Lecture 1 Professor Hicks General Chemistry (CHE132) Definition of rate Rate is how much a quantity changes in a given period of time The speed you drive your car is a rate distance speed time Reaction rate in solution Rate of a chemical reaction of dissolved substances in solution is often measured in terms of how much the molarity of a reactant or product changes in a given period of time molarity reactant or product Rate time For example [I2] rate change I2 in solution time Rates can be positive or negative 1
Reaction rate - gases Rate of a chemical reaction involving gases often expressed in terms of how much the partial pressure of a reactant/product changes in a given period of time partial pressure Rate time Reactions involving gases Sometimes directly measuring the partial pressure of a gas is not convenient Instead volume of gas produced or consumed can be measured to analyze the kinetics of the reaction In lab you measure volume of O 2 gas produced in the reaction: 2H 2 O 2 (aq) 2H 2 O(l) + ½ O 2 (g) volume O Rate O2 time 2 Instantaneous rate The instantaneous rate is the change in concentration at any one particular time Determined by taking the slope of a line tangent at one particular point instantaneous rate is like the speed displayed on your car s speedometer 2
H2 concentration, (M) H2 concentration, (M) 1.200 1.000 Rates are negative when the concentrations are decreasing Concentration vs. Time for H 2 + I 2 --> 2HI Slopes going down from left to right Instantaneous rate H 2 = slope tangent at one point 0.800 0.600 Instantaneous rate H 2 at 40 s = (0-0.80) M / (89-0) s = -0.80 M / 89 s = - 0.0090 M/s [H2], M 0.400 0.200 Instantaneous rate H 2 at 10 s = (0-1.0) M / (60-0) s = -1.0 M / 60 s = -0.017 M/s 0.000 0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 time, (s) Initial rate The Initial Rate is defined as the instantaneous rate at time=0 Initial rates are often measured because the initial concentrations or partial pressures are also known - they are the calculated / prepared concentrations pressures (more on this later) 1.200 Concentration vs. Time for H 2 + I 2 --> 2HI Initial rate H 2 = Slope of tangent at t=0 1.000 0.800 = (0.0 1.0) M / (39 0) s = -1.0/39 M/s = -0.026 M/s 0.600 [H2], M 0.400 0.200 0.000 0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 time, (s) 3
Average rate The Average Rate is the change in concentrations/partial pressure in any particular time period Measured by taking the slope of a line between two points on a concentration or partial pressure vs time curve average rate of a reaction like the average speed in your car distance traveled = 110 miles, time = 2 hours average rate = 55 mph Average rate x time = amount of reactant or product consumed or produced during that time Change in molarity Change in moles The quantity changing in rate descibed in M/s is the molarity (M or moles/liter) Rate time = change in molarity To calculate the number of moles of reactant that disappeared or product that formed use the change in molarity and Molarity x volume = number of moles a) Determine the average rate of reaction of HF between 0.10 and 0.90 seconds. b) How many moles of HF undergo reaction between 0.10 and 0.90 seconds in a 2.5 liters reaction vessel? [HF] 1.1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Average rate = -0.80 M / 0.95 s = -0.84 M/s Average rate HF x time = change in molarity of HF -0.84 M/s x 0.80 s = -0.67 M Change in moles HF is calculated using MV = number of moles -0.67 M x 2.5 L = 1.7 moles HF reacted 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time (s) 1
For the data in the following slide determine: a) The initial rate of reaction of O 2 b) The rate of reaction of O 2 at 30 s c) The average rate of reaction of O 2 between 10 and 90 seconds 5
Rate data for the reaction N 2 + ½ O 2 N 2 O 0.5 0.4 [O 2 ] 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 80 90 100 Time (s)
For the data in the following slide determine: a) The initial rate of reaction of N 2 O b) The rate of reaction of N 2 O at 45 s c) The average rate of reaction of N 2 O between 15 and 85 seconds d) The amount of N 2 O formed between 15 and 85 seconds 1
