Calculation of Enthalpies of Reaction from Enthalpies of Formation Delta H react

Question

Calculation of Enthalpies of Reaction from Enthalpies of Formation Delta H reaction = Sum of enthalpies of products - Sum of enthalpies of reactants Find the volume of the solid bounded by f(x, y) = 4 - xy and xy-plane over this region. H formation of any element in its pure state = 0 Find the enthalpy for the complete combustion of ethene: C_2H_4(g) + 3 O_2(g) rightarrow 2 CO_2(g) + 2 H_2O(g) Calculate the enthalpy change for the reaction: 2 Na(s) + 2 H_2O(l) rightarrow 2 NaOH(s) + H_2(g) Delta H_f H_2O(l) = -68 kcal/mol Delta H_f NaOH(s) = -102 kcal/mol

Explanation / Answer

Ans. 22. Stoichiometry of balanced reaction: 2 mol C2H2 reacts with 3 mol O2 to form 2 mol CO2 and 2 mol H2O.

Note that the enthalpy, dH is reported in terms of kJ / mol. So, for a chemical species present in more than 1 mol, multiply its molar enthalpy with its coefficient to get its total enthalpy contribution during the reaction.

So,

dH of reaction = Sum of enthalpies of products - Sum of enthalpies of reactants

                                    = [2 x (dH, CO2) + 2 x (dH, H2O) ] – [(dH, C2H4) + 3 x (dH, O2)]

Or, dH = [ 2 x (-393.5 kJ/ mol) + 2 x (-241.8 kJ/ mol)] - [ (52.30 kJ/ mol) + 3 x 0]

Or, dH = -1270.6 kJ/ mol – 52.30 kJ/ mol

Or, dH = -1322.9 kJ/ mol

Thus, enthalpy of combustion, dH = -1322.9 kJ/ mol

Ans. 23. dH = [ 2 x (dH, NaOH) + 0] – [0 + 2 x (dH, H2O)]

            Or, dH = [ 2 x (- 102 kcal/ mol) + 0] – [0 + 2 x (- 68 kcal/ mol)]

            Or, dH = - 204 kcal/ mol – (- 136 kcal/ mol)

            Or, dH = - 204 kcal/ mol + 136 kcal/ mol = - 68 kcal/ mol

Thus, dH of the reaction = - 68 kcal/ mol

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