Question

Given tha delta hydrogen = -393.5 kJ for the rxn.: C(s) +O2(g) -> CO2(g), and

given that delta Hydrogen = -285.8 kj for the rxn.: H2(g)+ 1/2O2(g) -> H2O(l), and given

that delta hydrogen = -84.7 kJ for the rxn .: 3H2(g) +2C +2C(graphite) -> C2H6(g),

calculate the dleta hydrogen for the rxn C2H6(g) + 3 1/2O2(g) -> 2CO2(g) + 3HO(l)

Delta hydrogen = kJ

Answer 1

Multiply rxn 1 by 2, we get-

2C(s) + 2O2(g) ---> 2CO2(g) ------(1) deltaH = -393.5kJ*2 = -787kJ

Multiply rxn 2 by 3-

3H2(g) + 3/2O2(g) ----> 3H2O(l) -----(2) deltaH = -285.8kJ*3 = -857.4kJ

Switch the sides of reaction 3, we get-

C2H6(g) ---> 3H2(g) + 2C(s) --------(3) delta H = 84.7kJ

Adding all the above 3 reactions we get -

C2H6(g) + 7/2O2(g) ---> 2CO2(g) + 3H2O(l)

which is the reaction for which we have to calculate deltaH.

DeltaH = -787kJ + (-857.4kJ) + 84.7kJ = -1559.7kJ