\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{ [Substane| $S^{\circ}\left(\frac{J}{\mathrm{~mol}^{*} \mathrm{~K}} ight.$} \
\hline C(s graphite) & 5.7 \
\hline C(s diamond & d) 2.4 \
\hline $\mathrm{H}(\mathrm{g})$ & 114.7 \
\hline $\mathrm{H} 2(\mathrm{~g})$ & 130.7 \
\hline $\mathrm{Cl}(\mathrm{g})$ & 165.2 \
\hline $\mathrm{Cl} 2(\mathrm{~g})$ & 223.1 \
\hline $\mathrm{C} 3 \mathrm{Cl}(\mathrm{g})$ & 234.6 \
\hline
\end{tabular}

Part A
Find the deltaSdegree for the formation of $\mathrm{CH3Cl}(\mathrm{g})$ from its elements in their standard states
deltaSrxn $=J / K$

Question

\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{ [Substane| $S^{\circ}\left(\frac{J}{\mathrm{~mol}^{*} \mathrm{~K}}ight.$} \
\hline C(s graphite) & 5.7 \
\hline C(s diamond & d) 2.4 \
\hline $\mathrm{H}(\mathrm{g})$ & 114.7 \
\hline $\mathrm{H} 2(\mathrm{~g})$ & 130.7 \
\hline $\mathrm{Cl}(\mathrm{g})$ & 165.2 \
\hline $\mathrm{Cl} 2(\mathrm{~g})$ & 223.1 \
\hline $\mathrm{C} 3 \mathrm{Cl}(\mathrm{g})$ & 234.6 \
\hline
\end{tabular}

Part A
Find the deltaSdegree for the formation of $\mathrm{CH3Cl}(\mathrm{g})$ from its elements in their standard states
deltaSrxn $=J / K$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the balanced chemical equation for the formation of CH3Cl(g)**: The formation of CH3Cl(g) from its elements in their standard states can be represented by the following equation: $C_{(s, graphite)} + 3H_2(g) + \frac{1}{2}Cl_2(g) ightarrow CH_3Cl(g)$. ‖ ‖ 2 ⋮ **Calculate the change in standard entropy (ΔS°)**: The change in standard entropy for the reaction (ΔS°rxn) is calculated using the standard molar entropies (S°) of the reactants and products. The formula to calculate ΔS°rxn is: $\Delta S^\circ_{rxn} = S^\circ_{products} - S^\circ_{reactants}$. Using the values from the table: $\Delta S^\circ_{rxn} = S^\circ_{CH3Cl(g)} - \left( S^\circ_{C(s, graphite)} + 3S^\circ_{H_2(g)} + \frac{1}{2}S^\circ_{Cl_2(g)} ight)$. Plugging in the values: $\Delta S^\circ_{rxn} = 234.6\ J/(mol\cdot K) - \left( 5.7\ J/(mol\cdot K) + 3 imes 130.7\ J/(mol\cdot K) + \frac{1}{2} imes 223.1\ J/(mol\cdot K) ight)$. ‖ ‖ 3 ⋮ **Perform the calculation**: Now we perform the arithmetic: $\Delta S^\circ_{rxn} = 234.6 - (5.7 + 392.1 + 111.55) = 234.6 - 509.35 = -274.75\ J/(mol\cdot K)$. ‖ ∻Answer∻ ⚹ -274.8 J/(mol*K) ⚹ ∻Key Concept∻ ⚹ Entropy change for a reaction ⚹ ∻Explanation∻ ⚹ The entropy change for a reaction is calculated by subtracting the sum of the standard molar entropies of the reactants from the sum of the standard molar entropies of the products. ⚹◊What is the formula to calculate the change in standard entropy (delta S°) for the formation of a compound from its elements in their standard states in Chemistry?⍭ Generate me a similar question◊

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