Cinética Química Exercícios

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14 Chemical Kinetics Solutions to Exercises 14.30 (a) rate = k[H₂][NO]² (b) rate = (6.0 X 10⁴ s⁻¹)(0.035 M)²(0.015 M) = 1.1 M/s (c) rate = (6.0 10⁴ (0.10 (0.010 M) = 6.0 M/s (d) rate = (6.0 X 10⁴ (0.010 M)² (0.030 M) = 0.18 M/s 14.31 Analyze/Plan. Write the rate law and rearrange to solve for k. Use the given data to cal- culate k, including units. Solve. (a, b) rate = k[CH₃Br][OH⁻]; k = [CH₃Br][OH⁻] rate 0.0432 M/s at 298 K, k = (5.0 10⁻³ M)(0.050 M) = 1.7 X 10² (c) Since the rate law is first order in if [OH⁻] is tripled, the rate triples. (d) If [OH⁻] and [CH₃Br] both triple, the rate increases by a factor of (3)(3) = 9. rate 14.32 (a, b) rate = k = at 298 K, k = [0.0477 1.7 M][0.100 10⁻⁷ M/s M] = 3.6 X 10⁻⁵ (c) Adding an equal volume of ethyl alcohol reduces both and by a factor of two. new rate = (1/2)(1/2) = 1/4 of old rate 14.33 Analyze/Plan. Follow the logic in Sample Exercise 14.6. Solve. (a) From the data given, when doubles, rate doubles. When [Γ] doubles, rate doubles. The reaction is first order in both and rate = (b) Using the first set of data: 1.36 10⁻⁴ M/s k = rate = (1.5 10⁻³ X X 10⁻³ M) = 60.444 = 60 (c) rate = 60.444 M-s (2.0 X 10⁻³ M)(5.0 X 10⁻⁴ M) = 6.0444 10⁻⁵ = 6.0 X 10⁻⁵ M/s 14.34 (a) From the data given, when increases by a factor of 3 (experiment 2 to ex- periment 1), the rate increases by a factor of 9. When increases by a factor of 3 (experiment 2 to experiment 3), the rate increases by a factor of 3. The reac- tion is second order in and first order in [OH⁻]. rate = (b) Using data from Expt 2: k = rate [OH⁻] = (0.020 0.00276 (0.030 M/s = 2.3 10² (c) rate = 2.3 10² (0.100 M) = 0.115 = 0.12 M/s 14.35 Analyze/Plan. Follow the logic in Sample Exercise 14.6 to deduce the rate law. Rearrange the rate law to solve for k and deduce units. Calculate a k value for each set of concen- trations and then average the three values. Solve. 403