Chemistry 1202 Final Exam for John Hogan practice one

Practice One



Chemistry 1202 Final
Exam for John Hogan
Practice One
1.
2. For the gas-phase reaction shown below, choose the answer which indicates how the
rate of change of N2 is related to the rate of change of O2.
4 NH3(g) + 3 O2(g) 2 N2(g) + 6 H2O(g)
a) - 1/2 [N2] /t = - 1/3 [O2] /t
b) 1/2 [N2] /t = 1/3 [O2] /t
c) 1/2 [N2] /t = - 1/3 [O2] /t
d) 1/3 [N2] /t = - 1/2 [O2] /t
e) 3 [N2] /t = 2 [O2] /t
(-1/Coefficient for O2) and (1/Coefficient for N2)
3. Consider the reaction of tetrafluorohydrazine with hydrogen: 1 N2F4(g) + 5 H2(g) 2 NH3(g)
+ 4 HF(g). If NH3 is produced at the rate of 0.29 mol/s, what is the rate of production of HF?
a) The rate of production of HF is 2.3 mol/s.
b) The rate of production of HF is -0.58 mol/s.
c) The rate of production of HF is 0.15 mol/s.
d) The rate of production of HF is 0.58
mol/s.
e) The rate of production of HF is -0.15 mol/s.
(0.29 mol/s NH3) * (4 mol HF/ 2 mol NH3) = 0.58 mols/s HF




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,Practice One


4. Consider the following reaction: 2 HgCl2(aq) + C2O42-(aq) 2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s)
The rate law for this reaction is first order in HgCl2(aq) and second order in C2O42-(aq). Pick
the correct rate law for this reaction from the multiple choices.
a) Rate = k [HgCl2(aq)] [C2O42-(aq)]
b) Rate = k [HgCl2(aq)] [C2O42-
(aq)]2
c) Rate = k [HgCl2(aq)] [C2O42-(aq)]4
d) Rate = k [HgCl2(aq)]2 [C2O42-(aq)]
e) Rate = k [HgCl2(aq)]4 [C2O42-(aq)]
First Order = X^1 Second Order = X^2 and so on
5. Consider the following reaction: 2 ClO2(aq) + 2 OH-(aq) ClO3-(aq) + ClO2-(aq) +
H2O(l) The rate law for this reaction is: Rate = k [ClO2(aq)]2 [OH-(aq)]
If the rate constant for this reaction at a certain temperature is 4.05e+02 M-2 s-1, what is the
reaction rate when [ClO2(aq)] = 0.0509 M and [OH-(aq)] = 0.0373 M?
a) The reaction rate is 3.91e-02 M s-1.
b) The reaction rate is 2.39e-07 M s-1.
c) The reaction rate is 5.83e+03 M s-1.
d) The reaction rate is 2.56e+01 M s-1.
e) The reaction rate is 2.81e+01 M s-1.
Use the given rate law and plug in the numbers Rate = (4.05e+2) * (0.0509)^2 *(0.0373)




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,Practice One


6. Consider the following reaction: 4 HBr(g) + O2(g) 2 H2O(g) + 2
Br2(g) The rate law for this reaction is: Rate = k [HBr(g)] [O2(g)]
At a certain temperature when [HBr(g)] = 0.00474 M and [O2(g)] = 0.00575 M, the rate of this
reaction is 1.68e-01 M s-1. What is the reaction rate when the concentration of O2(g) is
doubled, to 0.0115 M, while the concentration of HBr(g) is 0.00474 M?
a) The reaction rate is 3.35e-01 M s-
1. b) The reaction rate is 6.16e+03 M
s-1.
c) The reaction rate is 1.67e-01 M s-1.
d) The reaction rate is 6.70e-01 M s-1.
e) The reaction rate is 3.08e+03 M s-1
Use the given rate and concentrations to find constant k using the given rate law then use
that constant and the given rate law to find the new rate with the new concentrations
(1.68e-01) = k * (0.00474) * (0.00575) Find k using this then plug k into rate law with new
concentrations like Rate = (k that you found) * (0.00474) * (0.0115)
7. For a reaction of the form, A + B + C Products, the following observations are made:
doubling the concentration of A increases the rate by a factor of 4, doubling the
concentration of B has no effect on the rate, and doubling the concentration of C increases
the rate by a factor of 2. Write down the rate law for this reaction.
By what factor will the rate of the reaction described above change if the concentrations of
A, B, and C are all halved (reduced by a factor of 2)?
a) The rate will change by a factor of 0.0625.
b) The rate will change by a factor of 0.5.
c) The rate will change by a factor of 0.25.
d) The rate will change by a factor of
0.125.
e) The rate will change by a factor of 0.03125.
Doubling A: 2A = 4A Doubling B: 2B = B Doubling C: 2C = 2C
Rate = k * (A)^2 * (B) So then you would do (1/2)^2 * (1/2) = 1/8 which is answer



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, Practice One


8. The reaction, 2 HgCl2(aq) + C2O42-(aq) 2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s), was studied at
a certain temperature with the following results:
Experiment HgCl2 C2O4 Rate
1 0.106 0.106 1.75e-05
2 0.106 0.212 7.00e-05
3 0.212 0.106 3.50e-05
4 0.212 0.212 1.40e-04


What is the rate law for this reaction?
a) Rate = k [HgCl2(aq)]2 [C2O42-
(aq)]2 b) Rate = k [HgCl2(aq)] [C2O42-
(aq)]2
c) Rate = k [HgCl2(aq)]4 [C2O42-(aq)]
d) Rate = k [HgCl2(aq)] [C2O42-(aq)]3
e) Rate = k [HgCl2(aq)] [C2O42-(aq)]
9. The reaction, S2O82-(aq) + 3 I-(aq) 2 SO42-(aq) + I3-(aq), was studied at a
certain temperature with the following results:
Experiment S2O8 I Rate
1 0.0480 0.0480 1.77e-05
2 0.0480 0.0960 3.55e-05
3 0.0960 0.0480 3.55e-05
4 0.0960 0.0960 7.10e-05


If the rate law for this reaction is, Rate = k [S2O82-(aq)] [I-(aq)], what is the value of the rate
constant?
a) The value of rate constant is k = 7.70e-03 M-1 s-1.
b) The value of rate constant is k = 2.58e-03 M-1 s-1.
c) The value of rate constant is k = 1.06e-03 M-1 s-1.
d) The value of rate constant is k = 6.82e-02 M-1 s-1.
e) The value of rate constant is k = 1.78e-01 M-1 s-1.
Choose one row of numbers to use and write out the rate law and solve for
k 1.77e-05 = k * (0.0480) * (0.0480) solve for k

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