Rate data for the reaction N 2 + ½ O 2 N 2 O 1.1 1 0.9 0.8 0.7 P N2O (atm) 0.6 0.5 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 80 90 100 110 Time (s)
Converting rates of reaction A + 2B + 3C 4 D Rate of reaction of all substances is proportional to their stoichiometric number Rate of reaction of B is double rate of reaction of A with same sign Rate of reaction of C is triple rate of reaction of A with the same sign Rate of reaction of D is double rate of reaction of B with the opposite sign Converting rates of reaction A + 2B + 3C 4 D Rates can be determined by conversion work using the balanced reaction Say rate of reaction of C is -0.18 M/s -0.18 M s -0.18 moles C liter * s Rate of reaction of C 4 moles D 3 moles C Conversion Factor = 0.24 M/s Rate of reaction of D Adjust the sign to be positive since C was a reactant and D was a product Based upon the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) what would be the rate of reaction of H 2 when the rate of reaction of N 2 was -0.15 M/s? 1
Based upon the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) what would be the rate of reaction of NH 3 when the rate of reaction of H 2 was -0.21 M/s? 2
Lecture 2 Professor Hicks General Chemistry (CHE132) 1 Kinetics: the study of reaction rates Kinetics is the study of the factors that affect the rate of a reaction and the steps by which a reaction happens. 4 factors affect the rate of a reaction: 1) Identity of the reactants 2) Concentration 3) Temperature 4) Presence of catalysts Rate laws Equation to calculate reaction rate from given concentrations/partial pressures Rate law is a summary of everything you could get out of the rate graphs without needing a graph k called the rate constant Rate laws must be determined experimentally Some plausible rate laws Rate = k [A] [B] k [A] 2 k [A] A+B C or in terms of partial pressure Some plausible rate laws Rate = k P A P B k P A 2 k P A 1
Reaction order Rate law Order. Rate = k[a] 2 [B] 3 (third) Rate = k[a][b] 2 (second) Rate = k[a] 1 (first) termolecular bimolecular unimolecular reaction order = sum of exponents in rate law aka the molecularity of the reaction Rate laws Involve concentration/partial pressure because the rate is proportional to the chance of reactants colliding When either or both concentrations get smaller the chance of reactants colliding gets smaller Third order (termolecular) rate laws are rare Determining Rate Laws - Overview rate = k[a] a k =? a =? 1) The reaction is run several times with different initial molarities or partial pressures of A 2) Graphs of [A] vs time plotted and the initial rates determined - initial rates are used b/c to a good approximation we know the molarity of the reactants is the initial molarity 3) Order of each reactant is determined from rate data 4) Substitute orders and rates into rate law and solve for k 6 2
Determining initial rates initial P N2O2 = 1.0 atm rate = 1.0 atm / 23 min = 0.043 atm/min P N2O2 (atm) initial P N2O2 = 0.50 atm rate = 0.50 atm/ 22 min =.023 atm/min initial P N2O2 = 0.25 atm rate = 0.25 atm/ 24 min =.010 atm/min 1.1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 N 2 O 2 (g) 2NO (g) these are rates for separate experiments not different reactants 0 0 5 10 15 20 25 30 35 40 time (min) 7 Determining Reaction Order Trial Initial P N2O2 (atm) Initial Rate (atm/s) 1 1.00 0.0434 2 0.50 0.0227 3 0.25 0.0104 N 2 O 2 (g) 2NO (g) Rate = kp N2O2 a What is value of a? First approach - by inspection P N2O2 is halved, rate is about halved- this means it is first order If the reaction was second order halving P N2O2 would quarter the rate (½) 2 = ¼ If the reaction was third order halving P N2O2 would 1/8 the rate (½) 3 = 1/8 8 Determining Reaction Order (if the data is not so good, or the concentrations are not simple multiples of each other) Trial Initial P N2O2 (atm) Initial Rate (atm/s) 1 1.00 0.0434 2 0.50 0.0227 3 0.25 0.0104 N 2 O 2 (g) 2NO (g) Rate = kp N2O2 a What is value of a? Rate trial 1 = 0.434 = k1.0 a 0.0434 = 1.0a Rate trial 2 = 0.0227 = k0.50 a 0.0227 0.50 a this method may be necessary for analysis of your lab data 1.92 = 2.0 a ln (1.92) = ln(2.0) a 0.652 = a*0.693 a = 0.652/0.693 = 0.94 round to nearest integer 1.0 9 3
