University:
Glendale Community College
Course:
CHM130 | Fundamental Chemistry
Academic year:

2019
Views:

407

Pages:

1
Author:

Vincent Reid

Kinetics Key 1. Calculate the average speed of reaction if at 10.0 seconds the reactant concentration is 1.25M but at 55.0 seconds the concentration is 1.02M. Ave. rate = ΔM / Δt = (Mf - Mi) / (tf - ti) = (1.02 – 1.25 M) / (55.0 – 10.0 s) = -5.1 x 10-3 M/s 2. If the rate of disappearance of H2 is 3.00 M/s, calculate the rate of NH3 appearance. Circle your final answer. Given: N2 + 3 H2 → 2 NH3 ΔNH3 / 2 Δt = ΔH2 / 3 Δt so rearrange ΔNH3 / Δt = (2/3) ΔH2 / Δt = (2/3) 3.00 M/s = 2.00 M/s 3. If a hypothetical rate law is rate = k[A]2[B]2 what is the overall reaction order? ____4th___ 4. The rate law for the following reaction is found to be rate = k[H2][I2]. H2 (g) + I2 (g) → 2 HI(g) If the rate = 4.24 x 10-3 M/s when [H2] = 0.150 M and [I2] = 0.250 M, calculate the rate when [H2] = 0.300 M and [I2] = 0.200 M. (Hint: first find k) and now solve for k = 0.1131 M-1s-1 4.24 x 10-3 M/s = k (0.150M) (0.250M) -1 -1 then rate = 0.1131 M s (0.300M)(0.200M) = 0.00678 M/s 5. If the rate of a reaction increases by a factor of 9 when the concentration of a reactant is tripled, then the reaction is __2nd___order with respect to that reactant. 6. If 25% of a sample, which obeys first order kinetics, remains after 45 seconds, calculate the rate constant. ln (25 / 100) = - k (45s) solve k = 0.031 s-1 7. If plutonium-246 decays with a first order half-life of 200.0 years, what percentage of a sample would remain after 600.0 years? 600 years would be 3 half lives, so 100% to 50% in one half life, 50 to 25% in 2 half lives, 25 to 12.5% in 3 half lives. So the answer is 12.5%. Or you could do the calculation the long way: 200 y = 0.693/ k solve for k = 0.003465 y-1 then ln At = (-0.003465 y-1)(600 y) + ln 100 and solve for At = 12.5 % 8. The half-life for a radioactive element that follows first-order decay is 534 years. If an 8.5 M solution is buried at a waste site, how long will it take for the solution to reach a concentration of 0.50 M? 534 y = 0.693 / k thus k = 0.001298 y-1 then ln (0.50 / 8.5) = (-0.001298 y-1) t and t = 2180 years 9. Given the following reaction: CO (g) + Cl2 (g) COCl2 (g) Determine the orders for CO and Cl2 and then write the rate law for this reaction. Then calculate the rate constant, k, and include the proper units. [CO] (M) 1.00x10-2 3.00x10-2 1.00x10-2 [Cl2] (M) 2.00x10-2 2.00x10-2 4.00x10-2 CO order = 1 k = Rate [CO][Cl 2 ] 2 Rate (M/s) 2.64 x 104 7.92 x 104 1.06 x 105 Rate = k[CO][Cl2]2 Cl2 order = 2 = 2.64x10 4 M/s (1.00x10 −2 M)(2.00x10 −2 M) 2 = 6.60x109 M-2s-1