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The enthalpy of formation of AB2(l) from A2(g) and B2(g) at 545.5*K is -642.4 kJ/mol. Determine...

The enthalpy of formation of AB2(l) from A2(g) and B2(g) at 545.5*K is -642.4 kJ/mol. Determine (delta)E for this process at 543.5*K (4 significant figures)

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Answer #1

A2(g) + 2 B2 (g) ---------------------> 2 AB2 (l) , H   = -642.4 kJ / mol

n = products - reactants

       = 0 -(1 + 2 )

        = -3

H = E + n RT

-642.4 = E - 3 x 8.314 x 10^-3 x 545.5

E = -628.8 kJ / mol

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