Solving for k N 2 O 2 (g) 2NO (g) trial initial P N2O2 Initial Rate (atm/s) 1 1.00 0.0434 2 0.50 0.0227 3 0.25 0.0104 Rate = kp N2O2 use all trials - a sample with trial 3 0.104 = k * 0.25 1 k = 0.0104/0.25 = 0.0416 s -1 The complete rate law is Rate = 0.0416 P N202 Repeat all trials and average the k values 10 Rate constant (k) Given symbol lower case k Has a different value for every reaction Always increases with temperature Rate law rate = k molarities or = k partial pressures 11 Units on k Rate = k [A] a first order rate law M s = (?) M 1 M s = M s For a first order reaction in partial pressure or molarity k has units of 1 s or 1/s or s -1 12 4
Units on k Rate = k[a][b] (M/s) = (?) M*M a second order rate law For a second order reaction expressed in M, k has units of 1/(M*s) or M -1 s -1 13 Units on k rate = k[a][b] 2 (M/s) = (?) M*M 2 (?) M 3 a third order rate law For a third order reaction expressed in M, k has units of 1/(M 2 *s) or M -2 s -1 14 15 5
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Volume Oxygen Formed (ml) The graph below was constructed for the trials described in the following table. The reaction being studied is H 2 O 2 (aq) H 2 O (l) + ½ O 2 (g) Kinetics of Hydrogen Peroxide Decomposition 20 18 16 14 12 10 8 6 4 2 0 0 5 10 15 20 Time (min) Mixture #1 Mixture #2 Mixture #3 18
a) Determine the rates from the graph and enter them in the table below. b) Determine the order with respect to H 2 O 2 and I 2. c) Write a complete rate law with rate in units of ml of O 2 including the numerical value of the rate constant. d) Determine the units of the rate constant in this rate law. Mixture Rate (ml/min) [H 2 O 2 ] [I - ] 1 0.75 0.25 2 0.75 0.50 3 1.50 0.25 19
Lecture 3 Professor Hicks General Chemistry (CHE132) 1 How chemical reactions occur: The collision model In order for a reaction to occur reactant molecules must collide with: 1) Correct orientation, and 2) Enough energy (the Activation Energy or more) Effective collisions: The orientation effect 1
% students with a score % molecules with a speed Effective collisions Collisions that have the correct orientation to allow reaction and have enough energy are called Effective Collisions The higher the frequency of effective collisions, the faster the reaction rate During an effective collision, a temporary, high energy chemical species is formed called a Transition state Boltzmann speed/energy distribution Increasing temperature increases average speed Higher speeds higher energy - kinetic energy = ½ mv 2 Ludwig Boltzmann lower temperature 50% 50% average speed higher temperature 0 300 600 900 1200 speed (m/s) Boltzmann energy distribution like the Hicks grade distribution superior teaching increases students understanding better understanding higher MCAT scores class averages on MCAT other classes medical school MCAT requirement 20% 4% Hicks class 0 10 20 30 40 45 MCAT Score class Hicks all others Charles Hicks % students with required grade 20% 4% 2
% molecules with a speed % molecules with a speed % molecules with a speed Boltzmann speed/energy distribution Higher speeds = higher energy Reactions have energy requirements lower temperature higher temperature 0 300 600 900 1200 speed (m/s) every speed has a kinetic energy KE = ½mv 2 Ludwig Boltzmann At the higher temperature larger % of molecules have enough kinetic energy to react 80% don t have enough energy or greater lower temperature 80% 20% 0 300 600 900 1200 speed (m/s) 45% don t have enough energy Reaction requires this speed (energy 45% higher temperature 55% 20% have enough energy 55% have enough energy 0 300 600 900 1200 speed (m/s) Boltzmann energy distribution Increasing temperature increases rate two ways: Ludwig Boltzmann 1) Increasing the number of molecules with activation energy or greater 2) Increasing the frequency of collisions Activation Energy and the Transition State high point in PE is like the transition state highest potential energy lowest potential energy 3
