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)
(\(H_{mix}\) is the enthalpy of mixing and \(S_{mix}\) is the entropy of mixing)
Show Hint
- Both \(H_{mix}\) and \(S_{mix}\) are finite
- \(H_{mix}\) is zero and \(S_{mix}\) is finite
- \(H_{mix}\) is finite and \(S_{mix}\) is zero
- Both \(H_{mix}\) and \(S_{mix}\) are zero
The Correct Option is A
Solution and Explanation
In an ideal solution, the two kinds of atoms mix completely at random, with no preference for like or unlike neighbours. Random mixing fixes the entropy of mixing at the standard configurational value
\[ S_{mix} = -R(x_1 \ln x_1 + x_2 \ln x_2) \]
which is a nonzero, finite number for any composition between the two pure ends. Since there is no extra interaction energy between unlike atoms in an ideal solution, the enthalpy of mixing is zero:
\[ H_{mix} = 0 \]
Step 2: Recall the regular solution model.
A regular solution keeps the same random mixing assumption as the ideal solution, so \(S_{mix}\) is worked out with the very same formula and stays finite and nonzero. What changes is the energy term. In a regular solution, an unlike atom pair (A-B) has a different bond energy than the like pairs (A-A, B-B), so forming the solution releases or absorbs a nonzero amount of heat:
\[ H_{mix} = \Omega x_1 x_2 \]
where \(\Omega\) is the interaction parameter, a nonzero constant for a given system at a given temperature.
Step 3: Compare the three solution types.
An ideal solution has \(H_{mix}=0\) and finite \(S_{mix}\). A regular solution has finite (nonzero) \(H_{mix}\) and finite \(S_{mix}\), since it borrows the ideal entropy expression but adds a real interaction energy. There is no physically meaningful solution model here where the entropy of mixing itself becomes zero, since mixing two distinct species randomly always increases configurational disorder.
Step 4: Analyze the options.
(A) Both Hmix and Smix are finite: matches the regular solution definition exactly. Correct.
(B) Hmix is zero and Smix is finite: this is the ideal solution case, not the regular solution case. Incorrect.
(C) Hmix is finite and Smix is zero: the entropy of mixing cannot vanish once two components mix randomly. Incorrect.
(D) Both Hmix and Smix are zero: this would mean no real mixing has occurred at all. Incorrect.
Final Answer:
A regular solution assumes random (ideal-like) mixing, so its entropy of mixing is finite, but it also assumes a nonzero interaction energy between unlike atoms, so its enthalpy of mixing is finite too.
\[ \boxed{\text{Both } H_{mix} \text{ and } S_{mix} \text{ are finite}} \]
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