CHEM 26.1 MIDTERM EXAM REVIEW Chemical Kinetics Rates of chemical reactions are affected by several factors, most notably: 1. Conc Concen entra tratio tion n of react reactan ants ts 2. temper temperatu ature re at whic which h the the reacti reaction on occu occurs rs 3. prese presenc ncee of of a catal catalys ystt 4. surface surface area area of slid slid or liqui liquid d reactan reactants ts or cataly catalysts sts The Dependence of Rate on Concentration *Reaction rates diminish as the concentrations of reactants diminish. Conversely, rates generally increase when reactant concentrations are increased. * For example: A + B C + D Rate = k[A][B], rate law shows how the rate depends on the concentrations of reactants The constant k in k in the rate law is called the rate constant. Reaction Order * The rate laws for most reactions have the general form: Rate = k[reactant]m[reactant 2]n… The exponents m and n are called reaction orders, and their sum is the overall reaction order * The following are some further examples of rate laws: 2N2O5(g) 4NO2(g) + O2(g) Rate = k[N2O5] CHCl 3(g) + Cl2(g) CCl4(g) + HCl(g) Rate = k[CHCl 3][Cl2]1/2 H2(g) + I2(g) 2HI(g) Rate = k[H2][I2] Notice that the reaction orders do not necessarily correspond to the coefficients in the balanced chemical equation. The values of these exponents are determined experimentally. In most rate laws, reaction orders are 0, 1, or 2. Sample Exercise Run 1 2 3 4
[NO] (M) 0.10 0.25 0.10 0.35
[Br2] (M) 0.20 0.20 0.50 0.50
Rate = 1/(t in sec.) 24 150 60 735
* Determine the rate law Order with respect to [NO]
Order with respect to [Br2 ]
Rate (for run1) = k[NO] xrun1[Br 2]yrun1 Rate (for run1) = k[NO] xrun1[Br 2]yrun1 Rate (for run2) k[NO]xrun2[Br 2]yrun2 Rate (for run3) k[NO]xrun3 [Br 2]yrun3 24_ = k(0.10) x(0.20)y 150 = k(0.25) x(0.20)y 0.16 = (0.10/0.25) x(0.20/0.20)y 0.16 = (0.4) x
24_ = k(0.10) x(0.20)y 60 = k(0.10)x(0.50)y 0.40 = (0.10/0.10) x(0.20/0.50)y 0.40 = (0.40) y
log (0.16) = (x) log(0.4) -0.795880017 = (x)(-0.397940008)
log (0.40) = (y) log(0.40) -0.397940008 = (x)(-0.397940008)
x = 1.999 or 2 y=1 nd => the reaction is first order with respect to [Br 2] and 2 order with respect to [NO]
Rate constant, k Rate (for run 1) = k[NO] 2[Br 2]1 24 = k(0.10) 2(0.20)1 24 = k(0.002) k = 12,000 Rate law *rate = k[NO]2[Br 2]1, where k is equal to 12,000 *Over-all order of the reaction x + y = over-all order 2+1=3 Temperature and Rate
the rates of most chemical reactions increase as the temperature rises whether it is an exothermic or endothermic reaction; the faster rate at higher temperature is due to an increase in the rate constant with increasing temperature Activation Energy, Ea is the minimum energy required to initiate a chemical reaction -Ea/RT Arrhemius Equation, k = Ae In this equation k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/molK), and T is the absolute temperature (temperature in Kelvin). The term A is constant or nearly so, as temperature is varied. Called the frequency factor , A is related to the frequency of collisions and the probability that the collisions are favourably oriented for reaction. Taking the natural log of both sides of the above equation, ln k = -Ea/(RT) + lnA this equation has the form of a straight line; it predicts that a graph of ln k versus 1/T will be aline with a slope equal to –Ea/R and a y-intercept equal to ln A •
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Increasing the temperature of an elementary step reaction would cause an increase the rate of the reaction but this increase would be more pronounced in a reaction that initially has a lower temperature than in a reaction that already initially has a high temperature. As one increase the temperature the rate of the reaction increases. As a rough approximation, for many reactions happening at around room temperature, the rate of reaction doubles for every 10ºC rise in temperature. But you have to be careful not to take this too literally. It doesn’t apply to all reactions. Even where it is approximately true, it may be that the rate doubles every 9ºC or 11ºC or whatever. The number of degrees needed to double the rate will also change gradually as the temperature increases.
