Answer:
[tex]\Delta H_{dep}=-46.68kJ/mol[/tex]
Explanation:
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In this case, since vaporization is the processes by which a liquid goes to the gas phase, fusion by which a solid goes to liquid and deposition by which a gas goes to solid; we infer that the following set up relates the enthalpies associated to each process:
[tex]\Delta H_{fus}+\Delta H_{vap}=-\Delta H_{dep}[/tex]
Because deposition goes from a state with more energy to a state with less energy, therefore it is negative; in such a way, by plugging in we obtain:
[tex]\Delta H_{dep}=-(\Delta H_{fus}+\Delta H_{vap})\\\\\Delta H_{dep}=-(6.01kJ/mol+40.67kJ/mol)\\\\\Delta H_{dep}=-46.68kJ/mol[/tex]
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