Question

using the data below, calculate the entropy for the decomposition of no₂ into no and o₂. 2no₂(g) → 2no(g) + o₂(g) \

$$\begin{tabular}{|c|c|c|} \\hline compound & s° (j/mol k) & δh°_f (kj/mol) \\\\ \\hline no₂(g) & 240 & 34 \\\\ \\hline no(g) & 211 & 90 \\\\ \\hline o₂(g) & 205 & 0 \\\\ \\hline \\end{tabular}$$

δs_rxn = ? j/k enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for entropy change of reaction

The formula for the entropy change of a reaction (\(\Delta S_{rxn}\)) is \(\Delta S_{rxn}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})\), where \(n\) and \(m\) are the stoichiometric coefficients of products and reactants respectively, and \(S^{\circ}\) is the standard molar entropy.

Step2: Identify the stoichiometric coefficients and \(S^{\circ}\) values

For the reaction \(2NO_2(g)
ightarrow 2NO(g)+O_2(g)\):

• Reactants: \(NO_2(g)\) with \(m = 2\) and \(S^{\circ}(NO_2)=240\space J/mol\space K\)
• Products: \(NO(g)\) with \(n = 2\) and \(S^{\circ}(NO)=211\space J/mol\space K\); \(O_2(g)\) with \(n = 1\) and \(S^{\circ}(O_2)=205\space J/mol\space K\)

Step3: Calculate the sum of \(S^{\circ}\) for products

\(\sum nS^{\circ}(\text{products})=2\times S^{\circ}(NO)+1\times S^{\circ}(O_2)=2\times211 + 205=422+205 = 627\space J/K\)

Step4: Calculate the sum of \(S^{\circ}\) for reactants

\(\sum mS^{\circ}(\text{reactants})=2\times S^{\circ}(NO_2)=2\times240 = 480\space J/K\)

Step5: Calculate \(\Delta S_{rxn}\)

\(\Delta S_{rxn}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})=627 - 480=147\space J/K\)
Since the value is positive, we have \(+ 147\).

Answer:

\(+147\)