krit.club logo

Solutions - Ideal and non-ideal solutions

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

An Ideal Solution is a solution that obeys Raoult's Law at all concentrations and temperatures. For such solutions, the enthalpy of mixing ΔHmix=0\Delta H_{mix} = 0 and the volume of mixing ΔVmix=0\Delta V_{mix} = 0.

•

In an Ideal Solution, the intermolecular attractive forces between the solute and solvent (A−BA-B interactions) are nearly equal to those between the solute-solute (B−BB-B) and solvent-solvent (A−AA-A) molecules.

•

Non-ideal solutions deviate from Raoult's Law. These are classified into solutions showing positive deviation and negative deviation.

•

Positive Deviation occurs when A−BA-B interactions are weaker than A−AA-A and B−BB-B interactions. This leads to ΔHmix>0\Delta H_{mix} > 0 (endothermic) and ΔVmix>0\Delta V_{mix} > 0. The vapor pressure of the solution is higher than predicted by Raoult's Law.

•

Negative Deviation occurs when A−BA-B interactions are stronger than A−AA-A and B−BB-B interactions. This leads to ΔHmix<0\Delta H_{mix} < 0 (exothermic) and ΔVmix<0\Delta V_{mix} < 0. The vapor pressure of the solution is lower than predicted by Raoult's Law.

•

Azeotropes are binary mixtures having the same composition in liquid and vapor phase and boil at a constant temperature. Minimum boiling azeotropes are formed by solutions showing large positive deviation (e.g., 95%95\% Ethanol solution), while Maximum boiling azeotropes are formed by solutions showing large negative deviation (e.g., 68% HNO368\% \text{ } HNO_3).

📐Formulae

PA=PA0χAP_A = P_A^0 \chi_A

PB=PB0χBP_B = P_B^0 \chi_B

Ptotal=PA+PB=PA0χA+PB0χBP_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B

ΔHmix=0 (For Ideal Solutions)\Delta H_{mix} = 0 \text{ (For Ideal Solutions)}

ΔVmix=0 (For Ideal Solutions)\Delta V_{mix} = 0 \text{ (For Ideal Solutions)}

Ptotal>(PA0χA+PB0χB) (For Positive Deviation)P_{total} > (P_A^0 \chi_A + P_B^0 \chi_B) \text{ (For Positive Deviation)}

Ptotal<(PA0χA+PB0χB) (For Negative Deviation)P_{total} < (P_A^0 \chi_A + P_B^0 \chi_B) \text{ (For Negative Deviation)}

💡Examples

Problem 1:

Vapor pressure of pure benzene (C6H6C_6H_6) and toluene (C7H8C_7H_8) at 300 K300 \text{ K} are 50.71 mm Hg50.71 \text{ mm Hg} and 32.06 mm Hg32.06 \text{ mm Hg} respectively. Calculate the total vapor pressure of an ideal solution containing 80 g80 \text{ g} of benzene and 100 g100 \text{ g} of toluene.

Solution:

  1. Moles of Benzene (nBn_B) = 8078≈1.026 mol\frac{80}{78} \approx 1.026 \text{ mol}.
  2. Moles of Toluene (nTn_T) = 10092≈1.087 mol\frac{100}{92} \approx 1.087 \text{ mol}.
  3. Mole fraction of Benzene (χB\chi_B) = 1.0261.026+1.087≈0.485\frac{1.026}{1.026 + 1.087} \approx 0.485.
  4. Mole fraction of Toluene (χT\chi_T) = 1−0.485=0.5151 - 0.485 = 0.515.
  5. Ptotal=PB0χB+PT0χT=(50.71×0.485)+(32.06×0.515)≈24.59+16.51=41.10 mm HgP_{total} = P_B^0 \chi_B + P_T^0 \chi_T = (50.71 \times 0.485) + (32.06 \times 0.515) \approx 24.59 + 16.51 = 41.10 \text{ mm Hg}.

Explanation:

Since the solution is ideal, we apply Raoult's Law directly using the mole fractions and pure component vapor pressures.

Problem 2:

Predict the type of deviation shown by a mixture of Chloroform (CHCl3CHCl_3) and Acetone (CH3COCH3CH_3COCH_3).

Solution:

Negative Deviation.

Explanation:

Chloroform and Acetone molecules form intermolecular hydrogen bonds (C−Cl3H⋯O=C(CH3)2C-Cl_3H \cdots O=C(CH_3)_2). This A−BA-B interaction is stronger than the original A−AA-A and B−BB-B interactions, leading to a decrease in vapor pressure.

Problem 3:

A mixture of Ethanol (C2H5OHC_2H_5OH) and Acetone (CH3COCH3CH_3COCH_3) shows positive deviation. Why?

Solution:

In pure ethanol, molecules are hydrogen-bonded. On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds. This weakens the intermolecular attractions (A−B<A−AA-B < A-A), increasing the tendency of molecules to escape into the vapor phase.

Explanation:

Weakening of intermolecular forces leads to higher vapor pressure than predicted, characteristic of positive deviation.