What is $$\Delta {n_{gas}}$$ for the combustion of $$1\,mole$$ of benzene, when both the reactants and the products are at $$298\,K?$$
A.
$$0$$
B.
$$\frac{1}{2}$$
C.
$$\frac{3}{2}$$
D.
$$ - \frac{3}{2}$$
Answer :
$$ - \frac{3}{2}$$
Solution :
$$\eqalign{
& {C_6}{H_6}\left( l \right) + \frac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( l \right) \cr
& \Delta n = 6 - \frac{{15}}{2} \cr
& = - \frac{3}{2} \cr} $$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$