Question

The enthalpy of vaporization of CO₂(l) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS₂(l) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers.

Answer 1

The enthalpy of vaporization of CS₂(l) is expected to be higher than that of CO₂(l) due to its stronger intermolecular forces, making 28 kJ/mol the most plausible value. The 9.8 kJ/mol is less plausible and -8.4 kJ/mol is implausible.

Step-by-step explanation:

When considering the enthalpy of vaporization for liquids like CO₂ and CS₂, we must consider the strength of intermolecular forces and the effect of molecular weight. For CS₂, since it has a higher molecular weight than CO₂, it experiences stronger dispersion forces. Thus, more energy is required to vaporize CS₂, making the enthalpy of vaporization higher than that of CO₂.

Given that the enthalpy of vaporization of CO₂(l) is 9.8 kJ/mol, it would make sense that the enthalpy of vaporization for CS₂(l) would be higher, due to stronger attractive forces. An enthalpy of 28 kJ/mol seems reasonable and is the most likely correct value. The option of 9.8 kJ/mol, although the same as CO₂, is less plausible due to the stronger intermolecular forces in CS₂ necessitating more energy to vaporize it. The value of -8.4 kJ/mol is clearly implausible as it would suggest that energy is released upon vaporization, which is not the case as vaporization requires energy input.