Chapter 10: Problem 30

Given the entropy change for the first two reactions below, calculate the entropy change for the third reaction below. \begin{equation} \begin{array}{c}{\mathrm{S}_{8}(s)+8 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{SO}_{2}(g) \Delta S=89 \mathrm{J} / \mathrm{K}} \\ {2 \mathrm{SO}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) \Delta S=-188 \mathrm{J} / \mathrm{K}} \\ {\mathrm{S}_{8}(s)+12 \mathrm{O}_{2}(g) \rightarrow 8 \mathrm{SO}_{3}(g) \Delta S=?}\end{array} \end{equation}

Short Answer

Expert verified

The entropy change for the third reaction is -653 J/K

Step by step solution

Analyze the reactions

It's clear to see that the third reaction can be obtained by adding the first and second reactions together. When \(S_8(s)\) and \( 12 O_2(g)\) react, the product is \(8 SO_3(g)\). This is equivalent to \(S_8(s)\) reacting with \(8 O_2(g)\) to form \(8 SO_2(g)\), and then, \(2 SO_2(g)\) reacting with \(O_2(g)\) to form \(2 SO_3(g)\), which happens four times (since there are 8 SO2). This suggests that the entropy change for the third reaction equals the sum of the entropy changes for the first and second reactions after multiplying the second reaction by 4.

Calculation of the entropy change for the third reaction

The entropy change for the first reaction, \(\Delta S_1\), is given as 89 J/K. The entropy change for the second reaction, \(\Delta S_2\), is given as -188 J/K. So, the entropy change for the third reaction can be calculated as: \(\Delta S_3 = \Delta S_1 + 4 \times \Delta S_2 = 89 \, J/K + 4 \times -188 \, J/K = -653 \, J/K \)

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Short Answer

Step by step solution

Analyze the reactions

Calculation of the entropy change for the third reaction

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