For a regular solution, which one of the following statements is correct? (\(H_{mix}\) is the enthalpy of mixing and \(S_{mix}\) is the entropy of mixing)
Show Hint
A regular solution mixes atoms randomly like an ideal solution (finite entropy) but has a nonzero interaction energy between unlike atoms (finite enthalpy).
Concept: What makes a solution "regular".
A solution is built from two ideas: how the atoms are arranged (entropy) and how much energy is released or absorbed when they mix (enthalpy). An ideal solution assumes the atoms don't care about their neighbours at all, so mixing costs no extra energy, $H_{mix}=0$, but arranging two kinds of atoms randomly still increases disorder, so $S_{mix}$ is a definite, nonzero number given by $S_{mix}=-R(x_1\ln x_1+x_2\ln x_2)$.
Step 1: Add the regular solution twist.
A regular solution keeps the random-arrangement assumption of the ideal solution, so its entropy term is calculated the exact same way and stays finite. But it drops the "atoms don't care about neighbours" assumption: unlike atom pairs now have their own bond energy, different from the like-atom pairs. That difference in bond energy shows up as a real, nonzero heat of mixing, $H_{mix}=\Omega x_1 x_2$, with $\Omega \neq 0$.
Step 2: Check the extreme cases against the options.
If $H_{mix}=0$ the solution would be ideal, not regular, so an option pairing $H_{mix}=0$ with a finite $S_{mix}$ is really describing the ideal case.
If $S_{mix}=0$ the atoms would have to be fully ordered rather than randomly mixed, which contradicts the very assumption a regular solution is built on.
If both terms were zero, the two components would not really be mixing at all.
None of these three situations is what "regular solution" means.
Step 3: State the correct pairing.
A regular solution has a finite, nonzero enthalpy of mixing from the changed bond energies, and a finite entropy of mixing from the random arrangement of atoms. Both quantities are finite at the same time.
\[ \boxed{\text{Both } H_{mix} \text{ and } S_{mix} \text{ are finite}} \]