Activation energy (energy absorbed) reactants transition state (energy released) transition state products heat released overall reaction Effect of temperature on rate Increasing the temperature increases the rate of a reaction by increasing the rate constant k Svante Arrhenius investigated this relationship and showed that: k A e E a RT where T is the temperature in Kelvins R is the gas constant in energy units, 8.314 J/(mol K) A is called the frequency factor E a is the Activation Energy, the energy needed for the molecules to react 11 Arrhenius equation k A e E a RT Svante Arrhenius reflects the energy requirement much less 1 when E a >> RT close to 1 when E a << RT Reflects orientation factor that determines the efficiency of collisions for different reactions RT has units, J/mol it reflects the energy per mole of matter E a /RT compares the activation energy to the available energy 4
Arrhenius equation k A e E a RT Svante Arrhenius Activation energy best determined graphically Rate constant measured at different temperatures Graphs of ln (k) vs 1/T (T must be in Kelvins) ln (k) = ln(a) E a /RT Slope = -E a /R and Intercept = ln (A) 13 Arrhenius equation: Two-point form If you only have two k values at two temperatures this form of the Arrhenius equation can be used: k ln k 2 1 E R a 1 T 2 1 T1 the A factor is eliminated from this equation Temperatures must be in kelvins 14 Catalysts Catalysts are substances that affect the rate of a reaction without being consumed Catalysts work by lowering activation energy Example O 3 (g)+ O (g) 2O 2 (g) ozone Very slow without Cl catalyst Catalyzed by Cl Catalysts appear in rate law because they must get bumped into rate law rate = k[o 3 ][O][Cl] 5
Enzymes Proteins are molecules produced by living organisms (more on this later) Proteins that catalyze biological reactions are called Enzymes Most biological reactions require an enzyme to proceed at a reasonable rate Most enzymes work by binding reactants and orienting them for reaction Enzymatic hydrolysis of sucrose a catalyzed reaction in the digestion of table sugar 6
14.31 Sketch a potential-energy-versus-reaction-progress plot for the following reactions: a) S(s) + O 2 (g) SO 2 (g) ΔH = 296.06 kj/mol b) Cl 2 (g) Cl(g) + Cl(g) ΔH = 242.7 kj/mol 14.33 Variation of the rate constant with temperature for the first-order reaction is given in the following table. Determine the activation energy for the reaction graphically or using the two-point form of the Arrhenius equation. T (K) k (s 1 ) 273 7.87 10 3 298 3.46 10 5 318 4.98 10 6 338 4.87 10 7 Given the same concentrations, the reaction A (g) + B 2 (g) AB 2 (g) at 150 C is 1.02 10 3 times as fast as the same reaction at 50 C. Calculate the energy of activation for this reaction. Assume that the frequency factor is constant. 7
For the reaction the frequency factor A is 8.7 10 12 s 1 and the activation energy is 63 kj/mol. What is the rate constant for the reaction at 275 C? The rate constant of a first-order reaction is 1.23 10 6 s 1 at 350 C. If the activation energy is 104 kj/mol, calculate the temperature at which its rate constant is 8.80 10 4 s 1. The rate constants of some reactions double with every 10-degree rise in temperature. Assume a reaction takes place at 273 K and 283 K. What must the activation energy be for the rate constant to double as described? 8
These data were collected for the reaction between hydrogen and nitric oxide at 700 C: ½ H 2 (aq) + NO 2 (aq) HNO 2 (aq) Experiment [H 2 ] [NO 2 ] Initial Rate (M/s) 1 0.010 0.025 4.8 10 6 2 0.0050 0.025 1.2 10 6 3 0.010 0.0125 2.4 10 6 Which is true? a) A catalyst increases the rate of a reaction b) A catalyst increases the activation energy of a reaction c) A catalyst increases the concentration of the reactants d) A catalyst increases the amount of energy released in a reaction 26 Which ones are false? a) Increasing the temperature decreases the activation energy b) Increasing the temperature always increases the rate of a reaction c) Increasing the temperature may in some cases decrease the rate of a reaction d) According to the Arrhenius equation rate constant always increases as the temperature increases 27 9
Which ones are false? a) At higher temperatures the average speed of gas molecules is larger b) When the average kinetic energy is equal to the activation energy all molecules have enough kinetic energy to undergo reaction c) At higher temperatures the average kinetic energy of gas molecules is larger d) In a gas sample all molecules have more kinetic energy at 300 K than at 200 K 28 10
