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Wize University Chemistry Textbook > Solutions and Colligative Properties

Solutions & Solubility

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Entropy

  • Spontaneous processes proceed to equilibrium (or completion) without continuous outside intervention
  • Enthalpy alone does not predict spontaneity (some processes are energetically favoured, but not spontaneous)
  • Equilibrium alone cannot determine spontaneity (some processes are favoured based on equilibrium, yet they are non-spontaneous
  • Entropy describes the number of equivalent ways to distribute energy within a system. In other words, entropy is a measure of disorder or randomness in a system resulting from dispersal of matter or energy.

  • Entropy is defined as:
∆Ssurroundings=qsurroundingsTUnits of S=J/(mol K)∆S_{surroundings}= \cfrac{q_{surroundings}}{T} \qquad \qquad \rm Units \ of\ S= J/(mol\ K)

qsurroundings=−qsystemq_{surroundings}=-q_{system}
  • At constant temperature and pressure:
qsystem=∆Hsystemq_{system}=∆H_{system}

∆Ssurroundings=−∆HsystemT∆S_{surroundings}= -\cfrac{∆H_{system}}{T}
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  • The second law of thermodynamics states that whenever a spontaneous event takes place in the universe, the total entropy of the universe increases
∆Suniverse=∆Ssystem+∆Ssurroundings>0∆S_{universe}=∆S_{system}+∆S_{surroundings}>0
  • Factors that affect entropy:
  1. State: solids < liquids < gas

  1. Temperature: entropy increases with temperature
  2. Moles of gas: the higher the amount of gas, the higher the entropy
  3. Volume occupied by a gas: entropy increases with volume
  4. Molecular size: larger molecules have higher entropy


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Example: Qualitatively Estimating Entropy Changes

Determine if the entropy change will be positive or negative for the following reactions:

(NH4)2Cr2O7(s)→Cr2O3(s)+4H2O(l)+N2(g)(NH_4 )_2 Cr_2 O_7 (s) → Cr_2 O_3 (s) + 4H_2 O(l)+N_2 (g)
  • On the left side we only have a solid
  • On the right side we have a solid, a liquid, and a gas
  • Since there is a liquid and gas on the right side, we automatically know that there is more disorder on the right side
  • There are also more molecules on the right than the left
  • The entropy change is positive (entropy increases going to the right)

2H2(g)+O2(g)→2H2O(g)2H_2 (g)+O_2 (g)→2H_2 O(g)
  • Each reactant and product are in the gaseous phase
  • On the left side, there are 3 moles of gas
  • On the right side, there are 2 moles of gas
  • The entropy change is negative since there are less moles of gas on the products side of the reaction

PCl5(g)→PCl3(g)+Cl2(g) PCl_5 (g)→PCl_3 (g)+Cl_2 (g)
  • 1 mole of gas on the left side
  • 2 moles of gas on the right side
  • Since there are more moles of gas on the right side it means that there is more disorder on the right side
  • The entropy change is positive (entropy increases going to the right)

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Colligative Properties

Solution properties which depend only on the concentration of the solute and not the type of solute are known as colligative property:
  1. Vapor pressure
  2. Boiling Point Elevation
  3. Freezing Point Depression
  4. Osmotic Pressure
Note: colligative properties are driven by entropic forces. Ignore intermolecular interactions.
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1. Vapor Pressure Lowering

  • The vapor pressure is the pressure produced by molecules in the gas phase above a liquid


Calculations of Vapor Pressure
  • PTotal= Pvapor1 +Pvapor2

  • Ideal Gas Law - PV=nRTPV=nRT
  • Boyle's Law - At constant temperature, the pressure of an ideal gas is inversely proportional to its volume. i.e. P1V1=P2V2P_1V_1=P_2V_2
  • Henry's Law - describes the solubility of a gas in a liquid, where solubility= k x P, where k = Henry's constant and P = vapor pressure of the gas over the solution
  • Vapor pressure is hindered by the presence of solute molecules. More solute → lower vapor pressure
  • How much does the pressure of the solution change as we add more solute? Raoult's Law - non-volatile solutes decrease the vapor pressure such thatΔPsolution=XsolventPsolvento \Delta P_{solution}=X_{solvent}P_{solvent}^o\ , where Xsolvent = mole fraction = # solvent particles / total # particles
  • Raoult's Law works for IDEAL solutions

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2. Boiling Point Elevation

  • Addition of solute increases the boiling point
  • ΔTb= Kb mi\Delta T_b=\ K_b\ mi where
  • Kb is a constant
  • m = molality = nsolute / kg solvent
  • i = van't Hoff factor
i = moles of particle in solution or moles of solute dissolved (assume complete dissociation)
i = sum of ions
For ionic solute, e.g. NaCl i=2 , CaCl2 → Ca Cl Cl i=3
For non-ionic solute, e.g. Ethylene Glycol, Methanol i=1

3. Freezing Point Depression

  • Addition of solute requires a lower temperature to freeze
ΔTf= Kf mi\Delta T_f=\ K_f\ mi

4. Osmotic Pressure



  • Rate of diffusion of water to the right is faster than rate of diffusion of water to the left because of the presence of a higher concentration of solute particles on the right side
  • The osmotic pressure is proportional to the molarity of the solution (M)
  • The osmotic pressure is the pressure that needs to be applied to prevent increased volume
Π=Mi RT\Pi=Mi\ RT

M = Molarity (mol/L)


Wize Tip
Remember the equations above refer to the change in VP, BP, or FP. To calculate the new VP, BP, or FP after the addition of solute to a solution, we have to ADD or SUBTRACT the change:

Freezing point goes down with the addition of solute, so we subtract the change in T:
normal freezing point - ΔTf\Delta T_f = new freezing point

Boiling point goes up with the addition of solute, so we add the change in T:
normal boiling point + ΔTb\Delta T_b= new boiling point

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An aqueous solution contains 25.0g of an ionic compound (i=2) dissolved in 250.0 g of water. The freezing point of the solution is -2.50 ̊C. Kf for water is 1.86 ̊C/m and normal freezing point of water is 0.00 ̊C. What is the molar mass of the substance?

Aim: Find the Molar mass after finding the molality: # of mols of solute/mass of solvent kg

Kf of water= 1.86 oC/m
Freezing point of solution= -2.50oC
Freezing point of water= 0.00oC
mass of solute= 25.0 g (ionic)
i=2
V of water= 250.0g= 0.2500 kg


ΔTf= −Kf mi\Delta T_f=\ -K_f\ mi
−2.50oC= −1.86oCm×m×2-2.50^oC=\ -1.86^o\frac{C}{m}\times m\times2
m = -2.50/-3.72 = 0.672 m

molality = 0.672 m = n of solute / mass of solvent (kg)
n of solute= molality x mass of solvent= 0.672 mol/kg x 0.2500 kg= 0.168 mol

m = nM
M = m/n = 25.0g / 0.168mol = 148.8 g/mol



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A sample of N2 gas was collected over water. The temperature of the vessel was 300.0 K and the volume of the sample was 18.0 L. The vapour pressure of water at 300.0 K is 3.53 kPa and the total pressure of the sample was 122.00 kPa. What is the mass of the N2 in this sample?

Estimate the molecular weight of a biological macromolecule, if a 0.2 g sample dissolved in 100g of water has an osmotic pressure of 80 torr at 25oC.