Question

The enthalpy of vaporization of $\mathrm{C}{\mathrm{O}}_2(\mathrm{l})$ is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of $\mathrm{C}{\mathrm{S}}_2(\mathrm{l})$ to be 28 kJ/mol, 9.8 kJ/mol, or -8.4 kJ/mol? Discuss the plausibility of each of these answers.

Short answer

The plausible answer is $28\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

Step-by-step solution

Definition of liquid

One of the states of matter is the liquid state. Because the particles in a liquid are free to move, it lacks a defined form even while having a defined volume.

Discussing the plausibility of each answer

The molecular weight of $\mathrm{C}{\mathrm{S}}_2$is higher than that of $\mathrm{C}{\mathrm{O}}_2$; consequently, the boiling point of $\mathrm{C}{\mathrm{S}}_2$will be higher. Now, since the boiling point of CS is higher, we can say that intermolecular forces are stronger in case of $\mathrm{C}{\mathrm{S}}_2$than in$\mathrm{C}{\mathrm{O}}_2$. As a result, the enthalpies of vaporisation of $\mathrm{C}{\mathrm{S}}_2$will be greater than in $\mathrm{C}{\mathrm{O}}_2$.

Because $\mathrm{C}{\mathrm{S}}_2$ has a larger molar mass than $\mathrm{C}{\mathrm{O}}_2$, its enthalpy will be greater than $9.8\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

The enthalpy of vaporisation in this case cannot be negative, since vaporisation implies the release of energy. As a result, $-8.4\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ is ruled out.

Hence, $28\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ seems to be the plausible answer.