Answer to Question #156588 in General Chemistry for Hafiz Faizal

Question

(a) Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Use the following information to calculate ΔH° in kilojoules for the combustion of methane: CH4 (g) + O2 (g) → CH2O (g) + H2O (g) ΔH° = –275.6 kJ CH2O (g) + O2 (g) → CO2 (g) + H2O (g) ΔH° = –526.7 kJ H2O (l) → H2O (g) ΔH° = 44.0 kJ

Answer 1

We are using the equations;

(i) CH_4(g) + O_2(g) $\to$ CH₂O_(g) + H₂O_(g) $\Delta$H^o = -275.6kJ

(ii) CH₂O_(g) + O_2(g) $\to$ CO_2(g) + H₂O_(g) $\Delta$H^o = -526.7kJ

(iii) H2O_(l) $\to$ H2O_(g) $\Delta$H^o = 44.0kJ

to calculate enthalpy change for;

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + 2H₂O_(l)

First, we need to combine the reactions that are similar, for instance, when we combine reaction (i) and (ii), we CH2O(g) will cancel. So we will have a combined equation and the enthalpy changes will add up.

(i) CH_4(g) + O_2(g) $\to$ CH₂O_(g) + H₂O_(g) $\Delta$H^o = -275.6kJ

(ii) CH₂O_(g) + O_2(g) $\to$ CO_2(g) + H₂O_(g) $\Delta$H^o = -526.7kJ

will be

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + 2H₂O_(g) $\Delta$H^o = -802.3kJ

We will need to reverse equation (iii) then multply it by 2 to be able to combine it with the above equation (enthalpy is also multiplied by 2);

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + 2H₂O_(g) $\Delta$H^o = -802.3kJ

2(H2O_(g) $\to$ H2O_(l) ) $\Delta$H^o = 2(-44.0kJ) (the sign changes to -ve when it is reversed)

It will become;

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + 2H₂O_(g) $\Delta$H^o = -802.3kJ

2H2O_(g) $\to$ 2H2O_(l) ) $\Delta$H^o = -88.0kJ

Combining the two and cancelling 2H₂O_(g) will give us the final equation and the enthalpy changes are added together to give;

CH_4(g) + 2O_2(g) $\to$ CO_2(g) + 2H₂O_(l) $\Delta$H^o =(-802.3) + (-88.0) kJ = -890.3kJ