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Chapter 19: Problem 63

A particular reaction is spontaneous at \(450 \mathrm{~K}\). The enthalpy change for the reaction is \(+34.5 \mathrm{~kJ}\). What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

Short Answer

Expert verified
We can conclude that the entropy change (螖S) for the reaction is positive and its magnitude is greater than 76.67 J/K.

Step by step solution

01

Write down the known variables and equation

We are given: - Reaction is spontaneous: 螖G < 0 - Temperature: T = 450 K - Enthalpy change: 螖H = +34.5 kJ We will be using the Gibbs Free Energy equation: \( 螖G = 螖H - T螖S \)
02

Plug in the known variables into the equation

We know that 螖G < 0, 螖H = +34.5 kJ, and T = 450 K, so the equation becomes: \( 0 > 螖H - T螖S \)
03

Solve for 螖S

Now, we want to isolate 螖S in the equation: \( 0 > 螖H - T螖S \) \( T螖S > 螖H \) \( 螖S > \dfrac{螖H}{T} \) Now, substitute the given values of 螖H and T: \( 螖S > \dfrac{+34.5 \mathrm{~kJ}}{450 \mathrm{~K}} \) \( 螖S > \dfrac{+34.5 \times 10^{3} \mathrm{~J}}{450 \mathrm{~K}} \) (converting kJ to J) \( 螖S > 76.67 \mathrm{~J/K} \)
04

Conclusion:

Since 螖S is greater than 76.67 J/K, we can conclude that the entropy change (螖S) for the reaction is positive and its magnitude is greater than 76.67 J/K.

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Key Concepts

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

Spontaneous Reactions
In chemistry, a reaction is deemed **spontaneous** if it occurs without an external input of energy upon reaching a certain state. Spontaneity is determined by the Gibbs Free Energy change (\( 螖G \) for the reaction. A reaction is spontaneous when \( 螖G < 0 \). This means the process releases free energy and can proceed on its own.
Understanding spontaneity is crucial for predicting whether chemical reactions will occur under specific conditions. For instance, if you're looking at a reaction at a given temperature and see that \( 螖G \) is less than zero, that means the reaction is spontaneous under those conditions. This prediction is based on the interplay of enthalpy change (\( 螖H \)) and entropy change (\( 螖S \)). To summarize, remember:
  • Spontaneous means \( 螖G < 0 \).
  • Spontaneity indicates a process can occur without needing more energy.
  • Depends on \( 螖H \), \( 螖S \), and the temperature.
Enthalpy Change
**Enthalpy change**, denoted as \( 螖H \), reflects the total heat content variation within a system during a reaction. It can be either positive or negative, impacting the spontaneity of the reaction. A positive \( 螖H \) indicates that the reaction absorbs heat (endothermic), while a negative \( 螖H \) signifies heat release (exothermic).
In the given problem, the reaction had an \( 螖H \) of +34.5 kJ, meaning it absorbs heat. This alone suggests the reaction is endothermic. However, for it still to be spontaneous (as it is given as so), the increase in entropy must compensate for the energy absorbed leading to a negative \( 螖G \).
It is crucial to analyze how both \( 螖H \) and temperature influence whether a reaction is favorable or not. In brief:
  • \( 螖H > 0 \): Endothermic (absorbs heat).
  • \( 螖H < 0 \): Exothermic (releases heat).
  • Spontaneity relies on balancing \( 螖H \) and \( T螖S \).
Entropy Change
Entropy change (\( 螖S \)) is a measure of the disorder or randomness in a system. It can influence whether a reaction will occur spontaneously at a particular temperature. Generally, an increase in entropy (\( 螖S > 0 \)) means that the disorder of the system is increasing, favoring spontaneity in a broader range of temperatures.
In this problem, to find out how the entropy change allows the reaction to be spontaneous, we used the formula:\[ 螖G = 螖H - T螖S\]Here, we know \( 螖G < 0 \), and having \( 螖H = +34.5 \, kJ \) implies a high \( 螖S \) is required to maintain spontaneity since the reaction is endothermic. After calculations, we found that:\[ 螖S > 76.67 \, J/K\]This positive entropy change helps offset the positive enthalpy, ensuring the overall \( 螖G \) remains negative under the given conditions, highlighting how crucial entropy is in the spontaneity equation. In summary:
  • \( 螖S > 0 \): Increases disorder, can aid in spontaneous reactions.
  • \( 螖S \) significantly impacts the Gibbs Free Energy.
  • Spontaneity is often the result of a balance between \( 螖H \) and \( 螖S \).

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Most popular questions from this chapter

From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at \(298 \mathrm{~K}\), at what temperature (if any) would the reaction become spontaneous? (a) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) \(\Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K}\) (b) \(2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g)\) $$ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} $$

Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of \(10 \mathrm{~atm}\) total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Ch芒telier's principle. \(\infty\) (Section 15.7)

The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?

Propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) melts at \(-126.5^{\circ} \mathrm{C}\) and boils at \(97.4^{\circ} \mathrm{C}\). Draw a qualitative sketch of how the entropy changes as propanol vapor at \(150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is cooled to solid propanol at \(-150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(\mathrm{C}\) to predict how \(\Delta G^{\circ}\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G^{\circ}\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?(\mathrm{c})\) Calculate \(\Delta G^{\circ}\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?
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