Thermodynamics and Thermochemistry - Result Question 18
####18. From the following data, calculate the enthalpy change for the combustion of cyclopropane at $298 K$. The enthalpy of formation of $CO _2(g), H _2 O(l)$ and propane $(g)$ are -393.5 , -285.8 and $20.42 kJ mol^{-1}$ respectively. The enthalpy of isomerisation of cyclopropane to propene is $-33.0 kJ mol^{-1}$.
(1998, 5M)
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Solution:
- The conversion of $1 kg$ of ice at $273 K$ into water vapours at 383 $K$ takes place as follows:
$$ \begin{aligned} \Delta S _1 & =\frac{\Delta H _{\text {Fusion }}}{\Delta T _{\text {Fusion }}}=\frac{334 kJ kg^{-1}}{273 K}=1.22 kJ kg^{-1} K^{-1} \ \Delta S _2 & =C \ln \frac{T _2}{T _1}=4.2 kJ K^{-1} kg^{-1} \ln \left(\frac{373 K}{273 K}\right) \ & =4.2 \times 2.303(\log 373-\log 273) kJ K^{-1} kg^{-1} \ & =4.2 \times 2.303(2.572-2.436)=1.31 kJ K^{-1} kg^{-1} \end{aligned} $$
$$ \begin{aligned} \Delta S _3 & =\frac{\Delta H _{\text {vap. }}}{\Delta T _{\text {vap. }}}=\frac{2491 kJ kg^{-1}}{373 K} \ & =6.67 kJ kg^{-1} K^{-1} \end{aligned} $$
$$ \begin{aligned} & \Delta S _4=C \ln \frac{T _2}{T _1}=2 kJ K^{-1} kg^{-1} \ln \left(\frac{383 K}{373 K}\right) \ &=2 \times 2.303(\log 383-\log 373) kJ K^{-1} kg^{-1} \ &=2 \times 2.303(2.583-2.572) kJ K^{-1} kg^{-1} \ &=0.05 kJ K^{-1} kg^{-1} \ & \begin{aligned} \Delta S _{\text {Total }} & =\Delta S _1+\Delta S _2+\Delta S _3+\Delta S _4 \ & =1.22+1.31+6.67+0.05 \ & =9.26 kJ kg^{-1} K^{-1} \end{aligned} \end{aligned} $$
