Question

The enthalpy of combustion of benzene from the following data will be

(i) $6C(s) + 3H_{2}(g) \rightarrow C_{6}H_{6}(l);\Delta H = + 45.9kJ$

(ii) $H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l);\Delta H = - 285.9kJ$

(iii) $C(s) + O_{2}(g) \rightarrow CO_{2}(g);\Delta H = - 393.5kJ$

• A.
  • 3172.8 kJ
• B. –1549.2 kJ
• C. –3172.8 kJ
• D. –3264.6 kJ

Answer 1

Answer: (D)

–3264.6 kJ

Answer 2

$C_{6}H_{6} + \frac{15}{2}O_{2} \rightarrow 6CO_{2} + 3H_{2}O;\Delta H = ?$

$\mathbf{\Delta}\mathbf{H =}\mathbf{\Delta}\mathbf{H}_{\mathbf{(}\text{Product}\mathbf{)}}\mathbf{- \Delta}\mathbf{H}_{\mathbf{(}\text{Reactant}\mathbf{)}}$

$\mathbf{= \lbrack(6 \times - 393.5) + (3 \times - 285.9)\rbrack - (45.9)}\mathbf{=}\mathbf{-}\mathbf{3264.6kJ}$