The reaction whose rate constant we are to find is the forward reaction in the following equilibrium. NH + 4 (aq) + OH (aq) K b.

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1 THE RATES OF CHEMICAL REACTIONS 425 E22.3a The reaction for which pk a is 9.25 is NH + 4 aq + H 2Ol NH 3 aq + H 3 O + aq. The reaction whose rate constant we are to find is the forward reaction in the following equilibrium NH 3 aq + H 2 Ol k k NH + 4 aq + OH aq K b. The equilibria are related by pk b = pk w pk a = = Therefore, K b = k k = = mol dm 3 and k = K b k = mol dm dm 3 mol s = s. Now proceed as in Example 22.4: τ = k + k [NH + 4 ] + [OH ] = k + 2k K b [NH 3 ] /2 [[NH + 4 ] = [OH ] = K b [NH 3 ] /2 ] = s dm 3 mol s [ ] /2 mol dm 3 = s. Hence τ = 7.6 ns. COMMENT. The rate constant k corresponds to the pseudo first-order protonation of NH 3 in excess water and hence has the units s. Therefore, K b in this problem must be assigned the units mol dm 3 to obtain proper cancellation of units. E22.4a Call the rate constant k at temperature T, and the rate constant k at temperature T : ln k = ln A E a RT, ln k = ln A E a RT [22.29]. Hence, E a = R ln k /k /T /T = 8.34 J K mol ln / /303 K /323 K For A, we rearrange eqn 22.3: = 64.9 kj mol. A = k e E a/rt = mol dm 3 s e / = mol dm 3 s. E22.5a If cleavage of a C D or D H bond is involved in the rate-determining step, use eqn 22.53: { kc D kc H = e λ, λ = hc νc } H /2 µch. 2k B T µ CD TRAPP-ISM-WH: CHAP /6/28 20:7 PAGE 425 #7

2 426 INSTRUCTOR S SOLUTIONS MANUAL Express the C H vibrational wavenumber in terms of its force constant: ν = kf 2πc µ CH /2 k /2 f so λ = 2k B T µ /2 CH Compute kc D and see if this accounts for the difference. kc H µ /2. CD µ CD u =.7 u and µ CH 2 u = 0.92 u J s 450 N m /2 λ J K 298 K 0.92 u /2 u /2.7 u / kg.85. Hence, kc D kc H = e.85 = That is, kc H 6.4 kc D, in reasonable accord with the data. E22.6a k = k a + [analogous to 22.67]. k a k a p A Therefore, for two different pressures we have k k = k a p p /p /p and hence k a = /k /k = 2 Pa Pa s = Pa s, s or.9 MPa s. Solutions to problems Solutions to numerical problems P22.2 The procedure is that described in solution to Problem 22.. Visual inspection of the data seems to indicate that the half-life is roughly independent of the concentration. Therefore, we first try to fit the data to eqn 22.2b: [A] ln = kt [A] 0 TRAPP-ISM-WH: CHAP /6/28 20:7 PAGE 426 #8

3 430 INSTRUCTOR S SOLUTIONS MANUAL.4 n[a] 0 /[A] t / s Figure 22.2 P22.0 If the reaction is first-order the concentrations obey [A] ln = kt [22.2b] [A] 0 and, since pressures and concentrations of gases are proportional, the pressures should obey ln p 0 p = kt and t ln p 0 p should be a constant. We test this by drawing up the following table p 0 /Torr t/s p 0 /Torr ln p t/s p The values in the last row of the table are virtually constant, and so in the pressure range spanned by the data the reaction has first-order kinetics with k = s P22.2 Using spreadsheet software to evaluate eqn 22.40, one can draw up a plot like that in Figure The curves in this plot represent the concentration of the intermediate [I] as a function of time. They are labeled with the ratio k /k 2, where k 2 = s for all curves and k varies. The thickest curve, labeled 0, corresponds to k 2 = 0 s, as specified in part a of the problem. As the ratio k /k 2 gets smaller or, as the problem puts it, the ratio k 2 /k gets larger, the concentration profile for I becomes lower, broader, and flatter; that is, [I] becomes more nearly constant over a longer period of time. This is the nature of the steady-state approximation, which becomes more and more valid as consumption of the intermediate becomes fast compared with its formation. P22.4 a First, find an expression for the relaxation time, using Example 22.4 as a model: d[a] dt = 2k a [A] [A 2 ] TRAPP-ISM-WH: CHAP /6/28 20:7 PAGE 430 #2

