Question

Calculate the enthalpy change for:
C2H4 (g) + H2 (g) ---> C6H6 (l)

given the following:
C2H4 (g) + 3O2 (g) ---> 2CO2 (g) + 2H2O(l)
deltaH= -845.2kJ

2H2(g) + O2 (g) ---> 2H2O(l)
deltaH= -573.2kJ

2C2H6(l) + 7O2(g) ---> 2CO2 (g) + 6H2O(l)
deltaH= 1.1987.4kJ

Answer 1

NOTE: I will assume th product is C2H6, since there is no correc tbalance...

so, for:

C2H4 (g) + H2 (g) ---> C2H6 (l)

Use Hess Law:

C2H4 (g) + 3O2 (g) ---> 2CO2 (g) + 2H2O(l)   H= -845.2 kJ

2H2(g) + O2 (g) ---> 2H2O(l) H= -573.2 kJ

2C2H6(l) + 7O2(g) ---> 2CO2 (g) + 6H2O(l) H= 11987.4 kJ

divide (2) by 2, since we need H2; invert (3) nad divide by 2

C2H4 (g) + 3O2 (g) ---> 2CO2 (g) + 2H2O(l)   H= -845.2 kJ

H2(g) + 1/2O2 (g) ---> H2O(l)   H= -573.2/2 kJ

CO2 (g) + 3H2O(l)--->C2H6(l) + 7/2O2(g) H= -11987.4/2 kJ

Add all

C2H4 (g) + 3O2 (g) + H2(g) + 1/2O2 (g) +CO2 (g) + 3H2O(l) --> 2CO2 (g) + 2H2O(l)+H2O(l) + C2H6(l) + 7/2O2(g)

cancel common terms

C2H4 (g) + 3O2 (g) + H2(g) + 1/2O2 (g) +CO2 (g) + 3H2O(l) --> 2CO2 (g) + 2H2O(l)+H2O(l) + C2H6(l) + 7/2O2(g)

C2H4 (g) + H2(g) --> C2H6(l)

so

HRxn =  -845.2 +  -573.2/2 + -11987.4/2 = -7125.5 kJ