Lecture 4 Professor Hicks General Chemistry (CHE132) Equilibrium Many reactions appear to stop before limiting reagent is used up They have reached Equilibrium Products and reactants both present Concentrations not changing Reaction quotient (Q) Q monitors progress of a reaction Q = 0 at moment reactants mixed increases as products form amount products Q = amount reactants 1
Reaction quotient for dissolved substances Pb 2+ (aq) +2Cl - (aq) PbCl 2 (s) 1 Q = [Pb 2+ ][Cl - ] 2 [Pb 2+ ] means molarity (moles per liter) of Pb 2+ For dissolved substances: 1) Molarity of dissolved (aq) substances appear in Q 2) Solids or liquids that are not dissolved never appear in Q 3) Molarities are raised to the power of the stoichiometric (balancing) number Reaction quotient (Q) for gases 2C 8 H 18 (g) +25O 2 (g) 16CO 2 (g) + 18H 2 O (g) (P CO2 ) 16 (P H2O ) 18 Q = (P C8H18 ) 2 (P O2 ) 25 For gases the partial pressures appear in Q Pressures must be in atm Solids or liquids do not appear in Q Each is raised to the power of the stoichiometric (balancing) number Reaction quotient for dissolved substances and gases CO 3 2- (aq) + 2H + (aq) H 2 O (l) + CO 2 (g) P CO2 Q = [CO 3 2- ][H + ] 2 Q can be a combination of partial pressures and molarities H 2 O does not appear in Q b/c it is a pure substance (a liquid) P CO2 must be in atm 2
Q, K eq and Equilibrium At the moment reactants are combined before amount products = 0 so Q=0 As reaction occurs Q increases Eventually Q stops changing amount of reactants and products becomes constant This is called equilibrium At equilibrium Q = K eq K eq is a natural constant that is differnet for every reaction 3
K eq K eq is a natural constant that is different for every reaction Larger values of K eq mean the reaction has more products, less reactants present at equilibrium Smaller values of K eq mean the reaction has less products, more reactants present at equilibrium Manipulating K eq Multiplying the amounts in a reaction by a constant raises K eq to that power A 2 (g) + B 2 (g) 2AB (g) K eq = 11 so 2A 2 (g) + 2B 2 (g) 4AB (g) K eq = 11 2 = 121 Manipulating K eq Reversing a reaction inverts K eq A 2 (g) + B 2 (g) 2AB (g) K eq = 11 so 2AB (g) A 2 (g) + B 2 (g) K eq = 1/11 = 0.091 4
15.12 The following diagrams represent the equilibrium state for three different reactions of the type A + X AX (X = B, C, or D) (a) Which reaction has the largest equilibrium constant? (b) Which reaction has the smallest equilibrium constant? Consider the following equilibrium process at 666 C: 2A 2 (g) + B 2 (g) 2A 2 B (g) Analysis shows that there are 1.23 moles of A 2, 2.34 10 5 mole of B 2, and 6.78 moles of A 2 B present in a 2.0-L flask at equilibrium. Calculate the equilibrium constant K c for the reaction. Consider this reaction: If the equilibrium partial pressures of N 2, O 2, and NO are 0.12 atm, 0.23 atm, and 0.34 atm, respectively, at a certain temperature, what is K P? 5
Ammonium carbamate, NH 4 CO 2 NH 2, decomposes as Starting with only the solid, it is found that at a certain temperature the total gas pressure (NH 3 and CO 2 ) is 0.48 atm. Calculate the equilibrium constant K P. 6
Lecture 5 Professor Hicks General Chemistry II (CHE132) Qualitative Interpretation of Equilibrium: Le Chateliers Principle Le Chateliers Principle states: If a system at equilibrium is disturbed it will move in a direction to counteract the disturbance Used to predict the direction a reaction will move in response to changes in temperature, pressure, or amounts of reactants/products The fine print solids and liquids that are not dissolved to a very good approximation do not affect the position of equilibrium. Only changi ng the molarity of dissolved reactants/products or their partial pressures if they are gases, significantly disturbs the equilibrium Le Chateliers Principle (changes in reactants/products) What effect will adding Cl- ions have? Ag + (aq) + Cl - (aq) AgCl (s) Disturbance = Cl- increased Response = system decreases [Cl - ] by moving towards products Le Chateliers Principle - If a system at equilibrium is disturbed it will respond by moving in a direction to counteract the disturbance 1
