General Chemistry for Pharmaceutical Sciences Pharm-101 PDF

 General Chemistry for Pharmaceutical Sciences PHARM-101 Presented by Dr. Azza H. Rageh Associate Professor of Pharmaceutical Analytical Chemistry  D- Stoichiometry, Solution Concentration and Chemical Reactions 1- Reaction Stoichiometry, Solution Concentration and Ty...

 General Chemistry for Pharmaceutical Sciences PHARM-101 Presented by Dr. Azza H. Rageh Associate Professor of Pharmaceutical Analytical Chemistry  D- Stoichiometry, Solution Concentration and Chemical Reactions 1- Reaction Stoichiometry, Solution Concentration and Types of Aqueous Solutions What is meant by Stoichiometry  Stoichiometry: calculations of the quantities of reactants and products in a chemical reaction.  Stoichiometry allows us to predict the amounts of products that will form in a chemical reaction based on the amount of the reactants.  Stoichiometry also allows us to determine the amount of reactants necessary to form a given amount of product. 4 Reaction Stoichiometry  The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction:  How Much CO2 is Produced?  Example: 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 2 molecules of C8H18 react with 25 molecules of O2 to form 16 molecules of CO2 and 18 molecules of H2O. Or: 2 moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O. 2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O 5 Reaction Stoichiometry From the balanced equation of the combustion of octane: 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) we can write the following stoichiometric ratio: 2 moles C8H18(l) : 16 moles CO2 (This ratio is called: The Conversion Factor) Suppose that we burn 22 moles of C8H18: the amount of CO2 produced can be calculated using the conversion factor, as follows: 22 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶8𝐻18 × 16 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2 = 176 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2 2 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶8𝐻18 6 Concentration of Solutions What Is a “Solution”? Solution: A homogenous mixture of two or more substances: - Solvent: material present in largest amount. - Solute: all other materials present. - Example: Consider sugar dissolved in water: - Water is the solvent. - Sugar is the solute. Concentrated solution has a relatively large proportion of solute to solvent. Dilute solution has a relatively smaller proportion of solute to solvent. 7 Concentration of Solutions  Concentration: is the amount of solute present in the solution. Concentration units: Name Units Symbol % Weight Gram solute/100 g solution %, w/w % Volume Milliliter solute/100 mL solution %, v/v % Weight per volume Gram solute/100 mL solution %, w/v Parts per million Gram solute/106 g solution ppm Parts per billion Gram solute/109 g solution ppb Gram molecular weight of solute Molarity M (Moles)/liter solution Gram formula weight of Formality F solute/liter solution Gram equivalent weight of solute Normality N /liter solution Molality Moles solute/1000 g solution m 8 Concentration of Solutions: Molarity  Molarity: is a method to express the concentration. It shows the relationship between the moles of solute and liters of solution.  is the No of gram molecular weight (moles) of solute per one litre of solution. No. of moles = weight (g)/molecular weight So, Molarity (M) = weight (g)/molecular weight × volume (L)  Unit of molarity (M) = moles of solute / liter of solution M = mol/L = mol.L-1 = molar 9 The States of Matter Concentration of Solutions: Molarity Example 1: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution Given: 25.5 g KBr, 1.75 L solution Find: molarity, M Plan: g KBr mol KBr M L sol’n Relationships: 1 mol KBr = 119.00 g, M = moles/L Solution: Check: because most solutions are between 0 and 18 M, the answer makes sense 10 Concentration of Solutions: Molarity Example 2: How many litres of 0.125 M NaOH solution would contain 0.255 mol NaOH? Given: 0.125 M NaOH, 0.255 mol NaOH Find: liters, L Plan: mol NaOH L sol’n Relationships: 0.125 mol NaOH = 1 L solution Solution: Check: because each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than112 L  Concentration of Solutions: Molarity Example 3: Preparing 1 L of a 1 M NaCl Solution 12 Types of aqueous solution and Solubility Consider two familiar aqueous solutions: salt water and sugar water: – Salt water is a homogeneous mixture of NaCl and H2O. – Sugar water is a homogeneous mixture of C12H22O11 and H2O. As you stir either of these two substances into the water, it seems to disappear. – How do solids such as salt and sugar dissolve in water? 