VII. Calculations:
II. Reaction Order with Respect to Hydrogen Peroxide- Calculate the initial reaction rate for each trial.
rate time s()I
s
==∆^3510 # -^4 M7 - A. ./
s
50 10 M 1 0638 10 Ms
47#^45
= #
- ^h
. ./
s
- ^h
50 10 M 1 0638 10 Ms
47#^45
= #
- ^h
. ./
s
- ^h
50 10 M 9 8039 10 Ms
51#^46
= #
- ^h
. ./
s
- ^h
50 10 M 8 3333 10 Ms
60#^46
= #
- ^h
. ./
s
- ^h
50 10 M 7 1429 10 Ms
70#^46
= #
- ^h
. ./
s
- ^h
50 10 M 5 5556 10 Ms
90#^46
= #
- ^h
III. Reaction Order with Respect to Hydronium Ion- Calculate the reaction rates as in step 2 of Part II.
log (rate)log [H 2 O 2 ]Graph
to find
x slope = order = xLaboratory Experiments- rate time s()
I
s
==∆7A^3510 # -^4 M..
s
50 10 M 1 1111 10
45#^45
= #
- ..
s
- ..
50 10 M 1 0000 10
50#^45
= #
- ..
s
- ..
50 10 M 6 66667 10
75#^46
= #
- ..
s
- ..
50 10 M 4 80769 10
104#^46
= #
- ..
s
- ..
50 10 M 3 84615 10
130#^46
= #Find z for each of the five reaction mixtures
and average the results
0.037761+0.0172+(−.193819)+(−.34387)+
(−.31940) = −.802128
−.80 21 28/5 = −.1604256
zero orderAlgebraic solution of order z.
log (rate 0 Part II) = log k+z log (1.00 × 10 –5)
log (rate 0 Part III) = log k+z log (3.16 × 10 –^5 )
log(1.00 × 10 −^5 ) = −5.00
log (3.16 × 10 −^5 ) = −4.50
log k = log rate (II) −z (5.0)
log k = log rate (III) −z (4.50)
log rate (II) −z (5.0) = log rate (III) −z (4.50)
z^05 =. h logcrate IIIrate IIm*formula to calcuate z.log
zrate IIrate III
= 05cm*...log.
z 05 0 0377611 0638 101 1111 10
55
##
==
dn
*...
.log
z1 000005 0 01729 8039 1010
65
##
==
dn
*...log.
z 05 0 1938198 3333 106 66667 10
66
##
==-
- dn
*...
log.
z7 14286
05 0 34387104 80769 10
66
##
==-
dn
*...
log.
z5 55556
05 0 31940103 84615 10
66
##
==-
dn