Question

For an ideal binary liquid mixture:
A $$\mathrm{\Delta }{\text{H}}_{\left(\text{mix}\right)}\text{ = 0; }\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}\text{ < 0}$$
B $$\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}\text{ > 0; }\mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ < 0}$$
C $$\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}\text{ = 0; }\mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ = 0}$$
D $$\mathrm{\Delta }{\text{V}}_{\left(\text{mix}\right)}\text{ = 0; }\mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ = 0}$$

Answer 1

Hint: Consider the various interactions, solute-solute, solute-solvent and solvent-solvent present in a binary solution. Consider the factors that affect enthalpy of mixing, entropy of mixing, Gibbs free energy change and the volume of mixing.

Complete Step by step answer:An ideal solution is the solution that obey Raoult's law for various concentration ranges.
For such solutions, the observed vapour pressure is equal to the vapour pressure calculated from Raoult’s law. In other words, the difference between the observed vapour pressure is equal to the vapour pressure calculated from Raoult’s law is zero.
For an ideal solution, the enthalpy of mixing is zero and the volume of mixing is also zero.
$$\mathrm{\Delta }{\text{H}}_{\left(\text{mix}\right)}\text{ = 0; }\mathrm{\Delta }{\text{V}}_{\left(\text{mix}\right)}\text{ = 0}$$
This is because the intermolecular forces of attraction between two solute particles and those between two solvent particles are equal to those between a solute particle and a solvent particle.
The entropy of mixing for an ideal solution is positive as the randomness or disorder increases upon formation of solution from solute and solvent. The disorder of solute and solvent in a solution is greater than if you have pure solute or pure solvent. In other words, when you mix solute and solvent to form an ideal solution, there is an increase in the entropy.
$$\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}\text{ > 0}$$
Write the relationship between the Gibbs free energy change of mixing, the enthalpy change
of mixing and the entropy change of mixing as shown below.
$$\mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ = }\mathrm{\Delta }{\text{H}}_{\left(\text{mix}\right)}-\text{T}\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}$$
But the enthalpy of mixing is zero. So rewrite the above expression as shown below:
$$\mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ = }0-\text{T}\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}$$
$$\Rightarrow \mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ = }-\text{T}\mathrm{\Delta }{\text{S}}_{\left(\text{mix}\right)}$$
$$\Rightarrow \mathrm{\Delta }{\text{G}}_{\left(\text{mix}\right)}\text{ < 0}$$
Since, entropy of mixing is positive, the Gibbs free energy change is negative as both have opposite signs.

Hence, the correct options are options (A) and (B).

Note: A binary solution is made from two components. The component present in minor quantity is the solute and the component present in excess is solvent. For an ideal binary solution, the intermolecular forces of attraction between two solute particles and those between two solvent particles are equal to those between a solute particle and a solvent particle.