Question

The enthalpy of combustion of benzene from the following data will be:
(i) $6C\left(s\right)+3H_2\left(g\right)\to C_6{H}_6\left(l\right);\Delta H=+45.9kJ$
(ii) $H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right);\Delta H=-285.9kJ$
(iii) $C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right);\Delta H=-393.5kJ$

• A. $+3172.8kJ$
• B. $-1549.2kJ$
• C. $-3172.8kJ$
• D. $-3264.6kJ$

Answer 1

The correct option is D $-3264.6kJ$

The combustion of benzene is represented by equation (iv)
(i) $6C\left(s\right)+3H_2\left(g\right)\to C_6{H}_6\left(l\right);\Delta H=+45.9kJ$
(ii) $H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right);\Delta H=-285.9kJ$

(iii) $C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right);\Delta H=-393.5kJ$

(iv) $C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

Reverse equation (i)

(v) $C_6{H}_6\left(l\right)\to 6C\left(s\right)+3H_2\left(g\right);\Delta H=-45.9kJ$

Multiply equation (ii) with 3.

(vi) $3H_2\left(g\right)+\frac{3}{2}{O}_2\left(g\right)\to 3H_{2}O\left(l\right);\Delta H=-285.9\times 3=-857.7kJ$

Multiply equation (iii) with 6.

(vii) $6C\left(s\right)+6O_2\left(g\right)\to 6C{O}_2\left(g\right);\Delta H=-393.5\times 6=-2361kJ$

Add equations (v), (vi) and (vii)

(iv) $C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)+3H_{2}O\left(l\right);\Delta H=-45.9-857.7-2361=-3264.6kJ$

The enthalpy of combustion of benzene is $-3264.6kJ$.