What is the standard entropy change for the following reaction?
$$\begin{array} { l l l l l }
& 4 \mathrm {~N} _ { 2 } ( \mathrm {~g} ) & + 12 \mathrm { H } _ { 2 } ( \mathrm {~g} ) & \rightarrow & 8 \mathrm { NH } _ { 3 } ( \mathrm {~g} ) \\
S _ { 298 } ^ { 0 } & 191.5 & 130.6 && 192.3
\end{array}$$ (J/mol?K)
A) -794.8 J/K?mol-rxn
B) 794.8 J/K?mol-rxn
C) 2948.8 J/K?mol-rxn
D) -2948.8 J/K?mol-rxn
E) -129.8 J/K?mol-rxn

Question

What is the standard entropy change for the following reaction? 
$$\begin{array} { l l l l l } 
& 4 \mathrm {~N} _ { 2 } ( \mathrm {~g} ) & + 12 \mathrm { H } _ { 2 } ( \mathrm {~g} ) & \rightarrow & 8 \mathrm { NH } _ { 3 } ( \mathrm {~g} ) \\
S _ { 298 } ^ { 0 } & 191.5 & 130.6 && 192.3
\end{array}$$ (J/mol?K)
A) -794.8 J/K?mol-rxn
B) 794.8 J/K?mol-rxn
C) 2948.8 J/K?mol-rxn
D) -2948.8 J/K?mol-rxn
E) -129.8 J/K?mol-rxn

Quizplus · Accepted Answer

The standard entropy change for a reaction is calculated by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. In this case, the standard entropy change is:$$\Delta S ^ { 0 } = 8 	imes 192.3 - 4 	imes 191.5 - 12 	imes 130.6 = -794.8 \mathrm {~J/K~mol-rxn}$$

What Is the Standard Entropy Change for the Following Reaction