# CE 541 Solutions.

## Presentation on theme: "CE 541 Solutions."— Presentation transcript:

CE 541 Solutions

“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

Types of Solutions Gas in a Liquid Liquid in a Liquid
Solid in a Liquid

Solute Solvent Examples Gas Liquid Carbonated beverages (CO2 in water) Anti-freeze in car radiators (ethyl glycol in watr) Solid Dental fillings (mercury in silver) Sugar in water Solder (tin in lead)

Examples of Solutions Solute Gas Liquid Solid Solvent Oxygen and other gases in nitrogen (air) Water vapor in air (humidity) Iodine sublimates into air Carbon dioxide in water (carbonated water) Ethanol (common alcohol) in water; various hydrocarbons in each other (petroleum) Sucrose (table sugar) in water; sodium chloride (table salt) in water; gold in mercury, forming an amalgam Hydrogen dissolves rather well in metals; platinum has been studied as a storage medium. This effect was used in the cold fusion experiments. Hexane in paraffin wax, mercury in gold. Steel, duralumin, other metal alloys

Factors Affecting Solubility and Rate of Solution
1. Factors that affect the actual solubility of a given solute in a solvent: Properties of solute Properties of solvent Temperature Pressure 2. Factors that affect the rate (how fast) at which a given solute dissolves in a given solvent: Particle size of solute Rate of stirring

Assignment 2: Discuss the two sets of factors.

Example 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 g/l at a total pressure of 1.00 atm and 0 C.

Solution Using Dalton's Law of Partial Pressure Ptotal = Pgas Pwater
Pwater at 0 C is atm Pgas = 1.00 – = 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 / atm) = 1.87 g/l

Saturated, Unsaturated and Supersaturated 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
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.

Concentrations of Solutions
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 3. molarity M = molarity = (moles of solute / liter of solution) 4. Molality m = molality = (moles of solute / kilogram of solvent) 5. Normality N = normality = (equivalents of solute / liter of solution)

Reaction Rates 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  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: 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.

Example 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.

Factors Affecting the Rate of a Chemical Reaction
Nature of Reactants Concentration of Reactants Temperature Catalysts

Assignment 3: Discuss the Four Factors

Chemical Equilibria (Reversible and Irreversible Reactions)
Some reactions are irreversible in practice, meaning that the chemical equilibrium is not established and that the reaction is complete. When can this happen? Products are removed Rate of reverse reaction is very slow (negligible) What products act as a driving force for a reaction to go irreversibly? Gas Precipitate Non-ionized or partially ionized substance, such as water

MgCO3 + 2HCl  MgCl + H2O + CO2
GAS 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

PRECIPITATE 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

WATER NaOH + HBr  NaBr + H2O
The equilibrium is established but the reaction strongly favors the products

Reversible Reactions A + B  C + D

The system is in chemical equilibrium when:
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.

Rate Forward  [A][B] = kf [A][B] Similarly, Rate Reverse = kr [C][D]
For the Law of Mass Action 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. At equilibrium: Rate of forward reaction = Rate of reverse reaction

then 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. Generally, If then,

Example Write the expression of K for the following reactions:

Solution 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.

Le Chatelier’s Principle
“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.”

Concentration When the concentration of one of the substance in a system at equilibrium is increased, the principle predicts that the equilibrium will shift so as to use up partially the added substance. Decreasing the concentration of one substance in a system at equilibrium will cause the equilibrium to shift so as to replenish partially the substance removed. In all cases, the equilibrium constant, K, will remain constant with the concentration of the reactants or products varying.

If we have: increase in concentration of either A or B will shift the equilibrium to the products side increase in concentration of either C or D will shift the equilibrium to the reactants side decrease in concentration of either A or B will shift the equilibrium to the reactants side decrease in concentration of either C or D will shift the equilibrium to the products side

Temperature “If the temperature of a system at equilibrium is changed, the equilibrium will shift so as to change the temperature towards its original value.” A. Exothermic Reactions increase in temperature will shift the equilibrium to reactants side decrease in temperature will shift equilibrium to products side

B. Endothermic Reactions
increase in temperature will shift the equilibrium to the products side decrease in temperature will shift equilibrium to the reactants side The equilibrium constant, K, will change when temperature is changed.

Pressure Increasing the pressure on a system at equilibrium will shift the equilibrium in the direction which will decrease the volume. Decreasing the pressure will have the opposite effect. If no change in volume in going from reactants to products, pressure will have no effect on the equilibrium. The equilibrium constant, K, does not change with change in pressure.

Examples

Concentration Changing the concentration of an ingredient will shift the equilibrium to the side that would reduce that change in concentration. This can be illustrated by the equilibrium of carbon monoxide and hydrogen gas, reacting to form methanol. CO + 2 H2 ⇌ CH3OH

Suppose we were to increase the concentration of CO in the system
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.

Temperature Let us take for example the reaction of nitrogen gas with hydrogen gas. This is a reversible reaction, in which the two gases react to form ammonia: N2 + 3 H2 ⇌ 2 NH3 ΔH = -92kJ This is an exothermic reaction when producing ammonia. If we were to lower the temperature, the equilibrium would shift in such a way as to produce heat. Since this reaction is exothermic to the right, it would favor the production of more ammonia.

Total Pressure Once again, let us refer to the reaction of nitrogen gas with hydrogen gas to form ammonia: N2 + 3 H2 ⇌ 2 NH3 ΔH = -92kJ Note the number of moles of gas on the left hand side, and the number of moles of gas on the right hand side. We know that gases at the same temperature and pressure will occupy the same volume. We can use this fact to predict the change in equilibrium that will occur if we were to change the total pressure.

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.

a. Effect of Adding an Inert Gas
An inert gas (or noble gas) such as helium is one which does not react with other elements or compounds. To add an inert gas into a closed system at equilibrium may or may not result in a shift. For example, consider adding helium to a container with the following reaction: N2 + 3H2 ⇌ 2NH3 The main effect of adding an inert gas to a closed system is that it will increase the total pressure or volume. An inert gas would not be directly involved in the reaction, but could result in a shift.

b. Volume Held Constant 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.

c. Volume Allowed to Increase
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.