Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions The reaction 2A P follows second-order kinetics The rate constant for the reaction is k350 0 4 Lmol s Determine the time required for the concentration of A to drop from its initial value of 0260 mol/l to 00 mol/l We first need to develop a relation for the integrated rate equation, since the stoichiometry involves 2 moles of A reacting The second-order rate law for this reaction is 2 d [ A] d t k [ A] 2 Rearranging and integrating yields d [ A] [ A] 2 2k dt d [ A] [ A] 2 2k dt A [ ] 2k t + C Using the initial condition at t0, [ A], the constant of integration is C the integrated rate law for this reaction becomes A [ ] 2kt + Solving for the time, we have t 2k [ A] t 2k [ A] 2 350 0 4 Lmol s 00mol/L 0260 mol/l t 24 0 5 s (or 35hours)
2 One of the hazards of nuclear explosions is the generation of 90 Sr and its subsequent incorporation into the bones in place of calcium Suppose mg of 90 Sr was absorbed into the body How much will remain after 0 years, 20 years, and 50 years if none was lost metabolically? Note that 90 Sr has a half-life of 28 years and that nuclear decay follows first-order kinetics For first-order kinetics, the half-life t / 2 is given by The rate constant k is therefore t / 2 ln2 k k ln2 t / 2 ln2 28 yr k 002467 yr 2 One form of the integrated rate law for first-order kinetics is [ A] e kt Substituting the initial amount of 90 Sr and the rate constant, we can determine how much 90 Sr remains after 0 years, Amt 90 Sr mg { } exp 002467 yr ( 0 yr) 078 mg Similarly, we can determine how much 90 Sr remains after 20 years, Amt 90 Sr mg { } exp 002467 yr ( 20 yr) 06mg And finally, we can determine how much 90 Sr remains after 50 years, Amt 90 Sr mg { } exp 002467 yr ( 50 yr) 029 mg
3 3 The reaction 2A P follows second-order kinetics Initially, the concentration of A was 0075 mol/l After hour, the concentration had fallen to 0020 mol/l Determine the rate constant for the reaction and the half-life For second-order kinetics with two moles of reactant, the integrated rate law is (see Problem ): A [ ] 2kt + Solving for the rate constant k, k 2t [ A] k 2t [ A] 2 3600s 0020 mol/l 0075mol/L k 509 0 3 L mol s Note that the time could have been left in units of hours but it is more common to use seconds To determine the half-life t / 2 of the second-order reaction, the definition is that at t t / 2, [ A] [ 2 A] o 2 A [ ] o 2kt / 2 + Solving for the half-life, 2 2kt [ A] / 2 o 2kt [ A] / 2, o or t / 2 2k t / 2 2k A [ ] o 0075molL 2 509 0 3 L mol s t / 2 309s (or 28 min)
4 A particular consecutive reaction, A k k B 2 C, has rate constants given by k 00s and 000s Sketch a graph of [A], [B], and [C] as functions of time and calculate the time at which the intermediate B reaches its maximum concentration We showed in class that the concentrations [A], [B], and [C] as functions of time for consecutive reactions are: [ A] e k t [ ] o [ B] k A k [ C] A [ ] o + e k t e t k e k t k e ( t ) These results also are given in the textbook A graph of concentration versus time for an initial concentration of [A] o 0 M is shown below Note that since k is ten times larger than, the reactant A dies away very quickly and the concentration of the intermediate B gets quite high before dropping off 4 From the graph, it appears that [B] reaches a maximum at about 250 seconds To determine the time more accurately, we must find the maximum in [B] From above, the concentration of B is given by [ ] o [ B] k A k e k t e ( t ) For a maximum, the slope d[b]/dt must equal zero So, taking the derivative and setting it equal to zero gives d [ B] d t k [ A] o k k e k t + e ( t ) 0 Since the expression equals zero, we can divide both sides by the constant terms out in front to give k e k t + e t 0
5 4 Continued Moving the constants and the rate constants to opposite sides yields k e k2t e k t e ( k )t Taking the natural log of both sides allows us to solve for the time at which [B] is a maximum, ln k k or t t ln k k t, ln k k 00s 00s ln 000s 000s t 256s
5 The gas phase decomposition of acetic acid at high temperature (89 K) proceeds by way of two parallel reactions, 6 CH 3 COOH CH 3 COOH k CH4 + CO 2 k 374 s k 2 H2 C C O + H 2 O 465 s Calculate the ratio of ketene (H 2 CCO) concentration to methane concentration and determine the maximum percent yield of ketene (H 2 CCO) In class, we showed that for a set of parallel reactions, the ratio of the two product channels was related to the ratio of rate constants That is, for the generic set of parallel reactions, A A k B k 2 C, the ratio of concentrations of products is [ B] C [ ] k, or [ C] B [ ] k This result also is given in the textbook For the reaction in this problem, therefore, [ H 2 C C O] [ CH 4 ] [ H 2 C C O] [ CH 4 ] k 465s 374 s 24 We also showed in class that the yield is related to the rate constant for formation of the specific product divided by the sum of the rate constants For ketene, therefore, the yield is %yield ketene k + 00 %yield ketene 554% 465s 374 s 00 + 465s
