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  • The entropy change for a chemical reaction is given by: <img alt="\m

The entropy change \mathrm{(\Delta S)} for a chemical reaction is given by:
\mathrm{\Delta S=\sum n_{i} \cdot S_{i, \text { products }}-\sum n_{j} \cdot S_{j, \text { reactants }}}

Calculate the entropy change for the reaction:

\mathrm{2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)}
Given the standard molar entropy \mathrm{\left(S^{\circ}\right) values: }
S^{\circ}\left(\mathrm{SO}_{2}\right)=248.15 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}
S^{\circ}\left(\mathrm{O}_{2}\right)=205.15 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}
$ $S^{\circ}\left(\mathrm{SO}_{3}\right)=256.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}

Option: 1

12.60 \mathrm{~J} / \mathrm{mo}


Option: 2

1.70 \mathrm{~J} / \mathrm{mo}


Option: 3

189.05 \mathrm{~J} / \mathrm{mo}


Option: 4

36.20 \mathrm{~J} / \mathrm{mo}


Answers (1)

best_answer

The entropy change \mathrm{(\Delta S)} for the reaction can be calculated using the given formula:
\mathrm{\Delta S=\sum n_{i} \cdot S_{i, \text { products }}-\sum n_{j} \cdot S_{j, \text { reactants }}}

Given the reaction:
\mathrm{2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)}

Substitute the given standard molar entropy values:
\mathrm{\Delta S=\left(2 \cdot S^{\circ}\left(\mathrm{SO}_{3}\right)\right)-\left(2 \cdot S^{\circ}\left(\mathrm{SO}_{2}\right)+S^{\circ}\left(\mathrm{O}_{2}\right)\right)}

\mathrm{\Delta S=(2 \cdot 256.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})-(2 \cdot 248.15 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+205.15 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})}

\mathrm{\Delta S=512.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}-701.45 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}
\mathrm{\Delta S=-189.05 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}

Therefore, the entropy change for the reaction is \mathrm{\Delta S=-189.05 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}.
Therefore, the correct option is 3.

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