13 What Happens When a Solute Dissolves? There are attractive forces between the solute particles holding them together. There are also attractive forces between the solvent molecules.  When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules.  If the attractions between solute and solvent are strong enough, the solute will dissolve. 14 Dissolving of Sodium Chloride in Water  Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal.  Compounds such as salt that dissociate into ions when dissolved in water are called electrolytes, and the resulting solutions are able to conduct electricity. 15 Dissolving of Sugar in Water  Table sugar (sucrose, C12H22O11) molecules homogeneously mixed with water molecules (H2O).  Compounds such as sugar that don’t dissociate into ions when dissolved in water are called nonelectrolytes, and the resulting solutions do not conduct electricity. 16 Electrolytes and Nonelectrolytes Substances that dissolve in water to form solutions that conduct electricity are called Electrolytes  Solution of salt (an electrolyte)  Solution of sugar (a nonelectrolyte) 17 Electrolytes and Nonelectrolytes  Strong Electrolytes:  substances that completely ionize when dissolve in water.  They can conduct electrical current strongly.  Important Examples: Soluble ionic salts (e.g. NaCl & MgBr2 …), strong acids (e.g. HCl & HNO3) and strong bases (e.g. NaOH & Mg(OH)2).  Weak Electrolytes:  Include substances that partially ionize when dissolve in water.  They can conduct electrical current weakly.  Important Examples: weak acids (e.g. HF & CH3COOH) and weak bases (e.g. NH4OH).  Nonelectrolytes:  Include substances that do not ionize when dissolve in water.  They don’t conduct electrical current.  Important Examples: molecular substances (e.g. sugar & alcohol). 18 Electrolytes and Nonelectrolytes: A Summary Complete Ionizing in Partial Ionizing in water No Ionizing in water (no water (full dissociation) (partial dissociation) dissociation) Examples: ionic salts, Examples: molecular Examples: weak acids strong acids & strong (covalent) compounds as bases & weak bases sugars & alcohols 19 Electrolytes Solutions in Pharmacy Electrolyte Use Magnesium sulfate A drug used to treat convulsions during pregnancy, nephritis in children, magnesium deficiency, and tetany. + Ammonium chloride Expectorant in cough syrups. The ammonium ion (NH4 ) in the body plays an important role in the maintenance of acid-base balance. Sodium chloride An ingredient found in a variety of nutritional products as a source of electrolytes and water. Sodium acetate A compound used for electrolyte replenishment and total parenteral nutrition (TPN) therapy. Magnesium chloride An ionic compound and source of magnesium used for electrolyte replenishment and conditions associated with magnesium deficiencies. Potassium chloride A potassium salt used to treat hypokalemia. Calcium chloride An ionic compound used for the treatment of hypocalcemia and hyperkalemia, and as an antidote to magnesium intoxication due to overdosage of magnesium sulfate. 20 Assessment All of the following compounds are soluble in water, indicate which of them is expected to produce strong, weak or non- electrolyte solution? 1. CsCl(aq) 2. CH3OH(aq) 3. Ca(NO2)2(aq) 4. C6H12O6(aq) 5.Acetic acid, vinegar, CH3COOH(aq) (weak acid) 6. HCl(aq) (strong acid) 7. NaOH(aq) (strong base) 8. KOH(aq) (strong base) 9. HF(aq) (weak acid) 10.NH4OH(aq) (weak base) 21 What is “Chemical Kinetics” It is the branch of physical chemistry that deals with the speeds or rates of the chemical reactions. e.g. How rapidly food spoils?… How rapidly a drug is decomposed?... What determines the rate at which steel rusts?. 44 Kinetics of Chemical Reactions  All chemical reactions require time for their completion; some reactions proceed rapidly whereas others proceed slowly.  