6 For the reaction A + B C + D, various initial rate measurements were made starting with A and B only and then C and D only From the data below, calculate the equilibrium constant for the reaction 7 [A] o (mol/l) [B] o (mol/l) Initial Rate (moll s ) 0660 23 08 0 5 40 23 6577 0 5 40 225 6568 0 5 [C] o (mol/l) [D] o (mol/l) Initial Rate (moll s ) 288 0995 7805 0 7 288 65 290 0 6 0 65 300 0 6 The first step is to determine the rate laws for the forward and reverse reactions using the method of initial rates Once this is done, the rate constants for the forward and reverse reactions can be calculated Finally, the equilibrium constant can be determined as the ratio of the forward and reverse rate constants For the forward reaction, the rate law can be written υ k [ A] α [ B] β The data from the first two experiments can be employed to determine the order of the reaction with respect to A since [B] remains constant Thus, from the method of initial rates, the order of the reaction with respect to A is α α ln υ 02 υ0 ln [ A] 02 [ A]0 ln 6577 0 5 moll s 08 0 5 moll s ln 40mol/L 0660 mol/l
8 6 Continued The data from experiments 2 and 3 can be employed to determine the order of the reaction with respect to B since [A] remains constant Thus, from the method of initial rates, the order of the reaction with respect to B is β β 0 ln υ 03 υ02 ln [ B] 03 [ B]02 ln 6568 0 5 moll s 6577 0 5 moll s ln 225mol/L 23mol/L The orders of the reaction with respect to A and B can be substituted into the rate law for the forward reaction and the rate constant can be solved for, or k υ k [ A] [ B] 0, υ [ A] Substituting the data from experiment, the forward rate constant k can be determined, k υ A [ ] 08 0 5 moll s 0660 mol/l k 638 0 5 s For the reverse reaction, the rate law can be written υ [ C] γ [ D] δ The data from experiments 4 and 5 can be employed to determine the order of the reaction with respect to D since [C] remains constant Thus, from the method of initial rates, the order of the reaction with respect to D is δ δ ln υ 05 υ04 ln [ D] 05 [ D]04 ln 290 0 6 moll s 7805 0 7 moll s ln 65mol/L 0995 mol/l
9 6 Continued The data from experiments 5 and 6 can be employed to determine the order of the reaction with respect to C since [D] remains constant Thus, from the method of initial rates, the order of the reaction with respect to C is γ γ 0 ln υ 06 υ05 ln [ C] 06 [ C]05 ln 300 0 6 moll s 290 0 6 moll s ln 0mol/L 288 mol/l The orders of the reaction with respect to C and D can be substituted into the rate law for the reverse reaction and the rate constant can be solved for, or υ [ C] 0 [ D], υ [ D] Substituting the data from experiment 4, the reverse rate constant can be determined, υ D [ ] 7805 0 7 moll s 0995 mol/l 7844 0 7 s The equilibrium constant is then defined as the ratio of the forward and reverse rate constants, K eq k 638 0 5 s 7844 0 7 s K eq 209
0 7 The second-order rate constant for the decomposition of a certain substance is 70 0 2 Lmol s at 30ºC and 20 0 2 Lmol s at 37ºC Determine the activation energy and pre-exponential factor The linear form of the Arrhenius rate law is At some temperature T, this equation is ln k ln A RT ln k ln A RT At some other temperature T 2, the equation is ln ln A RT 2 To obtain the activation energy, we can subtract the two equations, Now we can solve for the activation energy, E ln k ln a E + a RT RT 2 or ln k E a R T T 2 R ln k T T 2 R ln k T T 2 ( 834 J/molK) ln 70 0 2 L mol s 20 0 2 L mol s 3035K 305 K 8706 J/mol or 87 kj/mol
7 Continued Substituting the result for the activation energy into the Arrhenius Equation for temperature T, we can solve for the pre-exponential factor, ln k ln A or ln A ln k + RT RT ln A ln k + RT + 8706 J/mol ln 70 0 2 L mol s ( 834 J/molK) ( 3035K) ln A 33473 Therefore, the pre-exponential factor is A e 33473 A 2843L mol s Note that the pre-exponential factor carries the same units as the rate constant
8 Nitrous oxide, N 2 O, decomposes thermally at high temperatures The measured rate constants for the gas phase decomposition of N 2 O at different temperatures [S K Ross et al, J Phys Chem A 997, 0, 04] are given in the table below 2 T (K) k (cm 3 molecule s ) 2056 679 0 6 2095 838 0 6 232 03 0 5 273 39 0 5 Assuming Arrhenius behavior, determine the activation energy (in kj/mol) and the pre-exponential factor (in units of cm 3 molecule s ) The linear form of the Arrhenius rate law is ln k ln A RT This means that a plot with ln k on the y-axis and /T on the x-axis should be linear with a slope equal to R and an intercept equal to ln A A plot of the data is given below From the graph, we see that the slope is 27276 K Solving for the activation energy, R slope ( 27276 K) 834 J/molK 226770 J/mol or 2268 kj/mol From the graph, the intercept is 2683 Solving for the pre-exponential factor, intercept ln A 2683 or A e 2683 A 383 0 0 cm 3 molecule s
9 Many reactions double their rates with a 0 C rise in temperature Assume that the rate of a reaction is measured at 305 and 35 K What must the energy of activation be if the rate of reaction at 35 K is twice the rate of reaction at 305 K? The Arrhenius rate law gives the temperature-dependence of chemical reactions, At 305 K, the rate constant is k( T ) A e / RT 3 At 35 K, the rate constant is Ea 305 K ( k(305k) A e )R Taking the ratio of the two equations, Ea 35 K ( k(35k) A e )R k(35k) k(305k) A e A e ( 35 K)R E a ( 305 K)R Since the rate constant at 35 K is assumed to be twice the rate at 305 K, the ratio of the equations becomes k(305k) k(305k) 2 A e 2 e e 2 e A e ( 35 K)R E a ( 305 K)R ( 35 K)R E a ( 305 K)R E a ( 35 K)R + ( 305 K )R Taking the natural log of both sides, 35K ln 2 R + ( 305K)R R ln 2 305K 35K ( 96075K)R ln 2 55350 J/mol or 5535 kj/mol