Thermodynamics And Thermochemistry Result Question 18
18. From the following data, calculate the enthalpy change for the combustion of cyclopropane at $298 \mathrm{~K}$. The enthalpy of formation of $\mathrm{CO} _2(g), \mathrm{H} _2 \mathrm{O}(l)$ and propane $(g)$ are -393.5 , -285.8 and $20.42 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The enthalpy of isomerisation of cyclopropane to propene is $-33.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
(1998,5 M)
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Answer:
Correct Answer: 18. $(-2091.32 \mathrm{~kJ})$
Solution:
- Given: Cyclopropane $\longrightarrow$ Propene $\left(\mathrm{C}_3 \mathrm{H}_6\right) ; \Delta H=-33 \mathrm{~kJ}$
Propene $\left(\mathrm{C}_3 \mathrm{H}_6\right)+\frac{9}{2} \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$;
$\Delta H=-3(393.5+285.8)-20.42=-2058.32 \mathrm{~kJ}$
Adding :
$\text { Cyclopropane }+\frac{9}{2} \mathrm{O}_2(g) \longrightarrow 3 \mathrm{CO}_2(g)+3 \mathrm{H}_2(\mathrm{~g}) \text {; }$
$\begin{aligned} \Delta H & =H_1+H_2 \\ & =-33+(-2058.32) \mathrm{kJ} \\ \Delta H & =-2091.32 \mathrm{~kJ} \end{aligned}$
BETA
