1. Ch 14 #82
RATE = [NO] = [O2] = 9.3 * M/s = 4.7 * M/s
2 t t
PART B,C)
k = RATE / [NO]2[O2]
k = * 10-5 M/s = = / M2s
(0.040M)2(0.035)
The reaction is second order for NO, if you increase NO by 1.8 :
1.82 = 3.24, the rate increases 3.24 times

2. RATE = k [I-] [OCl-]/[OH-]
Ch 14 #83
Part a)
RATE = k [I-] [OCl-]/[OH-]
Part b) Tripling [I-] triples rate, I- is first order!
Part c) OH- is inverse to rate so doubling it halves the rate!

3. Ch 14 #88-a ln[c]t = -kt + ln[c]i BEER’S LAW: A=abc
Where a= asorbtivity, b = path length, A= absorbance, c = concentration.
1st ORDER INTEGRATED RATE LAW TO SOLVE FOR THE TIME
ln[c]t = -kt + ln[c]i
BEER’S LAW: A=abc
solve for concentration
Then use
BEER’S LAW: A=abc
Where a= asorbtivity (5.6 * 103/cm M), b = path length( 1.00 cm), A= absorbance (0.605), c = concentration.
A=abc
0.605 = (5.6 * 103/cm M)(1.00 cm)(c)
c = 1.08 * 10-4 M (at 30 min)

4. Ch 14 #88-a,b ln[c]t = -kt + ln[c]i
A=abc
0.605 = (5.6 * 103/cm M)(1.00 cm)(c)
c = 1.08 * 10-4 M (at 30 min = 1800s) a b
ln[c]t = -kt + ln[c]i
ln[1.08 * 10-4 M ]30 = -k(1800s) + ln[4.464 * 10-3]0
k = 4.91*10-4 s-1 c
FIRST ORDER HALF LIFE = t1/2 = /k
t1/2 = /k
t1/2 = /4.91*10-4 s-1 = 1.41* 103 s (23.5 min)

5. ln[1.786 * 10-5 M ] = -(4.91*10-4 s-1) t + ln[4.464 * 10-3]0
Ch 14 #88-d
d.1
A=abc
0.100 = (5.6 * 103/cm M)(1.00 cm)(c)
c = * 10-5 M
d.2
ln[c]t = -kt + ln[c]i
ln[1.786 * 10-5 M ] = -(4.91*10-4 s-1) t + ln[4.464 * 10-3]0
T = 3.666* 103 s = 61.1 min