For the reaction below, the enthalpy change is given as $\Delta H_r^\circ$. $2\text{C}_2\text{H}_6(g) + 3\text{O}_2(g) \rightarrow 2\text{CH}_4(g) + 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l)$. Which expression could be used to determine $\Delta H_r^ $?
- A$\Delta H_c^\circ(\text{C}_2\text{H}_6(g))$
- B$2\Delta H_c^\circ(\text{C}_2\text{H}_6(g)) - 2\Delta H_c^\circ(\text{CH}_4(g))$
- C$E(\text{C-C}) + 2E(\text{C-H}) - 4E(\text{C=O}) - 4E(\text{H-O})$
- D$\Delta H_f^\circ(\text{CH}_4(g)) + \Delta H_f^\circ(\text{CO}_2(g)) + \Delta H_f^\circ(\text{H}_2\text{O}(l)) - \Delta H_f^\circ(\text{C}_2\text{H}_6(g))$