Question

Given standard enthalpy of formation of $CO$ (-110 KJ ${mol}^{-1}$) and $C{O}_2$ (-394 KJ ${mol}^{-1}$ ). The heat of combustion when one mole of graphite burns is:

• A. $-110KJ$
• B. $-284KJ$
• C. $-394KJ$
• D. $-504KJ$

Answer 1

Answer: (B)

$-284KJ$

Enthalpy of formation of $CO=-110kJ/mol$

Enthalpy of formtaion of $C{O}_2=-394kJ/mol$

$C+O_2\to C{O}_2$ $△H_c=-394kJ/mol$

Formation of Carbon monoxide

$C+\frac{1}{2}{O}_2\to CO$ $△H_c^{{}^{\prime }}=-110kJ/mol$

$\Delta H=\Delta H_c-\Delta H_c^{{}^{\prime }}$

$=-394-(-110)$

$=-394+110$

$=-284kJ/mol$

Hence, the correct option is $B$