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Chapter 18: Problem 19

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : Reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

Short Answer

Expert verified
At \(25^{\circ} C\), both the reactions A and B are non-spontaneous. Reaction A becomes spontaneous at temperatures greater than roughly \(76.85^{\circ} C\). Reaction B will remain non-spontaneous at all temperatures due to the negative value of entropy.

Step by step solution

01

Convert to correct units

Before starting, ensure all the values are in the correct units for calculation. Here, \(\Delta H\) is given in kJ/mol, which is perfect, but \(\Delta S\) is in J/K.mol. The latter needs to be converted into kJ/K.mol by dividing by 1000. For Reaction A: \(\Delta S_{A} = \dfrac{30}{1000} = 0.03 \, kJ/K \cdot mol \), and for Reaction B: \(\Delta S_{B} = \dfrac{-113}{1000} = -0.113 \, kJ/K \cdot mol \).
02

Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for both reactions

Use the formula \(\Delta G = \Delta H – T\Delta S\) , where \(T\) is the temperature in Kelvin. Hence, first convert the temperature to Kelvin: \(T = 25^{\circ} C + 273.15 = 298.15 \, K\). Now calculate \(\Delta G\) for both reactions A and B:For Reaction A: \(\Delta G_{A} = \Delta H_{A} − T\Delta S_{A} = 10.5 - 298.15 \times 0.03 \approx 1.6 \, kJ/mol\).For Reaction B: \(\Delta G_{B} = \Delta H_{B} – T\Delta S_{B} = 1.8 - 298.15 \times -0.113 \approx 35.8 \, kJ/mol\).
03

Determine Spontaneity

A reaction is spontaneous if \(\Delta G < 0\), non-spontaneous if \(\Delta G > 0\), and at equilibrium if \(\Delta G = 0\). From Step 2, it's clear that both Reaction A and B are not spontaneous at \(25^{\circ} C\) since \(\Delta G_{A} > 0 \) and \(\Delta G_{B} > 0 \).
04

Calculate the temperatures for spontaneous reaction

For a reaction to be spontaneous, \(\Delta G\) must be less than 0. Hence, set \(\Delta G = 0\), and solve the equation \(\Delta H – T\Delta S = 0\) for \(T\), i.e., \(T = \dfrac{\Delta H}{\Delta S}\).For Reaction A: \(T_{A} = \dfrac{10.5}{0.03} \approx 350 \, K \,or\, 76.85^{\circ} C\).For Reaction B: Since \(\Delta S_{B}\) is negative , Reaction B will never be spontaneous.

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Most popular questions from this chapter

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a more ordered structure than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

What is a coupled reaction? What is its importance in biological reactions?

Crystallization of sodium acetate from a supersaturated solution occurs spontaneously (see p. 426). What can you deduce about the signs of \(\Delta S\) and \(\Delta H ?\)

A student looked up the \(\Delta G_{\mathrm{f}}^{\circ}, \Delta H_{\mathrm{f}}^{\circ}\), and \(S^{\circ}\) values for \(\mathrm{CO}_{2}\) in Appendix 2. Plugging these values into Equation \((18.10),\) he found that \(\Delta G_{\mathrm{f}}^{\circ} \neq \Delta H_{\mathrm{f}}^{\circ}-T S^{\circ}\) at \(298 \mathrm{~K}\). What is wrong with his approach?

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)
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