314 J K^-1 mol^-1(a) – 18.61 kJ(b) – 18.86 kJ(c) – 18.65 kJ(d) – 18.85 kJThe question was posed to me in homework.My doubt stems from Enthalpy Change Due to Reaction in section Energy Balance of Bioprocess Engineering

314 J K^-1 mol^-1(a) – 18.61 kJ(b) – 18.86 kJ(c) – 18.65 kJ(d) �

1.	314 J K^-1 mol^-1(a) – 18.61 kJ(b) – 18.86 kJ(c) – 18.65 kJ(d) – 18.85 kJThe question was posed to me in homework.My doubt stems from Enthalpy Change Due to Reaction in section Energy Balance of Bioprocess Engineering
Answer» Correct option is (a) – 18.61 kJ The best I can explain: The REACTION is NH4NO3(s) → N2O(g) + 2 H2O(g) Given 2 mol of NH4NO3 decomposes, hence multiplying equation by 2 2 NH4NO3(s) → 2 N2O(g) + 4 H2O(g) Δn = nproduct (g) – nreactant (g) = (2 + 4) -(0) = 6 Work done in chemical reaction is given by ∴ W = – Δn RT = – (6) mol × 8.314 J K^-1 mol^-1 × 373 K = – 18607 J ∴ W = – 18.61 kJ Negative sign indicates that work is done by the system on the surroundings Work done by the surroundings on the system in the reaction is – 18.61 kJ.

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314 J K^-1 mol^-1(a) – 18.61 kJ(b) – 18.86 kJ(c) – 18.65 kJ(d) – 18.85 kJThe question was posed to me in homework.My doubt stems from Enthalpy Change Due to Reaction in section Energy Balance of Bioprocess Engineering

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