For the reaction, $${C_3}{H_8}\left( g \right) + 5{O_2}\left( g \right) \to 3C{O_2}\left( g \right) + 4{H_2}O\left( l \right)$$ at constant temperature, $$\Delta H - \Delta E$$ is
A.
$$3RT$$
B.
$$-RT$$
C.
$$+RT$$
D.
$$-3RT$$
Answer :
$$-3RT$$
Solution :
For the reaction,
$${C_3}{H_8}\left( g \right) + 5{O_2}\left( g \right) \to $$ $$3C{O_2}\left( g \right) + 4{H_2}O\left( l \right)$$
$$\Delta {n_g} = $$ number of gaseous moles of products $$ - $$ number of gaseous moles of reactants
$$\eqalign{
& = 3 - 6 = - 3 \cr
& \therefore \,\,\Delta H = \Delta E + \Delta n\,RT \cr
& {\text{or}}\,\,\Delta H - \Delta E = \Delta nRT \cr
& \therefore \,\,\Delta H - \Delta E = - 3RT \cr} $$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$