Session #23: Homework Solutions

Session #23: Homework Solutions Problem #1 The decomposition of hydrogen peroxide, H2O2, can be represented by the following reaction: 2H2O2(aq) → 2H2O(l) + O2(g) The table below reports data taken at room temperature (300 K). Table 1. Decomposition of H2O2(aq) at 300 K. concH2O2 (mol/liter) time (seconds) 2.32 2.01 1.72 1.49 0.98 0.62 0.25 0 200 400 600 1200 1800 3000 (a) Show that the reaction is first order. (b) Calculate the value of the half-life of this reaction. (c) Suppose that the initial concentration of H2O2 were 3.5 M. How long would it take at 300 K to reduce the concentration of H2O2 to 25% of its initial value? Solution (a) To show that the reaction is first order, try fitting the logarithm of concentration versus time. Least-squares analysis gives: ln c = 0.831 – 7.21×10–4 t with a correlation coefficient of 0.998. (b) The half-life is given by t1/2 = ln 2 0.693 = 961 s = k 7.21 × 10-4 (c) To decrease the concentration to 25% of initial value would take 2 half-lives, since after t1/2 the concentration would be 50% and after 2t1/2 it would be 50% of 50%. So the answer is 2×961 s = 1922 s. Problem #2 A chemical reaction which has an activation energy of 167.0 kJ/mole is to proceed at T = 450 K with a very constant rate; the rate is allowed to vary at most by ±1%. How constant must the temperature be to achieve this required rate stability? (For T>> Δ T, T1 x T2 = T2) Solution EA ⎛ 1 1 ⎞ ⎜⎜ T − T ⎟⎟ ⎝ 2 1⎠ k1 R = 1.02 = e k2 ln 1.02 = ln 1.02 = ΔT = EA ⎛ 1 1 ⎞ EA ⎛ T1 − T2 ⎞ EA ⎛ ΔT ⎞ − ⎜ ⎟= ⎜ ⎟= ⎜ ⎟ R ⎝ T2 T1 ⎠ R ⎝ T1 × T2 ⎠ R ⎝ T2 ⎠ EA ΔT RT 2 R × T2 × ln 1.02 = 2.0 × 10-1 K = ± 0.1 K EA The required temperature stability is beyond the capabilities of conventional thermostats. Problem #3 Determine the diffusivity D of lithium (Li) in silicon (Si) at 1200°C, knowing that D1100°C =10–5 cm2/s and D695°C = 10–6 cm2/s. Solution D1 10-6 = = 10-1 = e D2 10-5 EA = EA ⎛ 1 1 ⎞ − R ⎜⎝ 968 1373 ⎟⎠ R ln 10 = 62.8 kJ/mole 1 1 − 968 1373 − D1100 =e D1200 EA ⎛ 1 1 ⎞ − R ⎜⎝ 1373 1473 ⎟⎠ -5 D1200 = 10 ×e EA ⎛ 1 1 ⎞ − R ⎜⎝ 1373 1473 ⎟⎠ = 1.45 × 10−5 cm2 / sec Problem #4 For a chemical reaction, the concentrations of reactant as a function of time are given below for 25°C and for 50°C. at 25°C time (h) .00 3.15 10.00 13.50 26.00 37.30 at 50°C conc. (mole/L) 0.1039 0.0896 0.0639 0.0529 0.0270 0.0142 time (min) 0 9 18 54 105 180 conc. (mole/L) 0.1056 0.0961 0.0856 0.0536 0.0270 0.0089 (a) Indicate schematically (in two different graphic presentations) how you could prove, given concentration data at certain times, that a reaction is of first order. (b) Determine, from graphic presentations, the rate constants (k) for the given reaction at 25°C and 50°C. (c) Determine the half-life (t1/2) for the reaction at 50°C. (d) Determine the half-life (t1/2) for the reaction at 70°C. (e) What is the time required for the reaction at 25°C to be completed to the extent of 42%? Solution (a) For the first order reactions: ∧ co straight line ln c co/2 > t t1/2 = constant (time) (b) at 25oC, k = 0.0533 hr–1 255C ±2 ±3 k = 0.0533 hr±1 ln c ±4 ±5 0 10 20 30 40 t (hr) at 50oC, k = 0.0138 min–1 505C ±2 ±3 k = 0.0138 min±1 ln c ±4 ±5 0 (c) t1/2 = 40 80 120 160 200 t (min) ln 2 ln 2 = = 50.2 min k + 0.0138 (d) k = Ae E − A RT We need to determine A (and EA): − k25 =e k50 EA R ⎛ 1 1 ⎞ − ⎜ ⎟ ⎝ T25 T50 ⎠ k25 = 0.0533 hr-1 × ln k25 ⎛ EA = ⎜− k50 ⎝ R 1 hr = 8.88 × 10-4 min-1 60 min ⎞⎛ 1 1 ⎞ − ⎟ ⎟⎜ ⎠ ⎝ T25 T50 ⎠ ⎛ 8.88 × 10-4 ⎞ ⎛k ⎞ -8.3 ln ⎜ -R ln ⎜ 25 ⎟ ⎜ 1.38 × 10-2 ⎟⎟ k50 ⎠ ⎝ ⎝ ⎠ = 87.7 × 103 J/mole EA = = 1 ⎞ ⎛ 1 ⎛ 1 1 ⎞ − ⎜ 298 − 323 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ T25 T50 ⎠ o At 25 C: k = A =k E − A Ae RT EA e RT = (8.88 × 10-4 ) ⎛ 8.77×104 ⎞ ⎜ ⎟ ⎜ 8.31×298 ⎟ ⎠ e⎝ A = 2.22 × 1012 m-1 So, at 70oC (343 K): k =A E - A e RT = (2.22 × 1012 ) ⎛ 8.77×104 ⎞ ⎟ -⎜ ⎜ 8.31×343 ⎟ ⎠ e⎝ = 9.28 × 10-2 min-1 and (finally!): t1/2 = ln 2 ln 2 = = 7.47 min k 9.28 × 10-2 min-1 (e) If the reaction is 42% complete, then 58% of the reactants remain. Therefore, c = 0.58 co. c = coe–kt → ln c = -kt co ⎛ c ⎞ c ln ⎜ 0.58 o ⎟ co ⎠ co = - ⎝ = 10.22 hr = 10 hr, 13 min, 12.1 sec k 0.0533 hr-1 ln t =Problem #5 In a chemical reaction the concentration of a rate-determining component is measured (in moles) at one minute intervals from zero to 5 minutes. The data are: 1.0 x 10–2, 0.683 x 10–2, 0.518 x 10–2, 0.418 x 10–2, 0.350 x 10–2, and 0.301 x 10–2. (a) Determine the order (n) of this reaction. (b) Determine the rate constant (k). (c) Determine the half-life of this reaction. Solution (a) In c 400 –4.5 In c –5.0 300 – 5.5 –6.0 1/c 200 time (min) 0 1 2 3 4 5 100 The graph indicates that the ln c vs time plot yields almost a straight line, but the 1/c plot does yield a straight line. This identifies the reaction as a 2nd order reaction. (b) The rate constant (k) is given by the slope of the 1/c vs time plot: k = 332-146 1 = 46.5 5-1 mole × min (c) The half life of a second-order reaction (given in class) can be simply obtained from the second order rate equation. It is: τ1/2 = 1 1 = = 2.15 min k × co 46.5 × 10-2