Thermodynamics and Thermochemistry 1 Question 2
2. The difference between $\Delta H$ and $\Delta U(\Delta H-\Delta U)$, when the combustion of one mole of heptane $(l)$ is carried out at a temperature $T$, is equal to
(2019 Main, 10 April II)
(a) $-4 R T$
(b) $3 R T$
(c) $4 R T$
(d) $-3 R T$
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Solution:
$$ \begin{aligned} & \text { Key Idea The relation between } \Delta H \text { and } \Delta U \text { is } \ & \qquad \Delta H=\Delta U+\Delta n_{g} R T \end{aligned} $$
$$ \text { where, } \quad \Delta n_{g}=\Sigma n_{p}-\Sigma n_{R} $$
$=$ number of moles of gaseous products - number of moles of gaseous reactants.
The general combustion reaction of a hydrocarbon is as follows :
$$ \mathrm{C}{x} \mathrm{H}{y}+\left(x+\frac{y}{4}\right) \mathrm{O}{2} \longrightarrow x \mathrm{CO}{2}+\frac{y}{2} \mathrm{H}_{2} \mathrm{O} $$
For heptane, $x=7, y=16$
$$ \begin{array}{cc} \Rightarrow & \mathrm{C}{7} \mathrm{H}{16}(l)+11 \mathrm{O}{2}(g) \longrightarrow 7 \mathrm{CO}{2}(g)+8 \mathrm{H}{2} \mathrm{O}(l) \ \therefore & \Delta n{g}=7-11=-4 \end{array} $$
Now, from the principle of thermochemistry,
$$ \begin{aligned} \Delta H & =\Delta U+\Delta n_{g} R T \ \Rightarrow \quad \Delta H-\Delta U & =\Delta n_{g} R T=-4 R T \end{aligned} $$
