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Chemical Thermodynamics question

2018 · Q57
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Chemical Thermodynamics question

2018 · Q57

IIT JAMChemistryChemical ThermodynamicsNumerical+2 / −0
For the reaction H2(g)+12O2(g)⟶H2O(l)H_2(g) + \frac{1}{2} O_2(g) \longrightarrow H_2O(l)H2​(g)+21​O2​(g)⟶H2​O(l), the following information is given T = 300 K ΔHˉ∘=−285\Delta \bar{H}^{\circ} = -285ΔHˉ∘=−285 kJ mol −1SˉH2O∘(l)=70^{-1}\bar{S}_{H_2O}^{\circ}(l) = 70−1SˉH2​O∘​(l)=70 J K −1^{-1}−1 mol −1SˉO2∘(g)=204^{-1}\bar{S}_{O_2}^{\circ}(g) = 204−1SˉO2​∘​(g)=204 J K −1^{-1}−1 mol −1SˉH2∘(g)=130^{-1}\bar{S}_{H_2}^{\circ}(g) = 130−1SˉH2​∘​(g)=130 J K −1^{-1}−1 mol −1ΔSˉuniverse∘^{-1}\Delta \bar{S}_{universe}^{\circ}−1ΔSˉuniverse∘​ for the reaction is ‾\underline{\hspace{2cm}}​ J K −1^{-1}−1 mol −1^{-1}−1.
Numerical answer
View written solutionFree

Correct answer: 786, 790

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