15506_e_8570

15506_e_8570 - 1 1. (6 pts) Select the rate law that...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
1. (6 pts) Select the rate law that corresponds to the data shown for the following reaction: A + B → C Experiment [A]/M [B]/M Initial rate(M/s) 1 0.012 0.035 0.10 2 0.024 0.070 0.80 3 0.024 0.035 0.10 4 0.012 0.070 0.80 a) Rate = k[B] 4 b) Rate = k[A][B] 3 c) Rate = k[A] 2 [B] 2 d) Rate = k[B] 3 e) Rate = k[B] 2. (6 pts) What is the value of the rate constant in question 1? a) 4.5 b) 340 c) 3.2 x10 3 d) 2.3 x 10 3 e) 2500 3. (6 pts) What are the units for the rate constant calculated in question 2? a) M 2 s b) M 2 s -1 c) M -1 s -2 d) M -2 s -1 e) M -1 s -1 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4.(4 pts) For a first-order reaction, a plot of ____ versus ____ is linear. a) ln [A] t , 1/t b) ln [A] t , t c) 1/[A], t d) [A], t e) t, 1/[A] 5. (6 pts) The rate constant for the decomposition of N 2 O 5 is 6.2 x 10 -4 min -1 . After 3354 min what percent of N 2 O 5 remains? The reaction, given below, is first-order in N 2 O 5 . N 2 O 5 → NO 2 + NO 3 a) 6.25% b) 10.0% c) 12.5% d) 20.0% e) 25.0% 2
Background image of page 2
6. (6 pts) The rate constant for a particular second-order reaction is 0.476 L/mol-s. If the initial concentration of reactant is s 0.25 mol/L, it takes _______ s for the concentration to decrease to 0.13 mol/L. a) 7.8 b) 1.4 c) 3.7 d) 1.7 e) 0.13 7.(7 pts) A reaction with activation energy of 123 kJ/mol has a rate constant of 0.200 s -1 at 311 K. At a temperature of ____K, the rate constant will be double that at 311 K. (R = 8.314 J/mol-K) a) 304 b) 316 c) 622 d) 349 e) 246 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
What would be the rate expression if the mechanism is correct? 2O 3 (g) → 3O 2 (g) (balanced equation) O 3 ↔ O 2 + O (fast) Step 1 O 3 + O → 2O 2 (slow) Step 2 a) rate = k[O 3 ] 2 b) rate = k[O 3 ] 2 [O] c) rate = k[O 2 ] 3 d) rate = k[O 3 ] 2 /[O 2 ] e) rate = k[O 3 ][O] 9. (6 pts) Given the following potential energy diagram for the one-step reaction: X + Y → Z + R.
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/10/2009 for the course CHEM 1111 at Colorado.

Page1 / 14

15506_e_8570 - 1 1. (6 pts) Select the rate law that...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online