Exam 1 (Green) - 2008

Exam 1 (Green) - 2008 - CHEM 188 Spring, 2008 Hour Exam 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEM 188 – Spring, 2008 Hour Exam 1 (Green) February 14, 2008 Instructions: Your scantron answer sheet must show your NAME , 7-DIGIT KU ID NUMBER , and LAB SECTION . (Begin these entries at the LEFT end of the space provided.) In answering the questions, be careful to fill in the corresponding circles on the answer sheet according to the number of the question on the exam. USE A SOFT (No. 2) PENCIL .  Useful information: Gas constant, R = 8.314 J/K mol = 0.08206 L atm/K mol Integrated Rate Laws: kt [A] 1 [A] 1 : order Second e [A] [A] : order - First kt - [A] [A] : order - Zero 0 kt 0 0 + = - = = - Arrhenius equation: /RT -E a Ae k =  - = 2 1 2 1 a 2 1 T T T T R E k k ln Relation of K P to K C : K P = K C (RT) n 1. For the reaction 4NO 2 (g) + O 2 (g) 2N 2 O 5 (g) the rate of loss of molecular oxygen, –d[O 2 ]/dt is 0.018 M/s at a particular time during the reaction. What is the rate, d[N 2 O 5 ]/dt , at which N 2 O 5 is being formed? A. 0.009 M/s B. 0.018 M/s C. 0.036 M/s D. 0.072 M/s E. 0.144M/s 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
2. The reaction of nitric oxide with hydrogen 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O is second-order in [NO] and first-order in [H 2 ]. At 1280 o C, the rate constant is 25 M - 2 s - 1 . Calculate the rate of the reaction at this temperature if [NO] = 0.25 M and [H 2 ] = 0.080 M . A. 0.13 M/s B. 0.25 M/s C. 0.50 M/s D. 1.0 M/s E. 2.0 M/s 3. The reaction of nitric oxide with hydrogen at 1280 o C is 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O From the following data collected at this temperature, determine the rate law . Expt. # [NO] 0 (M) [H 2 ] 0 (M) Initial rate (M/s)  1 0.0050 0.0020 1.3 x 10 - 5 2 0.010 0.0020 5.2 x 10 - 5 3 0.010 0.0040 1.0 x 10 - 4 A. rate = k[NO] B. rate = k[NO] 2 C. rate = k[NO][H 2 ] D. rate = k[NO] 2 [H 2 ] E. rate = k[NO] 2 [H 2 ] 2 4. If concentration is expressed in units of moles per liter (mol/L) and time in units of seconds (s), the
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/08/2008 for the course CHEM 188 taught by Professor Hierl during the Spring '08 term at Kansas.

Page1 / 6

Exam 1 (Green) - 2008 - CHEM 188 Spring, 2008 Hour Exam 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

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