The rate of chemical change is expressed in terms of the quantities of substances, which are transformed in a given interval of time; this is called the reaction rate.  A study of many reactions reveals that some substances have a greater tendency to react than others (e.g. magnesium reacts more rapidly with dilute acids than iron under the same conditions). 45 Speed or Rate of Reaction Definition: It is the change in the concentration of a reactant or a product with time (M/s). Consider a reaction: A B The concentration of the reactant A decreases  with time, while that of the product B increases  with time. 46 Factors affecting Reaction Rate: Temperature and Catalyst 1. Temperature:  Changes in temperature produce corresponding changes in the speed of the moving molecules or ions.  This causes changes in the number of collisions per unit of time, and consequently changes in reaction rate.  In general, an increase in temperature of 10°C approximately doubles the reaction rate. 2. Catalyst:  The reaction rate may be altered by use of a catalyst (It is a substance that alters rate of reaction without including in the reaction). 47 Factors affecting Reaction Rate: Surface Area and Physical Nature 3. Surface Area and Physical Nature:  When two reactants form a heterogeneous mixture as a solid and a liquid, a solid and a gas, a liquid and a gas, or two immiscible liquids, the reaction takes place at the boundary between the two substances.  By increasing the surface of one of the reactants, the reaction rate is increased. For example, finely divided zinc reacts much more rapidly with dilute hydrochloric acid than does a single large piece of zinc.  When two solids (e.g. sodium chloride and silver nitrate) are mixed, they react very slowly; but they react very rapidly when a solvent (e.g. water), which dissolves both, is added. 48 Factors affecting Reaction Rate: Concentration of Reactants 4. Concentration of Reactants:  According to law of mass action; when the concentration of all the reactants increases, more molecules or ions interact to form new compounds, and the rate of reaction increases. Rate Law: The rate of a reaction is directly proportional to the product of molar concentrations of reactants, each concentration being raised to an appropriate power. Equation: aA + bB cC + dD m n Rate  [A] [B] m n  Reaction Rate = k [A] [B] m = order of the reaction with respect to A. n = order of the reaction with respect to B. m+n = overall order of the reaction. k = the rate constant for the reaction, its value does not change with change in concentration, but changes with temperature. (/X question) 49 Order of the Reaction m n Rate = k [A] [B] Definition: It is the sum of the powers of concentrations in the rate law. The exponents n and m are called “reaction orders with respect to reactant”, and their sum (m+n) is “the overall reaction order”. It describes the manner by which the rate of reaction is affected by the concentration of reactants. N.B. The order of the reaction is not related to the stoichiometric coefficient in the balanced equation. i.e. m,n ≠ a,b 50 Order of the Reaction Example: For the reaction NH4+ + NO2- N2 + 2H2O Experimentally, the rate law is: Rate = k [NH4+] [NO2-] - The concentration of ammonium ions is raised to the power 1; thus the reaction is said to be first order with respect to ammonium ion or first order in ammonium ion. - Similarly, it is also first order in nitrite ion. - The overall order of the reaction is 2 (second order). 51 Order of the Reaction Notes:  The rate law for any chemical reaction must be determined experimentally.  Type of order: Zero, 1st, 2nd, 3rd ….. etc, and may be also a fraction.  If the reaction is ZERO order in a particular reactant, changing its concentration will have no effect on the reaction rate (i.e. the rate is independent of the concentration of the reactant).  If the reaction is FIRST order in a particular reactant, changing its concentration will produce proportional changes in the reaction rate (i.e. doubling the concentration will double the rate, and so on...).  If the reaction is SECOND order in a particular reactant, doubling its concentration causes an increase in the reaction rate by a factor of 4 and tripling its concentration causes an increase in the rate by a factor of 9. 