Sample Exercise
The rate of the reaction CH3COOC2H5(aq) + OH-(aq) CH3COO-(aq) + C2H5OH(aq) was measured at several temperatures, and the following data were collected: Temperature (K) 1/T k = 1/(t in sec.) ln k 288.15 0.00347 0.0521 -2.95459 298.15 0.003354 0.101 -2.292635 308.15 0.0032452 0.184 -1.69282 318.15 0.0031432 0.332 -1.10262 Using these data, construct a graph of ln k versus 1/T. Using your graph, determine the value of Ea and A. Solution: Equation of the line: y = (-5,651.184676)x + 16.65576185 ; r 2 = 0.9999 Value of Ea Slope = -Ea/R -5,651.184676 = -Ea/(8.314) Ea = 679.719
Value of A y-intercept = ln A 16.65576185 = ln A A = 17, 120,067.13
Rate Mechanisms The process by which a reaction occurs is called the reaction mechanism . *Elementary Steps e.g. NO(g) + O3(g) NO2(g) + O2(g) The given reaction occur in a single event or step and is called elementary step (or elementary process) *Multi-step Mechanisms A multi-step mechanism consists of a sequence of elementary steps. e.g. NO2(g) + CO(g) NO(g) + CO2(g) the given reaction actually occurs in a series of elementary steps, namely: NO2(g) + NO2(g) NO3(g) + NO(g) step 1 NO3(g) + CO(g) NO2(g) + CO2(g) step 2
The elementary steps in a multi-step mechanism must always add to give the chemical equation of the overall process. Because NO3 is neither a reactant nor a product in the over-all reaction – it is formed in one elementary step and consumed in the next – it is called an intermediate. Multi-step mechanisms involve one or more intermediates. Rate Laws of Elementary Steps The rate law of any elementary step is based directly on its molecularity. For example: Elementary Step Rate Law Rate = k[A] A products Rate = k[A]2 A + A products Rate = k[A][B] A + B products Rate = k[A]3 A + A + A products
2A + B products A + B + C products
Rate = k[A]2[B] Rate =k[A][B][C]
Rate Laws of Multi-step Mechanisms The over-all rate of a rate of a reaction cannot exceed the rate of the slowest elementary step of its mechanism. Because the slow step limits the over-all reaction rate, it is called the rate-determining step . e.g. Given: Over-all reaction NO 2(g) + CO(g) NO(g) + CO2(g) Step 1: NO2(g) + NO2(g) NO3(g) + NO(g) (slow) Step 2: NO3(g) + CO(g) NO2(g) + CO2(g) (fast) The rate law for the reaction is therefore: rate = k[NO2][NO2] or rate = k[NO 2]2 •
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In the experiment HCl reacts with thiosulfate to produce S. The time from mixing the reactants to the appearance of white precipitate is the time for a fixed amount of S to be formed. 1/t (where t = time in sec.) was used for rate constant since the reaction is 1st order. The reaction of thiosulfate with HCl was found to be 1 st order with respect to thiosulfate and zeroorder with respect to H+ The correct mechanism for the reaction of thiosulfate with HCl: 1st S2O32- SO32- + S (slow) nd 2+ 2 SO3 + H HSO3 (fast) 3rd HSO3- + H+ H2O + SO2 (fast) Note that the slow step for the mechanism of the reaction between thiosulfate and HCl is: S2O32- SO32- + S, that is why the rate law derived from the experiment is rate = k[S 2O32-] since for a multistep reaction the rate law is dependent only on the slow step.
