Question

Consider the sudden evaporation of 86.2 g of hexane (C6H14) at a temperature of 346 K and 1.0 bar pressure. Thermodynamic data for hexane: Boiling point = 342 K; melting point = 178 K, molar mass = 86.2 g/ mol; Cpo (C6H14(e)] - 143.3 J/(mol K), Cpo (C6H14(0] = 196.2 J/mol K), Standard enthalpy of fusion at 178 K, AfusHO(C6H14) = 9.57 kJ/mol Standard enthalpy of evaporation at 342 K, AvapH0[C6H14] = 28.85 kJ/mol a) Calculate AH associated with the evaporation. b) Calculate the AS associated with the evaporation.

Answer 1

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PART (a):

Evaporation process is as follows:

$C_{6}H_{14}_{(l)}\rightarrow C_{6}H_{14}_{(g)}\;\;\; ;\Delta H_{vap}$

Hence, heat of vapourisation is given by,

$\Delta H_{vap}=(\Delta H_{reactants}-\Delta H_{prodcuts})=C_{po}[C_{6}H_{14}_{(l)}]-C_{po}[C_{6}H_{14}_{(g)}]$

Given that,

$C_{po}[C_{6}H_{14}_{(g)}]=143.3\;J/mol.K$

$C_{po}[C_{6}H_{14}_{(l)}]=196.2\;J/mol.K$

$\therefore \Delta H_{vap}=196.2\;J/mol.K-143.3\;J/mol.K$

$\mathbf{\therefore \Delta H_{vap}=52.9\;J/mol.K}$

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PART (b):

Given that, Boiling point, $T_{b}=342\;K$

Heat of vapourisation, $\Delta H_{vap}=52.9\;J/mol.K$

At boiling point, Change in entropy is given by,

$\Delta S= \frac{\Delta H_{vap}}{T_{b}}=\frac{52.9\;J/mol.K}{342\;K}$

$\mathbf{\therefore \Delta S=0.1547\;J/mol}$
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