Kinetics Practice Problems Ex. 1: Consider the following reaction, NH4+(aq) + NO2−(aq) → N2(g) + H2O(l), at 25°C. [NH4+] (in M) [NO2−] (in M) Initial Rate (in M/s) a. Determine the rate law for the reaction. rate = k [NH4+]x [NO2−]y 0.24 0.12 0.12 0.10 0.10 0.15 7.2×10-6 3.6×10-6 5.4×10-6 M/s103.6M/s107.2[0.10M] [0.12M] k[0.10M] [0.24M] k][NO ][NH k][NO ][NH kraterate6-6-yx2yxy22x24y12x1421××===−+−+ → (2.0)x = 2.0 → x=1 M/s103.6M/s105[0.10M] [0.12M] k[0.15M] [0.12M] k][NO ][NH k][NO ][NH kraterate6-6-yx2yxy22x24y32x3423××===−+−+4. → (1.5)y = 1.5 → y=1 rate = k [NH4+] [NO2−] b. Determine the rate constant for the reaction. =×==−+(0.10M) (0.24M)M/s107.2][NO ][NHratek-6`21413.0×10-4 M-1 s-1 Ex. 2: At 600°C acetone (CH3COCH3) decomposes to form several compounds. Experiment [CH3COCH3] (in M) Initial Rate (in M/s)a. Determine the rate law for the reaction. rate = k [CH3COCH3]x 1 2 6.0×10-3 9.0×10-3 5.2×10-5 7.8×10-5 M/s105M/s10[0.0060M] k[0.0090M] k]HCOH[ k]HCOH[ kraterate5-5-xxx133x23312××===287..CCCC → (1.5)x = 1.5 → x=1 rate = k [CH3COCH3] b. Calculate the rate constant for the reaction. =×==.0060MM/s10]HCOH[ ratek-51331025.CC8.7×10-3 s-1 c. What is the reaction order for this reaction at 600°C? 1st-orderd. Calculate the concentration of acetone after 45 minutes if the initial concentration of acetone is 0.0100M at 600°C. kt[A] [A]ln0t−= → [A]t = [A]0 e-kt [A]t =(0.0100M)min) (45min 1s 601-s (0.008666kte (0.0100M)e⎟⎠⎞⎜⎝⎛−−=) [A]t =(0.0100M) e-23.4 =6.9×10-13M (or essentially 0 since 0 dec. pl.→ 0 s.f.) e. What is the half-life for this reaction at 600°C? ===1-s 0.008666 0.693k 0.693t2180. s Ex. 3: At 500°C cyclopropane (C3H6) rearranges to propene (CH3CHCH2). The reaction is first-order with a rate constant of 6.7×10-4 s-1. a. Calculate the molarity of cyclopropane after 25 minutes if the initial concentration is 0.25M. [A]t = [A]0 e-kt =(0.25M)min) (25min 1s 601-s 10(6.7kt4-e (0.25M)e⎟⎠⎞⎜⎝⎛×−−=) [A]t =(0.25M) e-1.005 =0.09M (1 dec. pl. = 1 s.f.) b. How many minutes does it take for the concentration of cyclopropane to drop from 0.150M to 0.050M at 500°C? kt[A] [A]ln0t−= → s106.7- k- 1-4-=××==s 60min. 1(0.150M) (0.050M)ln [A] [A]ln t0t27 min. c. What is the half-life for the reaction at 500°C? =××==s 60min. 1s 106.7 0.693k 0.693t1-4-2117 min. d. How long does it take for the concentration to drop to 25% of the original concentration at 500°C? kt[A] [A]ln0t−= → s106.7- k- 1-4-=××==s 60min. 1[A] (0.25)[A]ln [A] [A]ln t000t34 min.Ex 4. Consider the following reaction energy diagram: a. Indicate the activation energy (Ea) and the transition state on the diagram. b. Check all of the following statements that are true:  i. NO---O3 in the diagram above is the transition state. NO! ; ii. NO---O3 in the diagram above is the activated complex.  iii. The activated complex and transition state are two terms that mean the same thing. NO! ; iv. Given ΔH for this reaction, the reaction is exothermic. Eproducts < Ereactants  v. Given ΔH for this reaction, the reaction is endothermic. ; vi. Given the relative activation energies for the forward and reverse reactions, the rate for the forward reaction should be greater than the rate for the reverse reaction. A lower Ea for the forward reaction compared to the reverse reaction means more of the collisions occur with the required energy, so more of the collisions lead to successful conversion of reactants to products and a higher reaction rate. transition state EaEx. 5. Atmospheric chemistry involves highly reactive odd-numbered electron molecules, such as the hydroperoxyl radical, HO2, which decomposes to form oxygen, 2 HO2(g) → H2O2(g) + O2(g). Consider the following experimental data at 25°C: Time (μs) [HO2] (μM) a. Show work supporting that the reaction is first-order with respect to HO2. zero-order reaction slope 1 = =s 0 - 0.6 M 8.5- 5.1μμ -0.5666 M/s 0 0.6 1.0 1.4 1.8 2.4 8.5 5.1 3.6 2.6 1.8 1.1 too different – not zero order! slope 2 = =s 1.8 - 2.4 M 1.8- 1.1μμ -.1666 M/s 1st-order reaction slope 1 = =s 0 - 0.6 ln(8.5) - ln(5.1)μ -0.8513 μs-1 about the same – probably 1st-order! slope 2 = =s 1.8 - 2.4 ln(1.8)- ln(1.1)μ -0.8207 μs-1 2nd-order reaction slope 1 = =s 0 - 0.6 8.5 - 5.1μμμMM11 0.1307 μM-1 μs-1 slope 2 = =s 1.8 - 2.4 1.8 - 1.1μμμMM11 -0.5892 μs-1 too different – not 2nd-order! Thus, this is a 1st-order reaction. b. Calculate the rate constant and the half-life for the reaction. k = - slope = - =+−21 s (-0.8207) [-0.8513 ]μ 0.8 μs-1 === s .836 0.693k 0.693t1-21μ00.8 μs rate constant = 0.8 μs-1 and the half-life = 0.8 μs