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Use Hess's law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal tothe sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid.
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Step 1
We have a solid (S) that we want to convert into a gas (G). The enthalpy of sublimation (ΔH_sub) is the energy required for this process: S → G. Show more…
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The enthalpy of sublimation of iodine is $60.2 \mathrm{~kJ} / \mathrm{mol}$, and its enthalpy of vaporization is $45.5 \mathrm{~kJ} / \mathrm{mol} .$ What is the enthalpy of fusion of iodine?
Use the given standard enthalpies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) to determine the enthalpy of reaction of the following reaction : $$ \begin{gathered} \mathrm{NH}_{3}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{NF}_{3}(g)+3 \mathrm{HF}(g) \\ \Delta H_{f}^{\circ}\left(\mathrm{NH}_{3}, g\right)=-46.2 ; \quad \Delta H_{f}^{\circ}\left(\mathrm{NF}_{3}, g\right)=-113.0 ; \quad \Delta H_{f}^{\circ}(\mathrm{HF}, g)=-269.0 \end{gathered} $$ 1.T/mnl (h) $-8738 \mathrm{~kJ} / \mathrm{mol}$ (c) $-697.2 \mathrm{~kJ} / \mathrm{mol}$ (d) $-890.4 \mathrm{~kJ} / \mathrm{mol}$
Given the following information: Heat of sublimation of $\mathrm{Li}(s)=166 \mathrm{kJ} / \mathrm{mol}$ Bond energy of $\mathrm{HCl}=427 \mathrm{kJ} / \mathrm{mol}$ Ionization energy of $\mathrm{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}$ Electron affinity of $\mathrm{Cl}(g)=-349 \mathrm{kJ} / \mathrm{mol}$ Lattice energy of LiCl(s) $=-829 \mathrm{kJ} / \mathrm{mol}$ Bond energy of $\mathrm{H}_{2}=432 \mathrm{kJ} / \mathrm{mol}$ Calculate the net change in energy for the following reaction: $$2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g)$$
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