Answer :
Answer:
[tex]\Delta _RS=-287.6J/K[/tex]
[tex]\Delta _RS=972.4J/K[/tex]
Explanation:
Hello,
In this case, the entropy of reaction is computed via:
[tex]\Delta _RS=\Sigma\nu_i S_i^0_{product}-\Sigma\nu_i S_i^0_{reactant}[/tex]
In such a way, for the given reactions, we find:
- 3NO2(g) + H2O(l) ⟶ NO(g) +2HNO3(l)
[tex]\Delta _RS=2*146+210.65-69.940-3*239.9=-287.6J/K[/tex]
- C6H12O6(s) + 6O2(g) LaTeX: \longrightarrow⟶ 6H2O(g) +6CO2(g)
[tex]\Delta _RS=6*188.72+6*213.7-212.1-6*205.0=972.4J/K[/tex]
Best regards.
A chemical reaction occurs wherein one or more reagents, sometimes known as reactants, are converted into one or more products, also known as products.
Reaction calculation:
First reaction calculating value:
3NO₂(g) + H₂O(l) [tex]\long{\longrightarrow}[/tex] NO(g) + 2HNO₃(l)
[tex]\long{\longrightarrow}[/tex] ΔS°rxn = ΔS(NO) + 2ΔS(HNO₃) - 3ΔS(NO₂) - ΔS(H₂O)
= 210.65 + (2 × 155.6) - (3 × 239.9) - 69.94 [tex]\frac{J}{K}[/tex]
= 210.65 + 311.2 - 719.7 - 69.94 [tex]\frac{J}{K}[/tex]
= 521.85 - 789.64 [tex]\frac{J}{K}[/tex]
= -267.79 [tex]\frac{J}{K}[/tex]
Second reaction calculating value:
C₆H₁₂O₆(s) + 6O₂(g) [tex]\long{\longrightarrow}[/tex] 6H₂O(g) + 6CO₂(g)
[tex]\long{\longrightarrow}[/tex] ΔS°rxn = 6ΔS(H₂O) + 6ΔS(CO₂) - ΔS(C₆H₁₂O₆) - 6ΔS(O₂)
= (6 × 188.72) + (6 × 213.7) - 212.1 - (6 × 205) [tex]\frac{J}{K}[/tex]
= 1132.32 + 1282.2 - 212.1 - 1230 [tex]\frac{J}{K}[/tex]
= 2414.52 - 1442.1 [tex]\frac{J}{K}[/tex]
= 972.42 [tex]\frac{J}{K}[/tex]
Find out more about the reaction here:
brainly.com/question/17434463