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Evgesh-ka [11]
3 years ago
14

Calculate ΔS°rxn (J/k) for 3NO2(g) + H2O(l)LaTeX: \longrightarrow⟶ NO(g) +2HNO3(l) C6H12O6(s) + 6O2(g) LaTeX: \longrightarrow⟶ 6

H2O(g) +6CO2(g) Enter numbers to 1 decimal places. Substance or Ion S° (J/molLaTeX: \cdot⋅K) N2(g) 191.5 N2O(g) 219.7 NO(g) 210.65 NO2(g) 239.9 F2(g) 202.7 H2(g) 130.6 HNO3(l) 155.6 HNO3(aq) 146 H2O(l) 69.940 H2O(g) 188.72 C6H12O6(s) 212.1 O2(g) 205.0 CO2(g) 213.7 CO2(aq) 121 NF3(g) 260.6
Chemistry
1 answer:
Zigmanuir [339]3 years ago
3 0

Answer:

\Delta _RS=-287.6J/K

\Delta _RS=972.4J/K

Explanation:

Hello,

In this case, the entropy of reaction is computed via:

\Delta _RS=\Sigma\nu_i  S_i^0_{product}-\Sigma\nu_i  S_i^0_{reactant}

In such a way, for the given reactions, we find:

- 3NO2(g) + H2O(l) ⟶ NO(g) +2HNO3(l)

\Delta _RS=2*146+210.65-69.940-3*239.9=-287.6J/K

- C6H12O6(s) + 6O2(g) LaTeX: \longrightarrow⟶ 6H2O(g) +6CO2(g)

\Delta _RS=6*188.72+6*213.7-212.1-6*205.0=972.4J/K

Best regards.

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