What is the entropy change if 4.500 g of CaCO₃(s) is placed in a container and allowed to decompose to CaO(s) and CO₂(g) according to the following reaction?
CaCO₃(s) $\leftrightarrow$ CaO(s) + CO₂(g) $\begin{array} { c c } \text { Substance } & \boldsymbol { S } ^ { \circ } ( \mathbf { J } / \mathbf { m o l } \cdot \mathbf { K } ) \\\mathrm { CaCO } _ { 3 } ( s ) & 92.88 \\\mathrm { CaOO } ( s ) & 39.75 \\\mathrm { CO } _ { 2 } ( g ) & 213.6\end{array}$

Question

What is the entropy change if 4.500 g of CaCO₃(s) is placed in a container and allowed to decompose to CaO(s) and CO₂(g) according to the following reaction?
CaCO₃(s)

\leftrightarrow

CaO(s) + CO₂(g)

\begin{array} { c c } \text { Substance } & \boldsymbol { S } ^ { \circ } ( \mathbf { J } / \mathbf { m o l } \cdot \mathbf { K } ) \\\mathrm { CaCO } _ { 3 } ( s ) & 92.88 \\\mathrm { CaOO } ( s ) & 39.75 \\\mathrm { CO } _ { 2 } ( g ) & 213.6\end{array}

Quizplus · Accepted Answer

The entropy change for the reaction is calculated using the following equation:$$\Delta S ^\circ = \sum nS ^\circ (	ext{products}) - \sum nS ^\circ (	ext{reactants})$$where n is the stoichiometric coefficient of each reactant and product, and S° is the standard molar entropy of each reactant and product.For the given reaction, the entropy change is:$$\Delta S ^\circ = [S ^\circ (	ext{CaO}) + S ^\circ (	ext{CO}_2)] - [S ^\circ (	ext{CaCO}_3)]$$$$\Delta S ^\circ = [(39.75 	ext{ J/mol$\cdot$K}) + (213.6 	ext{ J/mol$\cdot$K})] - [92.88 	ext{ J/mol$\cdot$K}]$$$$\Delta S ^\circ = +160.5 	ext{ J/mol$\cdot$K}$$Therefore, the entropy change for the reaction is +160.5 J/K, which corresponds to choice A.

What Is the Entropy Change If 4 ↔\leftrightarrow↔ CaO(s) + CO2(g)

What Is the Entropy Change If 4 $\leftrightarrow$ CaO(s) + CO₂(g)