Calculate standard enthalpy of reaction,
`2C_2H_6 (g) + 7O_2(g) to 4CO_2(g) + 6H_2O(l)`
Given that
`Delta_f H^0 (CO_2) = -393.5 kJ mol^(-1)`
` Delta_f H^0 (H_2O) = -285.8 kJ mol^(-1) and `
`Delta_fH^0 (C_2H_6) = -84.9 kJ mol^(-1)`

Explanation

`Delta_fH^0 = SigmaDelta_f H^0 ("Products") - Sigma Delta_fH^0("reactants")`
`= [4 Delta_f H^0 (CO_2) + 6Delta_f H^0 (H_2O)]`
`-[2 Delta_f H^0 (C_2H_6) + 7 Delta_f H^0 (O_2)]`
`= (4 mol xx (-393.5 kJ mol^(-1)) + 6 mol xx (-285.8 kJ mol^(-1)]`
`-[2 mol xx (-84.9 kJ mol^(-1) + 0]`
`= -1574 kJ -1714.8 kJ + 169.8 kJ `
`= -3119 kJ`