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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

Expert verified
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

01

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: \[ \Delta H^{\circ}_{reaction} = \sum \Delta H^{\circ}_{products} - \sum \Delta H^{\circ}_{reactants} \] \[ \Delta S^{\circ}_{reaction} = \sum \Delta S^{\circ}_{products} - \sum \Delta S^{\circ}_{reactants} \]
02

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°: \[ \Delta H^{\circ} - T \Delta S^{\circ} < 0 \]
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The following processes were all discussed in Chapter 18 "Chemistry of the Environment." Estimate whether the entropy of the system increases or decreases during each ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalination of water by reverse osmosis.

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {\text { (b) } 2 \mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)} \\ {\text { (c) } 2 \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)}\end{array} $$

(a) Is the standard free-energy change, \(\Delta G^{\circ},\) always larger than \(\Delta G ?(\mathbf{b})\) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0\) ? (c) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process, necessarily has a low activation barrier?

(a) What sign for \(\Delta S\) do you expect when the volume of 0.200 mol of an ideal gas at \(27^{\circ} \mathrm{Cis}\) increased isothermally from an initial volume of 10.0 \(\mathrm{L} ?(\mathbf{b})\) If the final volume is 18.5 \(\mathrm{L}\) , calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

A certain constant-pressure reaction is barely nonspontaneous at \(45^{\circ} \mathrm{C}\) . The entropy change for the reaction is 72 \(\mathrm{J} / \mathrm{K} .\) Estimate \(\Delta H .\)

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