The enthalpies of combustion of carbon and carbon monoxide are $$- 393.5$$ and $$ - 283\,kJ\,mo{l^{ - 1}}$$ respectively. The enthalpy of formation of carbon monoxide per mole is
A.
$$-676.5 kJ$$
B.
$$676.5 kJ$$
C.
$$110.5 kJ$$
D.
$$-110.5 kJ$$
Answer :
$$-110.5 kJ$$
Solution :
$$\left( {\text{i}} \right)\,C + {O_2} \rightleftharpoons C{O_2},$$ $$\Delta H = - 393.5\,kJmo{l^{ - 1}}$$
$$\left( {{\text{ii}}} \right)\,CO + \frac{1}{2}{O_2} \rightleftharpoons C{O_2},$$ $$\Delta H = - 283.0\,kJmo{l^{ - 1}}$$
$${\text{Operating (i) - (ii), we have }}$$
$$C + \frac{1}{2}{O_2} \to CO\,\,\Delta H = $$ $$ - 110.5\,\,kJmo{l^{ - 1}}$$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$