Solutions. CE 541. “A solution is a homogeneous throughout and is composed of two or more pure substances. They are weakly bounded mixtures of a solute and a solvent” Solute: is usually the component in less quantity Solvent: is usually the component in greater quantity
“A solution is a homogeneous throughout and is composed of two or more pure substances. They are weakly bounded mixtures of a solute and a solvent”
Solute: is usually the component in less quantity
Solvent: is usually the component in greater quantity
The solute dissolves in the solvent and is considered soluble in the solvent
Aqueous solutions: are solutions where water is used as the solvent
Calculate the solubility in grams per liter of a certain gas in water at a partial pressure of 3.5 atm and 0 C. The solubility is 0.530 g/l at a total pressure of 1.00 atm and 0 C.
Using Dalton's Law of Partial Pressure
Ptotal = Pgas Pwater
Pwater at 0 C is 0.006 atm
Pgas = 1.00 – 0.006 = 0.994 atm
Solubility1 = 0.53 g/l P1 = 0.994
Solubility2 = ? P2 = 3.50 atm
Solubility2 = Solbility1 Pressure Factor
= 0.53 g/l (3.5 atm / 0.994 atm) = 1.87 g/l
1. Saturated Solutions
Are solutions which are in dynamic equilibrium () with undissolved solutes
They can be prepared by adding an excess of solute to a given amount of solvent and allowing sufficient time for a maximum amount of solute to dissolve with excess solute present
In this case:
Rate of dissolution (dissolved solute) = Rate of crystallization (undissolved solute)
2. Unsaturated Solutions two or more pure substances. They are weakly bounded mixtures of a solute and a solvent”
Are solutions in which the concentration of solute is less than that of the Saturated (equilibrium) Solutions, under the same conditions.
3. Supersaturated Solutions
Are solutions in which the concentration of solute is greater than that possible in Saturated (equilibrium) Solutions, under the same conditions.
1. percent by mass
% by mass = (mass of solute / mass of solution) 100
2. parts per million, ppm
ppm = (mass of solute / mass of solution) 1,000,000
M = molarity = (moles of solute / liter of solution)
m = molality = (moles of solute / kilogram of solvent)
N = normality = (equivalents of solute / liter of solution)
The law of mass action states that:
"the rate of a chemical reaction is proportional to the active mass of the reactants"
The active mass is related to relative molar concentration of the reactants in moles per liter for solutions
aA + bB two or more pure substances. They are weakly bounded mixtures of a solute and a solvent” cC + dD
For the General Reaction
The overall rate of reaction is proportional to the concentration of the reactants in moles per liter raised to certain power
Rate [A]x [B]y
[A] = concentration of A in moles / liter
[B] = concentration of B in moles / liter
x and y = whole number, fractional numbers, negative numbers, or zero as determined by experimentation
Then: two or more pure substances. They are weakly bounded mixtures of a solute and a solvent”
Rate = k [A]x [B]y
k = a proportionality constant, called the specific rate constant
Sometimes x and y are equal to the coefficients of the balanced equation; that is a and b. The values of x and y have to be determined experimentally. The value of x and y is the reaction order of each reactant. The sum of x and y is the overall reaction order.
Given the following chemical equation and rate equation, determine the reaction order of each reactant and the overall reaction order.
Cl2 + CHCl3 HCl + CCl4
Rate = [Cl2]0.5[CHCl3]
Chlorine + Chloroform Hydrochloric Acid + Carbon tetra-chloride
The reaction is half order for chlorine and first order for chloroform, with the overall reaction order being 1.5.
A + B C
Rate = k [A]2[B]3
The reaction is second order for A and third order for B, with overall reaction order being 5.
Some reactions are irreversible in practice, meaning that the chemical equilibrium is not established and that the reaction is complete. When can this happen?
What products act as a driving force for a reaction to go irreversibly?
Gas removed as soon as it forms
MgCO3 + 2HCl MgCl + H2O + CO2
If gas remains in contact with the reactants, as in a closed container, then a reversible reaction occurs and an equilibrium is established
The precipitation of a substance acts to remove it from the reaction
AgNO3 + HCl AgCl + HNO3
The reaction is reversible as long as the precipitate is in contact with the solution but equilibrium favors the products
NaOH + HBr NaBr + H2O
The equilibrium is established but the reaction strongly favors the products
The system is in chemical equilibrium when: two or more pure substances. They are weakly bounded mixtures of a solute and a solvent”
Rate at which C and D molecules react to form A and B molecules = Rate at which A and B molecules react to form C and D molecules
For any equilibrium reaction, a constant known as the equilibrium constant (K) can be obtained experimentally if all quantities in the expression can be determined.
For the Law of Mass Action two or more pure substances. They are weakly bounded mixtures of a solute and a solvent”
Rate Forward [A][B] = kf [A][B]
Similarly, Rate Reverse = kr [C][D]
kf and kr are the specific rate constants for the forward and reverse reactions, respectively.
Rate of forward reaction = Rate of reverse reaction
then two or more pure substances. They are weakly bounded mixtures of a solute and a solvent”
since kf and kr are constants, then (kf / kr) is also constant.
K is the equilibrium constant which has a certain value at a given temperature for a given reaction.
Write the expression of K for the following reactions:
in equation (3), since CaCO3 and CaO are solids, they are not considered in the equilibrium expression because their concentrations are constant at a given temperature and hence they are included in the value for the constant K.
“If an equilibrium system is subjected to a change in conditions of Concentration, Temperature, or Pressure, the system will change to a new equilibrium position, where possible, in a direction that will tend to restore the original conditions.”
CO + 2 H2⇌ CH3OH
Suppose we were to increase the concentration of CO in the system. Using Le Chatelier's principle we can predict that the amount of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to fill the “gap” and favor the side where the species was reduced.
N2 + 3 H2⇌ 2 NH3 ΔH = -92kJ
N2 + 3 H2⇌ 2 NH3 ΔH = -92kJ
Suppose we increase total pressure on the system: now, by Le Chatelier's principle the equilibrium would move to decrease the pressure. Noting that 4 moles of gas occupy more volume than 2 moles of gas, we can deduce that the reaction will move towards the products if we were to increase the pressure.
N2 + 3H2⇌ 2NH3
If volume is held constant, the individual concentrations of the above gases do not change. The partial pressures also do not change, even though we have increased the total pressure by adding helium. This means the reaction quotient does not change, so the system is still at equilibrium and no shift occurs.
If the volume is allowed to increase, the concentrations, as well as the partial pressures, all decrease. Because there are more stoichiometric moles on the lefthand side of the equation, the decrease in concentration affects the lefthand side more than the righthand side. Therefore, the reaction would shift to the left until the system is at equilibrium again.