Le Chateliers Principle (changes in reactants/products) What effect will adding AgCl have? Ag + (aq) + Cl - (aq) AgCl (s) Disturbance = AgCl is added Response = system is approximately unaffected Le Chateliers Principle - If a system at equilibrium is disturbed it will respond by moving in a direction to counteract the disturbance Le Chateliers Principle (changes in reactants/products) What will be the effect of reducing the amount of dissolved O 2? Hb (aq) + O 2 (aq) HbO 2 (aq) hemoglobin = Hb oxygenated hemoglobin = HbO 2 Disturbance = [O 2 ] decreased This occurs when the hemoglobin reaches a cell that has a lower O 2 concentration due to using it in metabolism. The shift in equilibrium is the release of oxygen to the cell Le Chateliers Principle - If a system Response = system responds at equilibrium is disturbed it will respond to raise [O 2 ] by moving towards by moving in a direction to counteract the reactants disturbance Le Chateliers Principle (changes in reactants/products) What will be the effect of increasing the amount of dissolved O 2? Hb (aq) + O 2 (aq) HbO 2 (aq) hemoglobin = Hb oxygenated hemoglobin = HbO 2 Disturbance = O 2 increased Response = system responds to decrease [O 2 ] by moving towards products this occurs when the hemoglobin depleted of O 2 reaches the lungs where O 2 concentration is higher than at the O 2 depleted cells. Le Chateliers Principle - If a system at equilibrium is disturbed it will respond by moving in a direction to counteract the disturbance 2
Hemoglobin, O 2 and Equilibrium lungs [O 2 ] high [O 2 ] hemoglobin = Hb oxygenated hemoglobin = HbO 2 Hb (aq) + O 2 (aq) HbO 2 (aq) high [O 2 ] shifts reaction towards HbO 2 lungs HbO 2 (aq) Hb (aq) + O 2 (aq) cell [O 2 ] low cell low [O 2 ] shifts reaction towards Hb + O 2 Le Chateliers Principle (changes in applied pressure) If pressure appled to the system is increased the system will shift towards the side that has smaller volume to relieve the pressure C(s, graphite) C(s, diamond) larger volume smaller volume Putting graphite under large pressures reduces the volume and causes it to turn into diamond Le Chateliers Principle (changes in applied pressure) If applied pressure is decreased the system will respond by shifting towards the side that has more moles of gas attempting to increase the pressure H 2 CO 3 (aq) CO 2 (g) + H 2 O (l) less moles gas more moles gas pressure is decreased when champagne is uncorked the system responds by trying to increase the pressure. This means shifting the equilibrium towards the side with more moles gas releasing CO 2 gas 3
Le Chateliers Principle (changes in temperature) Raising temperature can be thought of as adding heat in order to remove the heat the reaction will move in the direction that consumes heat NH 4 Cl (s) + heat NH 3 (g) + H + (aq) + Cl- (aq) heating will drive the reaction above towards products heating will drive a reaction towards products if it is endothermic cooling will drive a reaction towards products if it is exothermic 4
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Note this problem is not from the text. It is not part of the assigned HW that I will grade but a similar question could appear on the exam. 6
Lecture 6 Professor Hicks General Chemistry II (CHE132) Acids H + + anion H+ - anion Strong acids HCl HNO 3 H 2 SO 4 H + Cl - H + NO - 3 H + SO 2-4 hydrochloric acid nitric acid sulfuric acid Weak acids HC 2 H 3 O 2 HF H + C 2 H 3 O - 2 H + F - acetic acid Ionic compounds must separate into ions to dissolve Acids are molecular compound because they can dissolve without dissociating into ions hydrofluoric acid Weak acids have a small percentage of molecules separated into H + and an anion, the rest stay together as one particle HF ~ 95% H + and F - 5% Strong acids separate 100% into H+ and anion in water HCl ~ 0 % H + and Cl - ~ 100% molecular compounds Acids are molecular compounds ionic compounds all other molecular compounds dissolve HF ~ 95% (molecules) dissolve acids HCl H + + Cl - ~ 100% dissolve H + + F - ~5% separated ions (separated ions) Dissolved molecules cations (+ ions) anions (- ions) 1
