A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+52.4 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Short Answer

Expert verified
(a) The reaction is endothermic because \(\Delta H^{\circ}=+23.7\, \mathrm{kJ}\) is a positive value. (b) The reaction leads to an increase in randomness or disorder because \(\Delta S^{\circ}=+52.4\, \mathrm{J} / \mathrm{K}\) is a positive value. (c) \(\Delta G^{\circ}=8.0928\, \mathrm{kJ}\). (d) The reaction is not spontaneous at \(298\, \mathrm{K}\) under standard conditions because \(\Delta G^{\circ}>0\).

Step by step solution

01

(a) Determine if the reaction is exothermic or endothermic

In this reaction, \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\), which is a positive value. A positive value of \(\Delta H^{\circ}\) means that the reaction is endothermic because it absorbs heat from the surroundings.
02

(b) Determine if the reaction leads to an increase or decrease in randomness

The given value for \(\Delta S^{\circ}\) is \(+52.4 \mathrm{~J} / \mathrm{K}\). This is a positive value, which indicates that the reaction leads to an increase in randomness or disorder of the system.
03

(c) Calculate \(\Delta G^{\circ}\)

To calculate \(\Delta G^{\circ}\), we can use the following formula: \[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\] where, \(\Delta H^{\circ} = +23.7\,\mathrm{kJ} = 23700\, \mathrm{J}\) \(\Delta S^{\circ} = +52.4 \,\mathrm{J} / \mathrm{K}\) \(T = 298 \,\mathrm{K}\) Plugging in the given values, \[\Delta G^{\circ} = 23700\, \mathrm{J} - 298\, \mathrm{K} \times 52.4\,\mathrm{J} / \mathrm{K}\] \[\Delta G^{\circ} = 23700\, \mathrm{J} - 15607.2\, \mathrm{J}\] \[\Delta G^{\circ} = 8092.8\, \mathrm{J} = 8.0928\, \mathrm{kJ}\]
04

(d) Determine if the reaction is spontaneous

A reaction is spontaneous if \(\Delta G^{\circ} < 0\). In this case, we calculated \(\Delta G^{\circ} = 8.0928\, \mathrm{kJ} > 0\), which means the reaction is not spontaneous at \(298\, \mathrm{K}\) under standard conditions.

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Key Concepts

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

Endothermic and Exothermic Reactions
Chemical reactions that absorb energy from their surroundings are known as endothermic reactions. They require heat to proceed, and as a result, you can often feel a decrease in temperature if you touch the reaction vessel. In contrast, exothermic reactions release energy, which usually manifests as heat, making the surroundings warmer.

Consider a reaction where the change in enthalpy (\(\Delta H^\circ\)) is positive, like +23.7 kJ. Since energy is absorbed to drive the reaction, it is endothermic. Conversely, a negative \(\Delta H^\circ\) would indicate an exothermic reaction, as it produces energy. This distinction is crucial because it affects the temperature of surroundings and can influence the feasibility of reactions under certain conditions.
Entropy and Disorder
Entropy is a measure of the randomness or disorder within a physical system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time; it can only increase or stay the same. Systems naturally progress towards a state of higher entropy.

In the context of chemical reactions, when the change in entropy (\(\Delta S^\circ\)) is positive, it indicates the products are more disordered than the reactants. A \(\Delta S^\circ\) of +52.4 J/K suggests an increase in disorder. This increase in entropy can be visualized as the system’s components spreading out more at the end of the reaction than at the beginning, perhaps moving from being well-organized to more random and spread out.
Gibbs Free Energy
The concept of Gibbs free energy (\(\Delta G^\circ\)) is pivotal in thermodynamics. It represents the maximum reversible work that can be performed by a thermodynamic system at a constant temperature and pressure. The equation \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) links the concepts of enthalpy, entropy, and temperature to predict the spontaneity of a reaction.

Calculating \(\Delta G^\circ\) involves converting all units into a consistent system (Joules, in this case) and using the temperature in Kelvin. If \(\Delta G^\circ\) is negative, the system can do work and is spontaneous, while a positive \(\Delta G^\circ\) means the reaction is non-spontaneous under standard conditions. This indicator helps chemists predict whether a reaction will occur without external energy input.
Spontaneous Reactions
The term 'spontaneous reaction' may imply that such reactions occur immediately, but it actually refers to the direction in which a process will proceed without any input of additional energy. For a reaction to be spontaneous at a given temperature, \(\Delta G^\circ\) must be negative, indicating that the process can release excess free energy as it moves towards equilibrium.

In our example, \(\Delta G^\circ\) is calculated to be +8.0928 kJ, which is positive. This implies the reaction is not spontaneous at 298 K under standard conditions. It's essential to understand that spontaneity doesn't necessarily mean a reaction will occur rapidly; rather, it signifies that the reaction has the potential to proceed without external intervention such as the addition of energy.

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Most popular questions from this chapter

Use Appendix \(\mathrm{C}\) to compare the standard entropies at \(25^{\circ} \mathrm{C}\) for the following pairs of substances: (a) \(\mathrm{Sc}(s)\) and \(\mathrm{Sc}(g)\), \(\mathrm{NH}_{3}(g)\) and \(\mathrm{NH}_{3}(a q)\) (c) \(1 \mathrm{~mol} \mathrm{P}_{4}(g)\) and \(2 \mathrm{~mol} \mathrm{P}_{2}(g)\), (d) C(graphite) and C(diamond). In each case explain the difference in the entropy values.

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{~kJ} /\) mol. (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00 \mathrm{~mol}\) of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8{ }^{\circ} \mathrm{C}\).

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about $88 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$. (a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{t}\) for \(\mathrm{Br}_{2}\) using data from Appendix C. Assume that $\Delta H_{\text {ap }}^{a}$ remains constant with temperature and that Trouton's rule holds. (b) Look hal boiling point of Br\(r_{2}\) in a chemistry handbook or at the WebElements Web site (www,webelements. com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

(a) If you are told that the entropy of a certain system is zero, what do you know about the system and the temperature? (b) The energy of a gas is increased by heating it. Using \(\mathrm{CO}_{2}\) as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, will they have the same number of microstates? Explain.

Using the data in Appendix \(C\) and given the pressures listed, calculate \(\Delta G^{\circ}\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad P_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=1.2 \mathrm{~atm} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{~atm} \\ \quad P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{~atm} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{~atm} \end{array} $$

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