Use standard enthalpies of formation to determine ?Horxn for: 3NO2(g) + H2O(l) ?

Question

Use standard enthalpies of formation to determine ?Horxn for: 3NO2(g) + H2O(l) ? 2HNO3(aq) + NO(g) Enter in kJ. Find the change in internal energy for this reaction. Enter in kJ.

Explanation / Answer

2NH3(g) + 3O2(g) + 2CH4(g) --> 2HCN(g) + 6H2O(g)) .....enthaply of reaction, ?H° Work out enthalpies of formation of products & reactants: Formation Equations:- Products 1/2H2(g) + C(s, gr) + 1/2N2(g) --> HCN.....enthaply of formation ?H°f(1) H2(g) + 1/2O2(g) --> H2O(l) .....enthaply of formation, ?H°f(2) Formation Equations: Reactants 1/2N2(g) + 3/2H2(g) --> NH3(g) .....enthaply of formation, ?H°f(3) [O2(g) --> O2(g) .....enthaply of formation = 0 kJ mol-¹ by defintion] C(s, gr) + 2H2(g) --> CH4(g) enthaply of formation, ?H°f(4) where (from data tables): ?H°f(1) = 135.1 kJ mol-¹ ?H°f(2) = -285.83 kJ mol-¹ ?H°f(3) = -46.11 kJ mol-¹ ?H°f(4) = -74.81 kJ mol-¹ Apply Hess's Law, ?H° = S?H°f(products) - S?H°f(reactants) with the correct stoichiometry: ?H° = [2 x ?H°f(1) + 6 x ?H°f(2)] - [2 x ?H°f(3) + 2 x ?H°f(4)] Substituting gives ?H° = -1203 kJ mol-¹ (to 4 s.f.). (2) use the same approach for this to give ?H° = -138.2 kJ mol-¹

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