Hydrogen ion H+ has no electrons! Smallest ion or atom moves the fastest to bond other atom must provide both electrons (lone pair) H + curved arrows used to show electron pair movement H N H H H N + H 1) start at lone pair 2) end at electron acceptor H NH 3 (aq) + H + NH 4 + (aq) H carboxylic acids weak acids most common acids in nature R = benzene benzoic acid food preservative H O C=O CH 3 R = CH 3 acetic acid 5% solution = vinegar H O C=O R = C,H R H O - C=O CH 3 + H + curved arrows 1) start at lone pair 2) end at electron acceptor hydronium ion H 3 O + H 3 O + = H 2 O + H + + form H + takes in water reactions of acids in water can be written with H + or H 3 O + 2
Bases Bronsted-Lowry base = proton acceptor proton acceptors must have lone pair Examples of bases - ionic compounds with OH- ion hydroxide ion any hydroxide cation ion 3 lone pairs O-H H H C H methane H no lone pairs not a base + cation OH - both proton acceptors = both bases - molecular compounds with lone pairs lone pair H N H ammonia H Strong Bronsted-Lowry Bases commonly aka Strong Bases - Soluble ionic compounds that contain the hydroxide ion (OH-) any cation hydroxide ion + cation OH - Examples: NaOH, LiOH, KOH, Ca(OH) 2, Ba(OH) 2, Sr(OH) 2, Weak Bronsted-Lowry Bases commonly aka Weak Bases Substances that react with water to produce OH - Base + H 2 O Base-H + + OH - NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Uncharged compounds that contain nitrogen are weak bases Methylamine (CH 3 NH 3 ), amphetamine, cocaine, or heroin Free-base form is the form that has not accepted an H + becoming positively charged yet 3
Strong acid-strong base neutralization reactions acid H+ - anion metathesis reactions ions change partners + cation OH - strong base acid + base H 2 O (l) + salt 2 HCl (aq) + Ba(OH) 2 (aq) 2 H 2 O (l) + BaCl 2 (aq) acid-base reactions sometimes require balancing Conjugate acid/base pairs HC 2 H 3 O 2 (aq) + NaOH (aq) NaC 2 H 3 O 2 (aq) + H 2 O (l) acid base conjugate base conjugate acid Whenever an acid-base reaction occurs: 1) The product that has the formula acid minus H + is called the conjugate base of the acid 2) The product that has the formula base plus H + is called the conjugate acid of the base Water Autoionizes H 2 O (l) H + (aq) + OH - (aq) Q = [H + ][OH - ] or written with hydronium ion 2H 2 O (l) H 3 O + (aq) + OH - (aq) Q = [H 3 O + ][OH - ] K eq for this reaction called K w K w = 10-14 at room temperature neutral solution has [H + ] =10-7 and [OH - ] =10-7 4
Water is Amphoteric (an acid and a base) water acting as a base H 2 O (l) + HCl (aq) H 3 O + (aq) + Cl - (aq) water acting as an acid Conjugate acid of water H 3 O + = hydronium ion H 2 O (l) + NH 3 (aq) OH - (aq) + NH 4 + (aq) Conjugate base of water OH - = hydroxide ion 5
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Titrations Burette has solution of known molarity = titrant Flask has unknown solution = analyte Titrant added until the endpoint is reached At endpoint the titrant has been added in slight excess Excess added reacts with an indicator that changes color Calculations use the approximation endpoint = equivalence point Equivalence point is the point where the number of moles of added titrant exactly, completely, reacts with all moles of analyte titrant analyte Equivalents are a quantity that is sort of like Moles Mass sample / Molar Mass = Moles in sample Mass sample / Equivalent Mass = Equivalents in sample Molar Mass Equivalent Mass = # of H + or OH Example Molar mass for HCl = equivalent mass HCl = 36.45 grams per mole or equivalent Equivalents are sort of like Moles Another example Molar mass for H 2 SO 4 = 98 grams per mol Equivalent mass H 2 SO 4 = 98/2 grams per eq = 49 grams per eq Yet Another example Molar mass for Ca(OH) 2 = 74 grams per mol Equivalent mass H 2 SO 4 = 74/2 grams per eq = 37 grams per eq 7
Normality is sort of like Molarity Normality = # equivalents per liter of solution Example: Calculate the normality of an H 2 SO 4 solution containing 25 grams of H 2 SO 4 in 0.150 L Equivalent mass H 2 SO 4 = 49 grams per eq Equivalents of H 2 SO 4 = 25/49 eq = 0.51 eq Normality = equivalents / liters =.51/ 0.150 = 3.4 N In a titration Normality and Equivalents are VERY useful At the Equivalence Point exact amounts of reactants have been combined so that neither or both are the limiting reactant # of equivalents of each reactant is equal Note: moles of each reactant may not be equal if balancing numbers are not all 1 N 1 V 1 = N 2 V 2 equivalents reactant 1 = equivalents reactant 2 8
Barium 9