ThermochemistryHard

Question

Standard enthalpy of formation of gaseous ethane, ethene and benzene from gaseous atoms are −2839, −2275 and −5506 kJ/ mol, respectively. The bond enthalpy of C−H bond is 412 kJ/ mol. The magnitude of resonance energy of benzene compared with one Kekule structure is

Options

A.24 kJ
B.52 kJ
C.2524 kJ
D.152 kJ

Solution

$2C(g) + 6H(g) \rightarrow C_{2}H_{6}(g)$

$\Delta H = - 2839 = \lbrack 0 + 0\rbrack - \left\lbrack B.E_{C - C} + 6 \times 412 \right\rbrack \Rightarrow B.E._{C - C} = 367\text{ kJ}$

and, $2C(g) + 4H(g) \rightarrow C_{2}H_{4}(g)$

$\Delta H = - 2275 = \lbrack 0 + 0\rbrack - \left\lbrack B.E_{C \equiv C} + 4 \times 412 \right\rbrack \Rightarrow B.E._{C \equiv C} = 627\text{ kJ}$

Now, $6C(g) + 6H(g) \rightarrow C_{6}H_{6}(g);$

$\Delta H = - 5506 = \lbrack 0 + 0\rbrack - \lbrack 3 \times 367 + 3 \times 627 + 6 \times 412\rbrack - R.E. $$$\therefore R.E. = 52\text{ kJ/mol}$$

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