a)
∆H⁰ = ∑n∆H⁰sp. substr. - ∑n∆H⁰sp. prod. = 3 ∙ ∆H⁰sp.C3H8(g) - (∆H⁰sp.CH4(g) + 2 ∙ ∆H⁰sp.C4H10(g)) = 3 ∙ (-2220 kJ/mol) – (-890 kJ/mol + 2 ∙ (-2878 kJ/mol)) = -6660 kJ/mol + 6646 kJ/mol = -14 kJ/mol
b)
∆H⁰ = ∑n∆H⁰sp. substr. - ∑n∆H⁰sp. prod. = (∆H⁰sp.C(s) + 2 ∙ ∆H⁰sp.H2(g)) - ∆H⁰sp.CH3OH(c) = (-394 kJ/mol) + 2 ∙ (-286 kJ/mol)) – (-726 kJ/mol) = -966 kJ/mol + 726 kJ/mol = -240 kJ/mol
