CHEM 1214 Study Guide - Midterm Guide: Fokker E.Ii, Phenazopyridine, Sodium Hydroxide

62 views5 pages

fuchsiatortoise277

23 Mar 2016

School

Northeastern University

Department

Chemistry & Chemical Biology

Course

CHEM 1214

Professor

Lykourinou Vasiliki

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

2. rate=k[a] (s)-1: 2nd order- the plot of 1/[a] versus time is a straight line with k = slope of the line. 2. rate=k[a][b] (m*s)-1: 0 order- [a] versus time is a straight line with k = - slope of the line rate=k. [oh-: weak acids, ka<1, do not dissociate completely, example quesion with % ionizaion (smartworks quesion 10): a. In a 1. 01m aq sol"n of a monoproic acid, 3. 49% of the acid is ionized. What is the value of it"s ka? i. ii. Equaions: ph=-log[h+, poh=-log[oh, ph+poh=14, pka+pkb=14, ka=10-pka, kw=1x10-14=[h+][oh-, chapter 15 lecture 2, kw=ka*kb, acid base properies of salts i. textbook quesions 15. 77, Ch3cooh and 0. 0245m ch3coona; pka=4. 75: ph=pka+log([a-]/[ha], ph=4. 75+log(0. 0245/0. 0195, when you add a base, you"ll be subtracing some of the acid, ph=4. 75+log(0. 0245+[oh-]/0. 0195-[oh-]) is. 2. now the equaion the ph can now be factored in (and [oh-] can be called x) so: 5. 75=4. 75+log(0. 0245+[x]/0. 0195- [x]: 101=0. 0245+[x]/0. 0195-[x, 0. 0245 + x = 0. 195 10x, x= (0. 0195-0. 0245)/9=[oh-] f.

Related Questions

Learning Goal:

To understand reaction order and rate constants.

For the general equation aA+bBâcC+dD, the rate law is expressed as follows:

rate=k[A]m[B]n

where m and n indicate the order of the reaction with respect to each reactant and must be determined experimentally and k is the rate constant, which is specific to each reaction.

Order

For a particular reaction, aA+bB+cCâdD, the rate law was experimentally determined to be

rate=k[A]0[B]1[C]2=k[B][C]2

This equation is zero order with respect to A. Therefore, changing the concentration of A has no effect on the rate because [A]0 will always equal 1.(Figure 1)

This equation is first order with respect to B. This means that if the concentration of B is doubled, the rate will double. If [B] is reduced by half, the rate will be halved. If [B] is tripled, the rate will triple, and so on.(Figure 2)

This equation is second order with respect to C. This means that if the concentration of C is doubled, the rate will quadruple. If [C] is tripled, the rate will increase by a factor of 9, and so on.

Overall reaction order and rate-constant units

The sum of the individual orders gives the overall reaction order. The example equation above is third order overall because 0+1+2=3.

For the units of rate to come out to be M/s, the units of the rate constant for third-order reactions must be Mâ2â sâ1 since

M/s=(Mâ2â sâ1)(M3)

For a second-order reaction, the rate constant has units of Mâ1â sâ1 because M/s=(Mâ1â sâ1)(M2). In a first-order reaction, the rate constant has the units sâ1because M/s=(sâ1)(M1).

Analyzing a specific reaction

Consider the following reaction:

2H2+2NOâN2+2H2O,rate=k[H2][NO]2

Part A

What is the overall reaction order?

Express your answer as an integer.

What are the units of the rate constant k for this reaction?

choose one

What are the units of the rate constant for this reaction?

Mâ sâ1

sâ1

Mâ1â sâ1