The standard enthalpies of combustion of C6H6(l), C(graphite) and H2(g) are respectively -3270 kJ mol-1, -394 kJ mol-1 and -286 kJ mol-1. What is the standard enthalpy of formation of C6H6(l) in kJ mol-1?
(a) -48
(b) +48
(c) -480
(d) +480
(b) +48
Explanation:
C6H6(l) + 15/2O2(g) → 6CO2(g) + 3H2O(l)
ΔH = -3270 kJ mol-1 ....(i)
C(gr) + O2(g) → CO2(g)
ΔH = -394 kJ mol-1 ....(ii)
H2(g) + 1/2O2(g) → H2O(l)
ΔH = -286 kJ mol-1 ....(iii)
Formation of C6H6
C(gr) + 3H2(g) → C6H6(l); ΔH =? ....(iv)
By multiplying eq. (ii) with 6 and eq. (iii) with 3 and adding we get,
6C(gr) + 6O2(g) + 3H2(g) + 3/2O2(g) → 6CO2(g) + 6H2O(l)
ΔH = 6(-394) + 3(-286)
= (-2364) + (-858)
= -3222 kJ/mol
Now, by subtracting eq. (i) from (v) we get
C(gr) + 3H2(g) → C6H6(l)
ΔH = -3222 - (-3270) = +48 kJ/mol