Thermodynamics and Thermochemistry - Result Question 77
####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 :
$$ C _x H _y+\left(x+\frac{y}{4}\right) O _2 \longrightarrow x CO _2+\frac{y}{2} H _2 O $$
For heptane, $x=7, y=16$
$$ \begin{array}{cc} \Rightarrow & C _7 H _{16}(l)+11 O _2(g) \longrightarrow 7 CO _2(g)+8 H _2 O(l) \ \therefore & \Delta n _g=7-11=-4 \end{array} $$
Now, from the principle of thermochemistry,
$$ \begin{aligned} \Delta H & =\Delta U+\Delta n _g R T \ \Rightarrow \quad \Delta H-\Delta U & =\Delta n _g R T=-4 R T \end{aligned} $$
