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Solutions - Ideal and Non-ideal Solutions

Grade 12CBSEChemistry

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

πŸ”‘Concepts

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Ideal Solutions: These are solutions that obey Raoult's law over the entire range of concentration. For an ideal solution formed by components AA and BB, the intermolecular attractive forces between Aβˆ’AA-A and Bβˆ’BB-B are nearly equal to those between Aβˆ’BA-B.

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Thermodynamics of Ideal Solutions: For these solutions, the enthalpy of mixing is zero, i.e., Ξ”mixH=0\Delta_{mix}H = 0, and the volume of mixing is zero, i.e., Ξ”mixV=0\Delta_{mix}V = 0.

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Non-Ideal Solutions (Positive Deviation): Occurs when Aβˆ’BA-B interactions are weaker than Aβˆ’AA-A or Bβˆ’BB-B interactions. In this case, PA>PA0Ο‡AP_A > P_A^0 \chi_A and PB>PB0Ο‡BP_B > P_B^0 \chi_B. These solutions show Ξ”mixH>0\Delta_{mix}H > 0 and Ξ”mixV>0\Delta_{mix}V > 0. Example: Ethanol and Acetone.

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Non-Ideal Solutions (Negative Deviation): Occurs when Aβˆ’BA-B interactions are stronger than Aβˆ’AA-A or Bβˆ’BB-B interactions. In this case, PA<PA0Ο‡AP_A < P_A^0 \chi_A and PB<PB0Ο‡BP_B < P_B^0 \chi_B. These solutions show Ξ”mixH<0\Delta_{mix}H < 0 and Ξ”mixV<0\Delta_{mix}V < 0. Example: Chloroform and Acetone.

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Azeotropes: These are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. Minimum boiling azeotropes are formed by solutions showing large positive deviation, while maximum boiling azeotropes are formed by solutions showing large negative deviation.

πŸ“Formulae

PA=PA0Ο‡AP_A = P_A^0 \chi_A

Ptotal=PA+PB=PA0Ο‡A+PB0Ο‡BP_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B

Ptotal=PA0+(PB0βˆ’PA0)Ο‡BP_{total} = P_A^0 + (P_B^0 - P_A^0) \chi_B

yA=PAPtotalΒ (MoleΒ fractionΒ inΒ vapourΒ phase)y_A = \frac{P_A}{P_{total}} \text{ (Mole fraction in vapour phase)}

Ξ”mixV=0Β (ForΒ IdealΒ Solutions)\Delta_{mix}V = 0 \text{ (For Ideal Solutions)}

Ξ”mixH=0Β (ForΒ IdealΒ Solutions)\Delta_{mix}H = 0 \text{ (For Ideal Solutions)}

πŸ’‘Examples

Problem 1:

Vapour pressure of chloroform (CHCl3CHCl_3) and dichloromethane (CH2Cl2CH_2Cl_2) at 298Β K298 \text{ K} are 200Β mmΒ Hg200 \text{ mm Hg} and 415Β mmΒ Hg415 \text{ mm Hg} respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5Β g25.5 \text{ g} of CHCl3CHCl_3 and 40Β g40 \text{ g} of CH2Cl2CH_2Cl_2 at 298Β K298 \text{ K}.

Solution:

  1. Molar mass of CHCl3=119.5Β g/molCHCl_3 = 119.5 \text{ g/mol} and CH2Cl2=85Β g/molCH_2Cl_2 = 85 \text{ g/mol}.
  2. Moles of CHCl3(nA)=25.5119.5=0.213Β molCHCl_3 (n_A) = \frac{25.5}{119.5} = 0.213 \text{ mol}.
  3. Moles of CH2Cl2(nB)=4085=0.47Β molCH_2Cl_2 (n_B) = \frac{40}{85} = 0.47 \text{ mol}.
  4. Mole fraction Ο‡B=0.470.213+0.47=0.688\chi_B = \frac{0.47}{0.213 + 0.47} = 0.688.
  5. Using Ptotal=PA0+(PB0βˆ’PA0)Ο‡BP_{total} = P_A^0 + (P_B^0 - P_A^0) \chi_B: Ptotal=200+(415βˆ’200)Γ—0.688=200+215Γ—0.688=347.92Β mmΒ HgP_{total} = 200 + (415 - 200) \times 0.688 = 200 + 215 \times 0.688 = 347.92 \text{ mm Hg}.

Explanation:

The total vapour pressure is calculated using Raoult's Law for a binary mixture of volatile liquids, assuming ideal behavior. The mole fraction of the more volatile component (CH2Cl2CH_2Cl_2) determines the increase in total pressure relative to pure CHCl3CHCl_3.