text{ J/mol or } -3.44 \text{ kJ/mol} \].

Calculate Standard Enthalpy Change (Delta H°)

The standard enthalpy change \( \Delta H^{\circ} \) can be calculated from the van't Hoff equation:\[ \ln\left(\frac{K_2}{K_1}\right) = \frac{-\Delta H^{\circ}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]Here, \( K_1 = 4.00 \) at \( T_1 = 298 \) K, and \( K_2 = 8.00 \) at \( T_2 = 310 \) K.Substitute values:\[ \ln\left(\frac{8.00}{4.00}\right) = \frac{-\Delta H^{\circ}}{8.314} \left(\frac{1}{310} - \frac{1}{298}\right) \]\[ \Delta H^{\circ} = - \ln(2) \times 8.314 \times \left(\frac{298 \times 310}{12}\right) \approx -11.1 \text{ kJ/mol} \].

Calculate Standard Entropy Change (Delta S°)

Use the relation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) at 298 K:\[ -3440 \text{ J/mol} = \Delta H^{\circ} - 298 \times \Delta S^{\circ} \]Substitute \( \Delta H^{\circ} = -11100 \text{ J/mol} \):\[ -3440 = -11100 - 298 \times \Delta S^{\circ} \]Solving gives:\[ \Delta S^{\circ} = \frac{-11100 + 3440}{298} \approx -25.8 \text{ J/mol K} \].

Calculate Free Energy Change at Non-standard Conditions (at 298 K)

Use the formula for the actual free energy change:\[ \Delta G = \Delta G^{\circ} + RT \ln Q \]where \( Q \) is the reaction quotient. If both reactants are at 0.01 M:\[ Q = \frac{[\text{Products}]}{[\text{Reactants}]} = 1 \]Thus:\[ \Delta G = -3440 + 8.314 \times 298 \times \ln(1) = -3440 \text{ J/mol} \].Thus, L-malate will form since the reaction quotient is less than equilibrium.

Free Energy Change at Equilibrium

When the reaction is at equilibrium, \( Q = K \) and \( \Delta G = 0 \). Therefore, the free energy change for the reaction at equilibrium is 0 J/mol, meaning the system has no net change in free energy.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), plays a crucial role in understanding chemical reactions at equilibrium. It is a value that indicates the ratio of concentrations of products to reactants at equilibrium, showing whether a reaction favors products or reactants under a given set of conditions.

For the reaction catalyzed by fumarase, the equilibrium constant \( K \) tells how the system behaves at certain temperatures. In our specific example, \( K \) is given as \( 4.00 \) at \( 298 \text{ K} \) and \( 8.00 \) at \( 310 \text{ K} \). This change in \( K \) with temperature suggests that the reaction shifts towards more products (\( L\)-malate formation) when the temperature increases.

Understanding \( K \) allows us to calculate other thermodynamic properties using standard equations. Remember, the relationship of \( K \) and temperature helps us deduce the nature of the reaction and predict whether the reaction favors the formation of products or reactants.

• If \( K > 1 \), products are favored.
• If \( K < 1 \), reactants are favored.
• As temperature changes, \( K \) can increase or decrease depending on the enthalpy change.

Standard Enthalpy Change

The standard enthalpy change, \( \Delta H^{\circ} \), represents the heat absorbed or released during a reaction at constant pressure. This value helps in understanding whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

In the fumarase reaction, \( \Delta H^{\circ} \) is calculated using the van't Hoff equation. It takes into account the equilibrium constants at different temperatures. In this case, involving temperatures \( 298 \text{ K} \) and \( 310 \text{ K} \), the calculation reveals \( \Delta H^{\circ} \approx -11.1 \text{ kJ/mol} \).

This negative value indicates the reaction is exothermic, meaning it gives off energy as heat. Knowing the sign and magnitude of \( \Delta H^{\circ} \) aids in predicting how a reaction is affected by temperature changes:

• Negative \( \Delta H^{\circ} \): Reaction releases heat (exothermic).
• Positive \( \Delta H^{\circ} \): Reaction absorbs heat (endothermic).
• Magnitude shows how sensitive the reaction is to temperature changes.

Standard Entropy Change

The standard entropy change, \( \Delta S^{\circ} \), measures the change in disorder or randomness associated with a chemical reaction. Unlike energy, which is conserved, entropy tends to increase in spontaneous processes.

For the fumarase catalyzed reaction, the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) helps in calculating \( \Delta S^{\circ} \) by rearranging and substituting given known values. The solution points to \( \Delta S^{\circ} \approx -25.8 \text{ J/mol K} \).

A negative \( \Delta S^{\circ} \) suggests that the reaction results in lower randomness, usually indicating more ordered products. Understanding \( \Delta S^{\circ} \) is integral for interpreting the spontaneity of reactions hands in hands with other thermodynamic quantities like \( \Delta G^{\circ} \):

• Positive \( \Delta S^{\circ} \): Increased disorder, more random.
• Negative \( \Delta S^{\circ} \): Decreased disorder, more ordered.
• Combining \( \Delta H^{\circ} \) and \( \Delta S^{\circ} \) provides insights into \( \Delta G^{\circ} \) behavior.