Question

For an ideal binary liquid mixture:

• A. \( \Delta S_{\text{(mix)}} = 0; \Delta G_{\text{(mix)}} = 0 \)
• B. \( \Delta H_{\text{(mix)}} = 0; \Delta S_{\text{(mix)}} < 0 \)
• C. \( \Delta V_{\text{(mix)}} = 0; \Delta G_{\text{(mix)}} > 0 \)
• D. \( \Delta S_{\text{(mix)}} > 0; \Delta G_{\text{(mix)}} < 0 \)

Hint:

For any spontaneous mixing process (ideal or non-ideal), \( \Delta S_{\text{mix}} \) is always positive (\( > 0 \)) and \( \Delta G_{\text{mix}} \) is always negative (\( < 0 \)). These two parameters do not depend on ideal behavior.

Answer 1

The Correct Option is D

Step 1: Define an ideal solution.
An ideal binary mixture obeys Raoult's law over the whole range, with solute-solvent forces equal to the original like-like forces.
Step 2: Decide the enthalpy of mixing.
Because no new energy is gained or lost forming the mix, $\Delta H_{\text{mix}} = 0$.
Step 3: Decide the volume of mixing.
With identical interactions there is no contraction or expansion, so $\Delta V_{\text{mix}} = 0$.
Step 4: Decide the entropy of mixing.
Mixing scatters molecules into more arrangements, so disorder rises and $\Delta S_{\text{mix}} > 0$.
Step 5: Combine into Gibbs energy.
Using $\Delta G = \Delta H - T\Delta S$ with $\Delta H_{\text{mix}}=0$: \[ \Delta G_{\text{mix}} = 0 - T\,\Delta S_{\text{mix}} < 0 \] since $\Delta S_{\text{mix}}>0$ and $T>0$.
Step 6: Match the option.
The pairing $\Delta S_{\text{mix}}>0$ together with $\Delta G_{\text{mix}}<0$ is the correct one, consistent with spontaneous mixing. \[ \boxed{\Delta S_{\text{mix}}>0,\ \Delta G_{\text{mix}}<0} \]