Using the data (all values are in kilocalories per mole at 25 °C) given below, calculate the bond energy of C–C and C–H bonds.
ΔH° combustion (ethane) = −372.0
ΔH° combustion (propane) = −530.0
ΔH° for C(graphite) → C(g) = 172.0
Bond energy of H–H = 104.0
ΔfH° of H2O(l) = −68.0
ΔfH° for CO2(g) = −94.0
Verified Answer
We are given:
(i) C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(l), ΔH° = −372.0 kcal
(ii) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l), ΔH° = −530.0 kcal
(iii) C(s) → C(g), ΔH° = 172.0 kcal
(iv) H2(g) → 2H(g), ΔH° = 104.0 kcal
(v) H2(g) + 1/2O2(g) → H2O(l), ΔH° = −68.0 kcal
(vi) C(g) + O2(g) → CO2(g), ΔH° = −94.0 kcal
Suppose the bond energy of C–C bond = x kcal mol−1 and that of C–H bond = y kcal mol−1.
To get equation (vii): (i) + 2×(iii) + 3×(iv) − 3×(v) − 2×(vi) → ΔH = 676 kcal
To get equation (viii): (ii) + 4×(iii) + 4×(iv) − 3×(v) − 3×(vi) → ΔH = 956 kcal
Thus, equations are:
x + 6y = 676
2x + 8y = 956
On solving: x = 82, y = 99
Hence, C–C bond energy = 82 kcal mol−1 and C–H bond energy = 99 kcal mol−1.