Question

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

Answer 1

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.