Question

Consider two reactions for the production of ethanol: $$\begin{array}{l}\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \\\ \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{H}_{2}(g) \end{array}$$ Which would be the more thermodynamically feasible at standard conditions? Why?

Short answer

The more thermodynamically feasible reaction at standard conditions is Reaction 1 (C2H4 + H2O → CH3CH2OH) because it has a negative Gibbs free energy change (ΔG° = -14.5 kJ/mol), indicating that it is thermodynamically favorable.

Step-by-step solution

Identify the components of the reactions

The first reaction involves the reactants: ethene (C2H4) and water (H2O) and the product: ethanol (CH3CH2OH). The second reaction involves the reactants: ethane (C2H6) and water (H2O) and the products: ethanol (CH3CH2OH) and hydrogen (H2).

Gather standard enthalpy change and standard entropy change values

Consult a standard thermodynamic table to obtain the following standard enthalpy change (ΔH°) and standard entropy change (ΔS°) values for each substance: ΔHf° (C2H4) = +52.3 kJ/mol ΔS° (C2H4) = 219.24 J/mol⋅K ΔHf° (H2O) = -241.8 kJ/mol (as gas) ΔS° (H2O) = 188.7 J/mol⋅K ΔHf° (CH3CH2OH) = -277.7 kJ/mol ΔS° (CH3CH2OH) = 160.7 J/mol⋅K ΔHf° (C2H6) = -84 kJ/mol ΔS° (C2H6) = 229.32 J/mol⋅K ΔHf° (H2) = 0 kJ/mol ΔS° (H2) = 130.68 J/mol⋅K

Calculate the enthalpy and entropy changes for each reaction

For Reaction 1: ΔH° (reaction 1) = ΔHf° (CH3CH2OH) - ΔHf° (C2H4) - ΔHf° (H2O) = (-277.7) - 52.3 - (-241.8) = -88.2 kJ/mol ΔS° (reaction 1) = ΔS° (CH3CH2OH) - ΔS° (C2H4) - ΔS° (H2O) = 160.7 - 219.24 - 188.7 = -247.24 J/mol⋅K For Reaction 2: ΔH° (reaction 2) = ΔHf° (CH3CH2OH) + ΔHf° (H2) - ΔHf° (C2H6) - ΔHf° (H2O) = (-277.7) + 0 - (-84) - (-241.8) = 48.1 kJ/mol ΔS° (reaction 2) = ΔS° (CH3CH2OH) + ΔS° (H2) - ΔS° (C2H6) - ΔS° (H2O) = 160.7 + 130.68 - 229.32 - 188.7 = -126.64 J/mol⋅K

Compute Gibbs free energy change for each reaction at standard conditions

Standard conditions imply a temperature of 298 K. For Reaction 1: ΔG° (reaction 1) = ΔH° (reaction 1) - TΔS° (reaction 1) = (-88.2 kJ/mol) - (298 K)(-247.24 J/mol⋅K)/1000 = -88.2 + 73.7 = -14.5 kJ/mol For Reaction 2: ΔG° (reaction 2) = ΔH° (reaction 2) - TΔS° (reaction 2) = (48.1 kJ/mol) - (298 K)(-126.64 J/mol⋅K)/1000 = 48.1 + 37.7 = 85.8 kJ/mol

Compare the Gibbs free energy change values for both reactions

ΔG° (reaction 1) = -14.5 kJ/mol (negative, indicating a thermodynamically feasible reaction) ΔG° (reaction 2) = 85.8 kJ/mol (positive, indicating a thermodynamically unfavorable reaction) The more thermodynamically feasible reaction at standard conditions is Reaction 1 because it has a negative Gibbs free energy change (ΔG° = -14.5 kJ/mol).

Key concepts

Gibbs free energy

Gibbs free energy is a critical concept in determining the thermodynamic feasibility of a reaction. It combines both enthalpy and entropy changes and gives a clear picture of whether the reaction will proceed spontaneously under given conditions. This energy is represented by the symbol \( \Delta G \). When \( \Delta G \) is negative, the reaction is considered spontaneous and feasible; when positive, it is non-spontaneous.

The formula for calculating Gibbs free energy is:

• \( \Delta G = \Delta H - T \Delta S \)

Here, \( \Delta H \) is the enthalpy change, \( \Delta S \) is the entropy change, and \( T \) is temperature in Kelvin. In essence, if the decrease in enthalpy and the increase in entropy contribute more significantly than the temperature impacts, the reaction will have a negative \( \Delta G \). In the exercise above, the first reaction's \( \Delta G \) value of -14.5 kJ/mol suggests that it is thermodynamically feasible at standard conditions.

Enthalpy change

Enthalpy change, represented by \( \Delta H \), measures the total heat content in a system and helps determine whether a reaction absorbs or releases heat. It is the difference between the enthalpy of the products and the reactants.

Whether a reaction is endothermic or exothermic depends on the sign of \( \Delta H \):

• Negative \( \Delta H \): Exothermic reaction, releasing heat.
• Positive \( \Delta H \): Endothermic reaction, absorbing heat.

In the first reaction of the exercise, \( \Delta H \) is -88.2 kJ/mol, indicating this reaction releases heat, contributing positively to the thermodynamic feasibility under standard conditions. For the second reaction, \( \Delta H \) is 48.1 kJ/mol, making it endothermic and less likely to proceed spontaneously.

Entropy change

Entropy change, represented by \( \Delta S \), is a measure of the disorder or randomness in a system. The change in entropy is crucial for understanding the feasibility of reactions and is utilized in the Gibbs free energy equation.

- A positive \( \Delta S \) indicates an increase in disorder, favoring spontaneity.- A negative \( \Delta S \) implies a decrease in disorder, making unfavorable spontaneity.In the step-by-step solution within the exercise, the first reaction has a \( \Delta S \) of -247.24 J/mol⋅K. Though negative, the significant heat release (enthalpy change) outweighs the entropy decrease, resulting in a negative \( \Delta G \). This leads to a spontaneous reaction. On the contrary, the second reaction also shows a decrease in entropy with its \( \Delta S \) as -126.64 J/mol⋅K, contributing to the non-spontaneity demonstrated by the positive \( \Delta G \).

Standard conditions

Standard conditions refer to a set of typical conditions used to measure enthalpy, entropy, and Gibbs free energy, ensuring all reactions are assessed under a common benchmark. These conditions include:

• Temperature: 298 K (25°C)
• Pressure: 1 atmosphere
• Concentration: 1 Molar (for solutions)

These standardized conditions are crucial in accurately comparing the thermodynamic properties of reactions. They ensure that the values of \( \Delta H \), \( \Delta S \), and \( \Delta G \) are gathered and compared under consistent conditions, making it possible to determine which reaction is more feasible. In this exercise, both reactions are assessed under these defined settings, allowing for an objective comparison of their thermodynamic favorability.