The heats of combustion of carbon and carbon monoxide are $$-393.5$$ and $$ - 283.5\,kJ\,mo{l^{ - 1}}, $$ respectively. The heat of formation ( in $$kJ$$ ) of carbon monoxide per mole is :
A.
-676.5
B.
-110.5
C.
110.5
D.
676.5
Answer :
-110.5
Solution :
$${\text{Given}}$$
$$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);$$ $$\Delta H = - 393.5\,kJ\,mo{l^{ - 1}}...({\text{i}})$$
$$CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right);$$ $$\Delta H = - 283.5\,kJ\,mo{l^{ - 1}}...({\text{ii}})$$
$$\therefore \,\,{\text{Heat of formation of }}CO{\text{ = }}$$ $${\text{eqn (i) }} - {\text{ eqn (ii)}}$$
$${\text{ = }} - 393.5 - \left( { - 283.5} \right)$$
$$ = - 110\,kJ$$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$