a
\[\begin{aligned}∆H &= \sum ∆H^0_\mathrm{f, produkter} - \sum ∆H^0_\mathrm{f, reaktanter} = \\ &= \left(3∆H^0_\mathrm{f, CO_2} + 2∆H^0_\mathrm{f, H_2O}\right) - \left(∆H^0_\mathrm{f, C_3H_8} + 5∆H^0_\mathrm{f, O_2}\right) = \\ &= \left(\left(3 \cdot (-393,5) + 4 \cdot (-285,8)\right) -\left(-103,8 + 5 \cdot 0 \right)\right)\mathrm{kJ/mol} = \\ &= -2219,9\mathrm{kJ/mol} ≈ -2220\mathrm{kJ/mol}\end{aligned}\]
b
\[\begin{aligned}∆H &= \sum ∆H^0_\mathrm{f, produkter} - \sum ∆H^0_\mathrm{f, reaktanter} = \\ &= \left(∆H^0_\mathrm{f, Al_2O_3} + 2∆H^0_\mathrm{f, Fe}\right) - \left(∆H^0_\mathrm{f, Fe_2O_3} + 2∆H^0_\mathrm{f, Al}\right) = \\ &= \left((-1675,7 + 2 \cdot 0) -(-824,2 + 2 \cdot 0 )\right)\mathrm{kJ/mol} = -851,5\mathrm{kJ/mol} \end{aligned}\]
c
\[\begin{aligned} ∆H &= \sum ∆H^0_\mathrm{f, produkter} - \sum ∆H^0_\mathrm{f, reaktanter} = \\ &= \left(2∆H^0_\mathrm{f, NH_3}\right) - \left(∆H^0_\mathrm{f, N_2} + 3∆H^0_\mathrm{f, H_2}\right)= \\ &= \left(2 \cdot (-45,6) -(0 + 2 \cdot 0 )\right)\mathrm{kJ/mol} = -91,2\mathrm{kJ/mol} \end{aligned}\]
d
\[\begin{aligned} ∆H &= \sum ∆H^0_\mathrm{f, produkter} - \sum ∆H^0_\mathrm{f, reaktanter} = \\ &= \left(∆H^0_\mathrm{f, CaO} + ∆H^0_\mathrm{f, CO_2}\right) - \left(∆H^0_\mathrm{f, CaO_3}\right)= \\ &= \left(\left(-635,1 + (-393,5)\right) - (-1207,6)\right)\mathrm{kJ/mol} = -179,0\mathrm{kJ/mol} \end{aligned}\]
