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Chemistry 112 – Elementary Inorganic Chemistry
Morehouse College – Problem Set
DUE: Monday, February 2, 2009 in class
Answer all questions in the space provided, neatly written. Thanks.--BML
- (1 pt. each) For each of the following questions, circle the BEST answer (no explanation is required for full credit).
a. Consider the following reaction: 2NO(g) + 2H2(g) N2(g) + 2H 2 O(g) The experimentally determined rate law for the above process is rate = k[NO]^2 [H 2 ]. What is the reaction order for this process with respect to hydrogen gas (H2(g))?
0 1 2 3 none of these
b. Consider the following reaction: A(g) B(g) In consideration of the energy of activation ( Ea ) for this process, which of the following are true?
A catalyst will Ea α T(K) Ea α [A] A catalyst will none of these increase Ea decrease Ea
c. Consider the following chemical reaction: 2A + 2B 2C + 2D + E
If the reaction above is found to be first order in A and first order in B, what is the rate law for this reaction?
k[A][B] k[A]^2 [B]^2 k[2A][2B] k[2A]^2 [2B]^2 none of these
d. Radioactive decay of 212 Bi is observed to be a first-order process with a rate constant ( k ) of 7.21 x 10-3^ hr-1. It is noted that a 2.82 g sample of 212 Bi decays over 8 days to leave only 705 mg behind. What is the half-life ( t½ ) for this process?
12.0 days 8.00 days 4.00 days 2.00 days none of these
e. A certain reaction is first order in reactant “C” and has a rate constant ( k ) of 1.2 x 10-2^ s-1. If [C] 0 = 2.9 M, what is the [C] after 3.33 min?
0.18 M 0.55 M 1.0 M 6.0 x 10-3^ M none of these
- (5 pts) Dinitrogen tetroxide, N 2 O 4 , decomposes spontaneously at room temperature in the gas phase to give nitrogen dioxide. The rate law governing the disappearance of dinitrogen tetroxide obeys first-order kinetics. At 30 °C, the rate constant k is measured to be 5.1 x 10^6 s-1^ and the activation energy for the reaction is 54.0 kJ mol-1. Using this information to guide you, calculate the time it takes, in seconds, for the partial pressure of nitrogen dioxide in a sealed vessel to decrease from 0.10 atm to 0.010 atm at both 30 °C and 300 °C. Report your answer in the space provided with the correct number of significant figures. Show your work. A correct response without any justification (work) will receive no credit.
Since the decomposition reaction described above is 1st^ order, we know that…
[N 2 O 4 ]t = [N 2 O 4 ] 0 x e-kt^ …where k = 5.1 x 10^6 s-
From the ideal gas law, we also know that the pressure of a gas is proportional to it’s concentration. That is...
[ ] [ ]
NO NO 2 4 2 4
P P x e
Subsitutionofthisresultintotheintegratedratelaw/equationfromaboveyields...
RT
P
RT NO xRTorinsimplerterms,NO V
n V
nRT P
24 24
2 4 24
t
Since the initial and final pressures of N 2 O 4 are given, we can solve this equation for t.
4.5x 10 sor 451 ns 5.1x 10 s
0.10atm
0.010atm ln(
k
P
P
ln( time (t) NO( 0) 6 - 1 7
NO 24
24
= − t =^ =− =^ −
At 300 °C (573 K), the reaction rate constant is ex pected to be different. Using the Arrhenius equation, we will first solve for A at 30 °C (303 K ).
16 1 10
6 1
(8.314Jmol K)(303K)
6 1
RT
k 5.1x 10 s 5.1x 10 s A
− −
− − = = = = e e
If we assume that the Arrhenius factor (A) is constant at both temperatures (Question: Why can we do this?), we can use the Arrhenius equation to determine the new rate constant…
k A 1.0x 10 s x (8.314JmolK )(573K) 1.2x 1011 s^1 (at 300 o C)
54000 RT^161
E (^) a - 1 1 −
− −
− = = =
− e e
Lastly, we substitute this “new” rate constant into our expression from above for time…
1.9x 10 sor 19 ps 1.2x 10 s
0.10atm
0.010atm ln(
k
P
P
ln( time (t) NO( 0) 11 - 1 11
NO 24
24
= − t =^ =− =^ −