4 THE RATES OF CHEMICAL REACTIONS [I] / mol dm t/s 4 5 Figure 22.3 Rewrite the expression in terms of a difference from equilibrium values, [A] = [A] eq + x: d[a] = d[a] eq + x = dt dt dt = 2k a[a] eq + x [A 2 ] eq 2 x dt = 2k a[a] 2 eq 4k a[a] eq x 2k a x [A 2 ] eq x 4k a [A] eq + x Neglect powers of x greater than x, and use the fact that at equilibrium the forward and reverse rates are equal: to obtain k a [A] 2 eq = [A 2 ] eq dt 4k a[a] eq + x so τ 4k a[a] eq + To get the desired expression, square the reciprocal relaxation time, τ 2 6k2 a [A]2 eq + 8k a [A] eq + k 2 b * introduce [A] tot = [A] eq + 2[A 2 ] eq into the middle term, τ 2 6k2 a [A]2 eq + 8k a [A] tot 2[A 2 ] eq + k 2 b 6k 2 a [A]2 eq + 8k a [A] tot 6k a [A 2 ] eq + k 2 b = 8k a [A] tot + k 2 b and use the equilibrium condition again to see that the remaining equilibrium concentrations cancel each other. COMMENT. Introducing [A] tot into just one term of eqn * above is a permissible step, but not a very systematic one. It is worth trying because of the resemblance between eqn * and the desired expression: we TRAPP-ISM-WH: CHAP /6/28 20:7 PAGE 43 #3

5 THE RATES OF CHEMICAL REACTIONS 437 P A k a A 2 d[a] dt = 2k a [A] [A 2 ] Define the deviation from equilibrium x by the following equations, which satisfy the law of mass conservation. Then, [A] = [A] eq + 2x and [A 2 ] = [A 2 ] eq x d[a] eq + 2x dt = 2k a [A] eq + 2x [A 2 ] eq x dt = k a[a] eq + 2x 2 + [A 2 ] eq x = k a [A] 2 eq + 4[A] eqx + 4x 2 + [A 2 ] eq x } = {4k a x k a [A] eq x + k a [A] 2 eq [A 2 ] eq } = { + 4k a [A] eq x + k a [A] 2 eq [A 2 ] eq In the last equation the term containing x 2 has been dropped because x will be small near equilibrium and the x 2 term will be negligibly small. The equation may now be rearranged and integrated using the following integration, which is found in standard mathematical handbooks. dw aw + b = lnaw + b a + 4k a [A] eq x + k a [A] 2 eq [A 2 ] eq = dt + constant + 4k a [A] eq ln + 4k a [A] eq x + k a [A] 2 eq [A 2 ] eq = t + constant y ln = + 4k a [A] eq t where y = + 4k a [A] eq x + k a [A] 2 eq [A 2 ] eq y 0 y = y 0 e +4k a [A] eq t Comparison of the above exponential to the decay equation y = y 0 e t/τ reveals that τ = + 4k a [A] eq Note that this equation can be used as an alternate derivation of the equation discussed in problem The manipulations use the facts that K = [A 2 ] eq /[A] 2 eq = k a/ and [A] tot = [A] eq + 2[A 2 ] eq by conservation of mass, which can be used to show that [A] tot = [A] eq + 2k a [A] 2 eq or 2k a [A] 2 eq + [A] eq [A] tot = 0 This quadratic equation can be solved for [A] eq. [A] eq = + 8k a[a] tot 4k a TRAPP-ISM-WH: CHAP /6/28 20:7 PAGE 437 #9

6 438 INSTRUCTOR S SOLUTIONS MANUAL Substitution of this equation into τ 2 = + 4k a [A] eq 2 and some algebraic manipulation yields the result of problem 22.4: τ 2 = k2 b + 8k a [A] tot. Solutions to applications P22.30 The first-order half-life is related to the rate constant by eqn 22.3 t /2 = ln 2 k so k = ln 2 t /2 = ln y = y The integrated rate law tells us [ 90 Sr] = [ 90 Sr] 0 e kt so m = m 0 e kt where m is the mass of 90 Sr. a After 8 y: m =.00 µg exp[ y 8 y] = µg b After 70 y: m =.00 µg exp[ y 70 y] = 0.77 µg k 2 P22.32 a A k B C The peak concentration of B, [P] n, immediately after administration of the n th dose, each of which have been administered at the time interval τ, is given by the sum: [P] n = [B] 0 + [B] 0 e k 2 t 2k + [B] 0 e + + [B] 0 e n k 2 t = [B] 0 e mk 2 t 2 t... S n m=0 Conc. contribution of n th dose remainder of n th dose remainder of n 2 th dose remainder of st dose The residual concentration of B, [R] n, just before administration of the n + th dose results from the first-order elimination of [P] n : [R] n = [P] n e k 2τ [22.2 a,b], [P] = lim [P] n = [B] 0 n m=0 e mk2τ = [B] 0 + x + x 2 + where x = e k2τ < This may be simplified using the Taylor series: + x + x 2 + = x = e k 2 τ when x < We conclude that [P] = [B] 0 e k 2 τ. Furthermore, [R] n = [B] 0 e k 2τ n m=0 e mk 2τ = [B] 0 n e mk2τ. m= [R] = [P] e k2τ = [B] 0e k 2τ e k 2τ = [B] 0 e k2τ = [B] 0 e k2τ. TRAPP-ISM-WH: CHAP /6/28 20:7 PAGE 438 #20

Solutions to Problem Assignment 1 (Kinetics) Oxtoby Problems -- see Solutions Manual. Solutions to Lecture Problems I # 1-6

Solutions to Problem Assignment 1 (Kinetics) Oxtoby Problems -- see Solutions Manual Solutions to Lecture Problems I # 1-6 I-1. This problem illustrates how the differential rate law (DRL) can be simplified