This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: zero, first, or second. After this we recorded the slope and interception points and printed the graphs. Part B In this part we simply repeated the experiment with a different value of [OH-]; we used a 0.1 M NaOH. Results – It is a first order reaction Calculations – Egbuniwe Chinedum 2 Part A – k 1 = [OH-] 1 n Part B - k 2 = [OH-] 2 n Since [OH-] 2 =2 [OH-] 1 k 2 = 2 n [OH-] 1 n k 2 /k 1 = k 2 n [OH-] 1 n / k [OH-] 1 n k 2 /k 1 = 2 n .3699/.3511 = 2 n 1.054 = 2 n N = 1/13 Discussion – From our experimentation and the graph that we derived from it we determined that both solutions were of the first order. We knew this because the graph with the Y-axis as ln[A] was linear for both part A and B....
View Full Document
This note was uploaded on 11/10/2009 for the course CHEM 107 taught by Professor Generalchemforeng during the Spring '07 term at Texas A&M.
- Spring '07