Kinetics - Part 1 Solutions

Chem Worksheet - Week 5 - Solutions Kinetics - Part 1 Key Questions 1. Define the rate of the following gas phase reaction on the basis of each reactant and product, so that the measured value of Rate would be the same at any point in the reaction, regardless of which species was measured: 2 N2O5(g) 6 4 NO2(g) + O2(g) Rate = –½Δ[N2O5]/Δt = +¼Δ[NO2]/Δt = Δ[O2]/Δt 2. 3. At a particular point in the course of the reaction 2 N2O5(g) 6 4 NO2(g) + O2(g), the rate of disappearance of N2O5(g) was found to be 1.16 x 10–4 mol/L@s. a. What is the rate of appearance of NO2(g) at this point? Rate of appearance of NO2 is twice the rate of disappearance of N2O5, so Δ[NO2]/Δt = (2)(1.16 x 10–4 mol/L@s) = 2.32 x 10–4 mol/L@s b. What is the rate of appearance of O2(g)? Rate of appearance of O2 is half the rate of disappearance of N2O5, so Δ[O2]/Δt = (½)(1.16 x 10–4 mol/L@s) = 5.8 x 10–5 mol/L@s c. What is the Rate of the reaction at this point, regardless of the basis for its definition? From Rate = –½Δ[N2O5]/Δt = +¼Δ[NO2]/Δt = Δ[O2]/Δt Rate = –½(–1.16 x 10–4 mol/L@s) = +¼(2.32 x 10–4 mol/L@s) = +5.8 x 10–5 mol/L@s = 5.8 x 10–5 mol/L@s For the reaction CHCl3(g) + Cl2(g) 6 CCl4(g) + HCl(g), the experimentally observed rate law is Rate = k[CHCl3][Cl2]½. a. What are the orders of the rate law with respect to each reactant and the overall order? First order with respect to [CHCl3], half order with respect to [Cl2], and threehalves order overall. b. What effect on the rate of the reaction would occur if the concentration of CHCl3 were doubled while the concentration of Cl2 remained the same? Rate % [CHCl3], so doubling [CHCl3] will double the rate; i.e., 21 = 2. c. What effect on the rate of the reaction would occur if the concentration of Cl2 were doubled while the concentration of CHCl3 remained the same? Rate % [Cl2]½, so doubling [Cl2] will cause the rate to increase by the square root of 2; i.e., 2½ = 1.41. d. What effect on the rate of reaction would occur if both reactant concentrations were doubled? Rate = k[CHCl3][Cl2]½, so the rate will increase by (21)(2½) = 2 x 1.41 = 2.82. e. 4. If the rate of the reaction was determined with concentrations in mol/L and time in seconds, what are the units of k for this reaction? For the rate law Rate = k[CHCl3][Cl2]½, k would have units such as (mol/L)–½ s–1. Determine the rate law and value of k for the following reaction at 800 oC. 2H2(g) + 2NO(g) 6 2H2O(g) + N2(g) Exp. (atm) (atm) Initial Rate (atm/min) Order #1 0.13 0.39 0.050 #2 0.26 0.39 0.10 Rate % #3 0.13 0.78 0.20 Rate % #4 0.26 0.78 0.40 Y3rd order overall Rate = k Using data from the first line: 5. Determine the rate law for the following hypothetical reaction. H2(g) + A2(g) 6 2HA(g) Exp. [H2] (mol/L) [A2] (mol/L) Initial Rate (mol/L/sec) #1 0.30 0.30 0.020 #2 0.30 1.20 0.040 Rate % [A2]½ #3 1.20 0.30 0.080 Rate % [H2] Rate = k[H2][A2]½ Using the first line of data, the rate constant, k, is Order 6. Determine the differential rate law expression and value of k for the reaction A + B2 + C 6 AB + BC given the following kinetic data: Exp. [A], M [B2], M [C], M Rate, M@s–1 #1 0.128 0.111 0.702 1.07 x 10–3 #2 0.384 0.111 0.702 3.21 x 10–3 Rate % [A] #3 0.128 0.444 0.702 2.14 x 10–3 Rate % [B2]½ #4 0.128 0.444 0.351 2.14 x 10–3 Rate % [C]0 Order Rate = k[A][B2]½ The zero-order dependence of the rate on [C] means that the rate does not depend on the concentration of this species, and no term for it needs to be written in the differential rate law expression. Any change in the concentration of C will have no effect on the rate of the reaction. Using the first line of data, we calculate k as follows: