Question

Calculate the standard free energy of the reaction at \(27^{\circ} \mathrm{C}\) for the combustion of methane using the given data: \(\mathrm{CH}_{4}(\mathrm{~g})\) \(+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) $$\begin{array}{lcccc} \hline \text { Species } & \mathbf{C H}_{4}(\mathrm{~g}) & \mathrm{O}_{2}(\mathrm{~g}) & \mathrm{CO}_{2}(\mathrm{~g}) & \mathbf{H}_{2} \mathrm{O}(\mathrm{l}) \\ \hline \Delta_{\mathrm{f}} \boldsymbol{H}^{\circ} /(\mathrm{kJ} & -74.5 & 0 & -393.5 & -286.0 \\ \left.\mathrm{~mol}^{-1}\right) & & & & \\ \boldsymbol{S}^{\circ} /\left(\mathrm{JK}^{-1}\right. & 186 & 205 & 212 & 70 \\\ \left.\mathbf{m o l}^{-1}\right) & & & & \\ \hline \end{array}$$ (a) \(-891.0 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-240 \mathrm{~kJ} / \mathrm{mol}\) (c) \(-819 \mathrm{~kJ} / \mathrm{mol}\) (d) \(-963 \mathrm{~kJ} / \mathrm{mol}\)

Short answer

The standard free energy change ΔrG° for the combustion of methane at 27°C is -819 kJ/mol, which corresponds to option (c).

Step-by-step solution

Calculate the standard enthalpy change

Use the standard enthalpies of formation (ΔH°f) to calculate the standard enthalpy change (ΔrH°) for the reaction using the formula: ΔrH° = ΣΔH°f(products) - ΣΔH°f(reactants). Plug in the given values to obtain: ΔrH° = (-393.5 + 2×(-286.0)) - (-74.5 + 2×0) kJ/mol.

Calculate the standard entropy change

Use the standard molar entropies (S°) to calculate the standard entropy change (ΔrS°) for the reaction using the formula: ΔrS° = ΣS°(products) - ΣS°(reactants). Plug in the given values to obtain: ΔrS° = (212 + 2×70) - (186 + 2×205) J/(K·mol).

Convert ΔrS° from J to kJ

To match the units of ΔrH°, which is given in kJ, convert ΔrS° from J/(K·mol) to kJ/(K·mol) by dividing by 1000. This gives ΔrS° in kJ/(K·mol).

Calculate standard free energy change (ΔrG°)

Use the Gibbs free energy equation: ΔrG° = ΔrH° - TΔrS°. Here, T is the temperature in Kelvin. To convert Celsius to Kelvin, add 273.15 to the given temperature. Plug in the value of T (27 + 273.15 K), ΔrH°, and ΔrS° to obtain the standard free energy change ΔrG°.

Choose the correct answer

Match the calculated standard free energy change with the given options (a), (b), (c), or (d) to determine the correct answer.

Key concepts

Enthalpy of Formation

Enthalpy of formation, symbolized as (ΔH_f^∘), is a key concept in thermodynamics that refers to the heat change occurring when one mole of a compound is formed from its elements under standard conditions (25^∘C and 1 atm). For a chemical reaction, the difference between the sum of the enthalpies of formation for products and reactants determines the overall change in enthalpy (ΔrH^∘).

It's important to remember that the enthalpies of formation for pure elements in their standard states are defined to be zero. When solving problems like the one in the exercise, enthalpies of formation provide the essential data to calculate the heat released or absorbed during a reaction. By following the formula ΔrH^∘ = ∑ΔH_f^∘(products) - ∑ΔH_f^∘(reactants), you can predict whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). The calculation of ΔrH^∘ for the combustion of methane demonstrates how this data determines whether the process will release or absorb energy in the form of heat.

Entropy Change

Entropy, denoted by the symbol S, measures the degree of disorder or randomness in a system. The change in entropy (ΔrS^∘) is a vital concept to grasp since it contributes to understanding the spontaneity of chemical processes under standard reaction conditions. The formula ΔrS^∘ = ∑S^∘(products) - ∑S^∘(reactants) allows us to compute the standard entropy change of a reaction by considering the sum of standard molar entropies of the products subtracted from the reactants.

In the exercise provided, the entropy change was calculated using given entropy values of each species. The outcome indicates whether the product mixture is more or less orderly compared to the reactants. Factoring in this entropy change is crucial when predicting the feasibility of a reaction, especially when combined with enthalpy data to determine the overall free energy change.

Gibbs Free Energy Equation

The Gibbs free energy equation, ΔrG^∘ = ΔrH^∘ - TΔrS^∘, links the concepts of enthalpy and entropy, providing a valuable measure of the spontaneity of a reaction at constant temperature and pressure. This equation is pivotal since it allows us to predict whether a reaction will occur without the input of additional energy. A negative value of ΔrG^∘ indicates a spontaneous reaction, while a positive value suggests non-spontaneity.

For a reaction to be spontaneous, it must release free energy, which is often the case when the enthalpy change is negative (exothermic reaction) and the entropy change is positive (system becomes more disordered). The exercise includes calculating ΀rG^∘ using the standard enthalpy and entropy changes along with the given temperature, demonstrating how these fundamental thermodynamic properties interplay to assess the course of a reaction.

Standard Reaction Conditions

Standard reaction conditions refer to a set of predefined conditions usually set at 1 atm of pressure and 25^∘C or 298.15 K temperature. These conditions provide a common reference for measuring and comparing thermodynamic properties like enthalpy (ΔH), entropy (Sϴ), and Gibbs free energy (G). When calculations are performed under these conditions, they are marked with a degree symbol (∘) to denote their standard status.

Understanding these conditions is essential when performing thermodynamic calculations because they ensure that data from different sources are compatible. In the exercise, the standard free energy change is calculated at a different temperature (27^∘C), which requires adjusting the standard entropy value to the specific temperature used. This adjustment reminds students that while standard conditions are a basis, the actual conditions under which a reaction occurs can vary, and calculations must account for these differences.