Soda Ash Experiment •
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Consider a mixture of NaOH and Na 2CO3: Na2CO3 reaction with HCl takes place in two steps (1) HCl + Na2CO3 NaHCO3 (2) HCl + NaHCO3 NaCl + CO2 + H2O NaOH reaction with HCl takes place in a single step (3) HCl + NaOH NaCl + H2O (titration curve will have a total of three breaks) Phenolphthalein endpoint is reached once reactions number 1 and number are completed. While the end of reaction number is indicated by the methyl orange. (V phenolphthalein > VMO) Consider a mixture of Na 2CO3 and NaHCO3 Na2CO3 reaction with HCl (4) HCl + Na2CO3 NaHCO3 (5) HCl + NaHCO3 NaCl + CO2 + H2O NaHCO3 reaction with HCl (6) HCl + NaHCO3 NaCl + CO2 + H2O (The titration curve here will have a total of two breaks) Phenolphthalein endpoint is reached once reaction number 4 is completed
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while the methyl orange endpoint is reached when reactions number 5 and 6 are completed (V phenolphthalei < VMO) Consider a soda ash sample containing NaOH only NaOH reaction with HCl takes place in a single step (7) HCl + NaOH NaCl + H2O (titration curve will only have one break) Phenolphthalein endpoint is reached once reaction number 7 is completed. There is no need for the methyl orange. ( V MO = 0) Consider a soda ash sample containing Na 2CO3 only Na2CO3 reaction with HCl (8) HCl + Na2CO3 NaHCO3 (9) HCl + NaHCO3 NaCl + CO2 + H2O (titration curve will also have two breaks) Phenolphthalein endpoint is reached once reaction number 8 is completed while the methyl orange endpoint is reached once reaction number 9 is completed (V Phenolphthalein = VMO) Consider a soda ash sample containing NaHCO 3 only NaHCO3 reaction with HCl (10) HCl + NaHCO3 NaCl + CO 2 + H2O (titration curve or pH curve will only have one break) There is no need for phenolphthalein. Methyl Orange endpoint is reached once reaction number 10 is completed (VPhenolphthalein = 0) Consider a mixture of NaOH and NaHCO 3 NaOH + NaHCO3 Na2CO3 + H2O this reaction will take place that is why a mixture of NaOH and NaHCO3 is incompatible.
Substance NaOH Na2CO3 NaHCO3 NaOH + Na2CO3
Relation for qualitative identification VMO = 0 V phen = VMO VPhen = 0 VPhen > VMO
NaHCO3 + Na2CO3
VPhen < VMO
Millimoles of substance present MHCl x VPhen MHCl x VPhen or MHCl x VMO MHCl x VMO NaOH: MHCl x (V phen – VMO) Na2CO3: MHCl x VMO NaHCO3: MHCl x (VMO – V phen) Na2CO3: MHCl x V phen
Sample Problem: A soda ash sample was analyzed by dissolving 10.00g of the sample in 250.0 ml of water. Two separate 25.00 ml aliquot were titrated. With one portion, an endpoint with phenolphthalein is obtained in cold solution with 44.52 ml of 0.500 N HCl. The other 25 ml aliquot with methyl orange as indicator required 46.53 ml of the acid to reach the endpoint. Calculate the percentage composition of the original sample. Solution: 1st step determine the V phen and VMO relationship. V phen = 44.52 ml VMO is not actually equal to 46.53 ml. To solve for the actual V MO: VMO = V used to directly reach the methyl orange endpoint - V Phen
= 46.53 ml – 44.52 ml = 2.01 ml Therefore the relationship of V MO and VPhen; VPhen > VMO Based on the volume relationship the sample is therefore made up of NaOH and Na 2CO3. Moles Na2CO3 = MHCl x VMO = (0.500) x (2.01 ml/1000) = 0.001005 mol Moles NaOH = M HC x (V phen – VMO) = (0.500) x [(44.52 ml – 2.01 ml)/1000] = (0.500) x (0.04251) = 0.021255 % Na2CO3 = [(0.001005 mol x 106 g/mol) x (250 ml/25 ml)] / (10 g) = [(1.0653 g) /(10 g)] x 100% = 10.653 % % NaOH = [(0.021255 mol x 40 g/mol) x (250 ml/25 ml)] / (10g) = [(8.502g) / (10g)] x 100% = 85.02 %
Common Ion Effect and Buffers * Common-ion effect is an application of the Le Chatelier’s principle * a salt is generally less soluble in s solution containing an ion which is the same as one of the constituent ions of the salt. This is known as the common-ion effect * a buffer solution is one which resists drastic changes in pH when small quantities of of acid or base is added * Effectivity of a buffer: pH = pKa * The capacity of a buffer increases when the concentration of the acid and base component of the buffer is increased * dilution does not affect the pH of the buffer * To calculate the pH of a buffer solution: pH = pKa + log([base component]/[acid component]) or pH = pKa + log(moles of base component/moles of acid component) If an acid was added to the buffer solution: pH = pKa + log [(moles of base– moles of acid added)/ (moles of acid + moles of acid added)] If base was added to the buffer solution: pH = pKa + log [(moles of base+ moles of base added)/ (moles of acid - moles of base added)] * Maximum buffering capacity: pH = pKa +/- 1