52 Order of the Reaction: A Summary The following table summarizes the relation between concentration change and the change in the reaction rate: Order ∆ [Concentration] ∆ Rate The same (no Zero Double change) 1st Double Double Double x4 2nd Triple x9 53 Order of the Reaction: Practice Example: The initial rate of the following reaction: A + B C was measured for several different starting concentrations of A and B with the results given below: Experiment No. [A], M [B], M Initial Rate, (M/s) 1. 0.1 0.1 4 x 10-5 2. 0.1 0.2 4 x 10-5 3. 0.2 0.1 16 x 10-5 Using these data determine: 1. The rate law for the reaction and its order. 2. The magnitude of the reaction rate constant. 3. The rate of the reaction when [A] = 0.05 M and [B] = 0.3 M. Solution: (1) Rate = k [A]m [B]n From exp. 1 & 3: m = 2, from exp. 1 & 2: n = 0  Rate = k [A]2 [B]0 = k [A]2 so the order of reaction = 2 + 0 = 2  The reaction is second order (2) k = Rate/[A]2 = 4 x 10-5/(0.1)2 = 4 x 10-3 M-1.s-1 (3) Rate = k [A]2 = 4 x 10-3 [0.05]2 = 1 x 10-5 M/s 54  E- Chemical Equilibrium in Aqueous Solutions 1- Chemical Equilibrium and Le-Châtelier’s principle 55 Chemical Equilibrium: An Introduction Dynamic Equilibrium is the condition in which the rate of the forward reaction equals the rate of the reverse reaction: Rateforward Ratereverse  As H2 and I2 react, their concentrations decrease, which in turn decreases the rate of the forward reaction.  At the same time, the concentration of HI increases, increasing the rate of the reverse reaction.  At a certain point, the rate of the reverse reaction, Rr (which has been increasing) equals the rate of the forward reaction, Rf (which has been decreasing). At that point, dynamic equilibrium is reached. 56 The Concept of Dynamic Equilibrium Chemical Equilibrium  Equal Concentrations! At equilibrium, the rates of the forward (Rf) and reverse (Rr) reactions are equal:  Rf = Rr: DOES NOT mean that the concentrations of reactants and products are equal in value! 57 Law of Mass Action From equations (1) & (2) Rf = Kf [A][B] (1) or Rf  [A][B] (*) Rr = Kr [C][D] (2) or Rr  [C][D] (**) From (*) and (**) we can deduce that: “The rate of a chemical reaction is directly proportional to the molar concentrations of the reactants at constant temperature” This is the law of mass action The Equilibrium Constant: Kc or Keq Equilibrium Constant (Kc or Keq): expresses the relationship between the amounts of products and reactants of a reaction at equilibrium. Expressing Kc or Keq: Rf Rr Where A and B are reactants, C and D are products, and a, b, c, and d are the coefficients in the balanced chemical equation: At equilibrium, Rf = Rr Kf [A][B] = Kr [C][D] Notes: 1. Each square bracket […] means the “concentration”: Example: [A] = molar concentration of “A” (mol/L). 2. Equilibrium constant (Keq) has no units. 59 Expressing Equilibrium Constants (Keq) for Chemical Reactions: Example 1 Example 1: write the equilibrium expression for the following reaction: Notice that the stoichiometric coefficients in the chemical equation become exponents in the equilibrium expression.  Therefore, all equations must be balanced first! 60 Expressing Equilibrium Constants (Keq) for Chemical Reactions: Example 2 Example 2: Express the equilibrium constant for the following chemical equation: Answer: 61 Expressing Equilibrium Constants (Keq) for Chemical Reactions: Practice Write the equilibrium constant expression Keq for the following balanced chemical equations Answer: Keq = ……….. b) Answer: Keq = ……….. c) Answer: Keq = ……….. 62 Characteristics of Equilibrium Constants (Keq) 1. It has definite value for every chemical reaction at particular temperature. 2. Independent on initial concentrations of reactants. 3. It changes with change in the temperature. 4. Depends on nature of reaction. 5. Not affected by catalyst. as it always affect both forward and backward reaction rates to the same extent. 6. Independent of the change of pressure, volume and concentrations of reactants and products. Summarizing The Significance of (Keq) Values  When Keq >> 1: Forward reaction is favored (the reaction moves right).  The position of equilibrium favors products.  When Keq