Question

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: $$ \mathrm{CH}_{3} \mathrm{COOH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CO}(g) $$ By using data from Appendix \(\mathrm{C}\) , calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature.

Short answer

To find the minimum temperature at which the decarbonylation of acetic acid will be spontaneous under standard conditions, first calculate the values of enthalpy change (∆H°) and entropy change (∆S°) for the reaction using the data from Appendix C: \[ \Delta H^{\circ}_{reaction} = \sum \Delta H^{\circ}_{products} - \sum \Delta H^{\circ}_{reactants} \] \[ \Delta S^{\circ}_{reaction} = \sum \Delta S^{\circ}_{products} - \sum \Delta S^{\circ}_{reactants} \] Apply the Gibbs-Helmholtz equation and the condition for spontaneity: \[ \Delta H^{\circ} - T \Delta S^{\circ} < 0 \] Rearrange the inequality to solve for the temperature: \[ T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \] Plug in the values for ∆H° and ∆S° and calculate the minimum temperature: \[ T > \frac{(\text{Value of } \Delta H^{\circ})}{(\text{Value of } \Delta S^{\circ})} \]

Step-by-step solution

Calculate the values of ∆H° and ∆S° for the reaction

Begin by calculating the values of enthalpy change (∆H°) and entropy change (∆S°) for the reaction using the data from Appendix C of the textbook. For each component, find the respective values and use the stoichiometry of the reaction to determine the overall values: \[ \Delta H^{\circ}_{reaction} = \sum \Delta H^{\circ}_{products} - \sum \Delta H^{\circ}_{reactants} \] \[ \Delta S^{\circ}_{reaction} = \sum \Delta S^{\circ}_{products} - \sum \Delta S^{\circ}_{reactants} \]

Determine the equation relating ∆G° with the minimum temperature

Use the Gibbs-Helmholtz equation as shown before: \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \] Then, apply the condition for spontaneity: \[ \Delta G^{\circ} < 0 \] Substitute the equation for ΔG°: \[ \Delta H^{\circ} - T \Delta S^{\circ} < 0 \]

Solve for the minimum temperature

Rearrange the inequality to solve for the temperature: \[ T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \] Now, plug in the values for ∆H° and ∆S° you calculated in Step 1: \[ T > \frac{(\text{Value of } \Delta H^{\circ})}{(\text{Value of } \Delta S^{\circ})} \] Calculate the value of the minimum temperature, which is the answer to this problem.

Key concepts

Enthalpy Change

Enthalpy change, denoted as \(\Delta H^{\circ}\), represents the heat absorbed or released during a chemical reaction at constant pressure. In the context of decarbonylation reactions, it is essential to understand whether the process absorbs heat (endothermic) or releases heat (exothermic). For the given decarbonylation reaction of acetic acid:\[ \text{CH}_3\text{COOH}(l) \rightarrow \text{CH}_3\text{OH}(g) + \text{CO}(g) \]you start by determining \(\Delta H^{\circ}\) from standard enthalpy values of the products and reactants. Using the formula:\[ \Delta H^{\circ}_{reaction} = \sum \Delta H^{\circ}_{products} - \sum \Delta H^{\circ}_{reactants} \]helps you determine whether the reaction absorbs or releases energy. For reactions that release heat, the enthalpy change is negative, indicating an exothermic reaction. Conversely, a positive \(\Delta H^{\circ}\) suggests that the reaction requires heat input to proceed.

Entropy Change

Entropy change, symbolized as \(\Delta S^{\circ}\), indicates the change in disorder or randomness during a chemical reaction. In decarbonylation reactions, increased entropy often results from the production of gaseous products from liquid reactants, as is the case in this reaction:\[ \text{CH}_3\text{COOH}(l) \rightarrow \text{CH}_3\text{OH}(g) + \text{CO}(g) \]To calculate \(\Delta S^{\circ}\), use the expression:\[ \Delta S^{\circ}_{reaction} = \sum \Delta S^{\circ}_{products} - \sum \Delta S^{\circ}_{reactants} \]A positive entropy change often promotes reaction spontaneity, especially at higher temperatures. It means that the system has moved toward a state of higher disorder, which is generally more favorable from a thermodynamic perspective. Hence, determining \(\Delta S^{\circ}\) is crucial in evaluating how likely a reaction is to occur naturally.

Gibbs Free Energy

Gibbs Free Energy, denoted as \(\Delta G^{\circ}\), is a pivotal concept when evaluating reaction spontaneity. It combines enthalpy and entropy changes to determine whether a reaction will proceed spontaneously at a given temperature.The relationship is given by the Gibbs-Helmholtz equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]Where \(T\) is the absolute temperature in Kelvin. If \(\Delta G^{\circ} < 0\), the reaction is likely to be spontaneous under standard conditions. Calculating \(\Delta G^{\circ}\) provides insight into the energy changes driving the reaction, encompassing the trade-offs between enthalpy and entropy. In practice, a negative Gibbs free energy indicates the reaction can proceed without the need for external energy input. This balance is vital for many chemical processes, not just in a laboratory, but also in industrial and biological systems.

Spontaneity of Reactions

Determining the spontaneity of reactions is essential to understand how and why some chemical reactions occur without external intervention. Spontaneity is primarily influenced by the changes in Gibbs Free Energy.For a reaction to be spontaneous at a given condition, \(\Delta G^{\circ}\) must be less than zero. This is expressed mathematically as:\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} < 0 \]a rearrangement allows calculation of the minimum temperature \(T\) for spontaneity:\[ T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \]This formula tells us that for reactions with positive \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), a higher temperature may be required for spontaneity. Conversely, a negative \(\Delta H^{\circ}\) can often drive reactions spontaneous at lower temperatures due to the energy release. Understanding these principles is crucial for predicting whether decarbonylation, and other reactions, will happen under specified conditions.