CHEM 135: Spring 2025 Quiz 5(100 points)
Name: KE Y
SHOW ALL WORK TO RECEIVE CREDIT!
1. (15pts) The radioactive decay of 14-C into 14-N proceeds via first-order kinetics with a half-life of 5,730
years. This decay is the basis of radiocarbon dating, which requires only a very tiny sample and can be used
to date objects that are up to 30,000 years old!
The basis of radiocarbon dating is simple: the ratio of 14-C to 14-Nis constant in living beings and
radioactive decay only begins after death. Thus, the reaction begins at the time of death (to).
%
a. What is the rate constant for this reaction?
ls‘ deder 72 T4E & Al[:‘ - In 2 — KK n2 - 2693 - ).2097 x10 4
/y K +‘/2. 5730 years i
K= 121 210"
\\ 2 = \
,'fi: (Lln‘\fi> M
S 2
\ N ————
years
b. How old is a wooden figurine that contains only 64.2% of its original 14-C (in years)?
[, - % In [WCL b k'{' + |n E*C]o
FCL = 0.642 x
\n (et - Inf“C]o = =+
\n< [cl:e—
- ["cl =t
K
\n (0.642) ‘ :f?:ééo years ololj
\.2.097 x|0 " Jears
2. (6 pts) Plots are shown for the reaction NO2(g) > NO(g) + %2 02 (g). What is the rate law for the reaction?
I/{NG,]
In [NO,]
]
time time m
time
Lond Order K}ne}‘fc;
a. rate = A[NOz]1
b. rate = £[NO2]
(cj rate= KNO2]2 Rare = K G\‘Oljz
d. rate=k
, 3. (20 pts) Consider the following reaction and data:
2NO(g) + Clz(g) — 2NOCI(g)
Experiment NO] (W [CL] H“
n {T° T 0.0300 LB
xZ T 00100 ) 134 x 10}* =.3. "' < 10” o x4
M 2<93 AR T
0.0150 RS R A 0.0400
(K (1 85507=
105 34
5.5 x107"
10 3" 1
34x1
a. Determine the reaction orders with respect to NO and Cl,
Cl, increased bg 4. The Rate ALSO
”:X.&iii (o) constant
iNncreases L;{ 4 = 'Sf order in Cly .
RXN \e2 (C\fl constant, E\JO] dovbleo. The Rate increases /Lg )
= 2" order tn NO
e Overall 3:4 Order . B
A\‘\-&’hOHVE\j Rate = K [_NO] [C'ZJ
]
\03 (3.1x’0'4 )
| 3.4%16° )
5 = I\ figs sm
S.cx /p-5 uloz‘fxxr 8.5 x10°% Z.,,
OCL\»OM\ ,ajl-[‘ u g\ 03 (0.03 (}"I‘fi’
I (5 0100 J 5o
b. Whatis the rate constant for this reaction? S
Rade = k [no] [cla] ,’gl- Eome) M- M
e 3
3.4 x10™ 2 = | (0.0300M)(0.0100 M)
o —
P | N\ . 2
= 373
- 3.4x]0
k= 4.000 %10 "M3
%
|38%
Name: KE Y
SHOW ALL WORK TO RECEIVE CREDIT!
1. (15pts) The radioactive decay of 14-C into 14-N proceeds via first-order kinetics with a half-life of 5,730
years. This decay is the basis of radiocarbon dating, which requires only a very tiny sample and can be used
to date objects that are up to 30,000 years old!
The basis of radiocarbon dating is simple: the ratio of 14-C to 14-Nis constant in living beings and
radioactive decay only begins after death. Thus, the reaction begins at the time of death (to).
%
a. What is the rate constant for this reaction?
ls‘ deder 72 T4E & Al[:‘ - In 2 — KK n2 - 2693 - ).2097 x10 4
/y K +‘/2. 5730 years i
K= 121 210"
\\ 2 = \
,'fi: (Lln‘\fi> M
S 2
\ N ————
years
b. How old is a wooden figurine that contains only 64.2% of its original 14-C (in years)?
[, - % In [WCL b k'{' + |n E*C]o
FCL = 0.642 x
\n (et - Inf“C]o = =+
\n< [cl:e—
- ["cl =t
K
\n (0.642) ‘ :f?:ééo years ololj
\.2.097 x|0 " Jears
2. (6 pts) Plots are shown for the reaction NO2(g) > NO(g) + %2 02 (g). What is the rate law for the reaction?
I/{NG,]
In [NO,]
]
time time m
time
Lond Order K}ne}‘fc;
a. rate = A[NOz]1
b. rate = £[NO2]
(cj rate= KNO2]2 Rare = K G\‘Oljz
d. rate=k
, 3. (20 pts) Consider the following reaction and data:
2NO(g) + Clz(g) — 2NOCI(g)
Experiment NO] (W [CL] H“
n {T° T 0.0300 LB
xZ T 00100 ) 134 x 10}* =.3. "' < 10” o x4
M 2<93 AR T
0.0150 RS R A 0.0400
(K (1 85507=
105 34
5.5 x107"
10 3" 1
34x1
a. Determine the reaction orders with respect to NO and Cl,
Cl, increased bg 4. The Rate ALSO
”:X.&iii (o) constant
iNncreases L;{ 4 = 'Sf order in Cly .
RXN \e2 (C\fl constant, E\JO] dovbleo. The Rate increases /Lg )
= 2" order tn NO
e Overall 3:4 Order . B
A\‘\-&’hOHVE\j Rate = K [_NO] [C'ZJ
]
\03 (3.1x’0'4 )
| 3.4%16° )
5 = I\ figs sm
S.cx /p-5 uloz‘fxxr 8.5 x10°% Z.,,
OCL\»OM\ ,ajl-[‘ u g\ 03 (0.03 (}"I‘fi’
I (5 0100 J 5o
b. Whatis the rate constant for this reaction? S
Rade = k [no] [cla] ,’gl- Eome) M- M
e 3
3.4 x10™ 2 = | (0.0300M)(0.0100 M)
o —
P | N\ . 2
= 373
- 3.4x]0
k= 4.000 %10 "M3
%
|38%
