Question

Given the following data:

H2(g) + 1/2O2(g) → H2O(l)	ΔH° = -286.0 kJ
C(s) + O2(g) → CO2(g)	ΔH° = -394.0 kJ
2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g)	ΔH° = 1300.0 kJ

Calculate ΔH° for the reaction:
2C(s) + H₂(g) → C₂H₂(g)

Answer 1

H2(g) + 1/2O2(g) ---> H2O(l) deltaH = -286.0kj ...(1)

C(s)+O2(g)---> CO2(g). DeltaH= -394.0kJ

Multiplying the reaction by 2-

2C(s) + 2O2(g) --> 2CO2(g) deltaH = -788kJ ...(2)

Adding reaction 1 and 2 we get-

H2(g) + 1/2O2(g) + 2C(s)+2O2(g)--> H2O(l) + 2CO2(g)

H2(g) +5/2O2(g) + 2C(s)--> H2O(l) + 2CO2(g)...(3)

DeltaH = -286.0 +(-788) = -1074kJ

2CO2(g)+H2O(l)--->C2H2(g) + 5/2O2(g) deltaH = 1300kJ. (4)

Adding reaction 3 to 4, we get-

H2(g)+5/2O2(g)+2C(s)+2CO2(g)+H2O(l)

---> H2O(l)+2CO2(g)+C2H2(g)+5/2O2(g)

Cancelling common species-

H2(g)+ 2C(s) ---> C2H2(g)

DeltaH = -1074kJ + 1300kJ = 226kJ

DeltaH° = 226kJ