Even%20More%20Rate%20Law%20Answers

Even%20More%20Rate%20Law%20Answers - [0.010] 2 [1.50] 2 =...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Even More Rate Law Practice 1. Given the following experimental data, develop a rate law: Rate (mol/L-s) A (mol/L) B (mol/L) C (mol/L) 8.4342 * 10 -10 0.0718 0.0585 0.129 9.3714 * 10 -11 0.0718 0.0585 0.043 1.3495 * 10 -8 0.6462 0.234 0.129 8.4342 * 10 -10 0.6462 0.0585 0.129 Rate = k [B] 2 [C] 2 k = 1.481 * 10 −5 2. Solve for the units of the rate constant k: Fourth order overall, k = L 3 / (mol 3 s) 3. Solve for the rate with the following concentrations [A] = 0.015, [B] = 0.010, [C] = 1.50 Rate = 1.481 * 10 −5
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: [0.010] 2 [1.50] 2 = 3.33 * 10 9 mol / (L s) 4. Given that the chemical reaction is: 2 A + 2 B + 3 C 4 D, solve for the rate of A in the conditions described in question 3. 3.33 * 10 9 mol / (L s) = (1/2) Rate(A) Rate(A) = 6.66 * 10 9 mol / (L s) 5. Using the above information, solve for the rate of D, in the conditions described in question 3. 3.33 * 10 9 mol / (L s) = (1/4) Rate(D) Rate(D) = 1.33 * 10 8 mol / (L s)...
View Full Document

This note was uploaded on 11/29/2011 for the course CHEM 1212 taught by Professor Dockery during the Fall '08 term at Kennesaw.

Ask a homework question - tutors are online