Answer to Question #233205 in Organic Chemistry for Jese Junior

Question

The standard enthalpy change of three combustion reactions is given below in kJ. 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) ∆HO = –3120

2H2(g) + O2(g) → 2H2O(l) ∆HO = –572

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) ΔHO = –1411

Based on the above information, calculate the standard change in enthalpy, ∆HO, for the following reaction. C2H6(g) → C2H4(g) + H2(g) 

Answer 1

Standard change in enthalpy, ∆HO=$-1137kJ$