CHM 1046 Spring 2016
Test 2
Name (print)_________________________
Show all work for complete (and partial) credit. Report answers to the correct number of
significant figures, and use units where appropriate. All equations should balance;
indicate phases.
ln([A]t/[A]0) = -kt
1/[A]t = kt + 1/[A]0 t½=1/(k[A]0) t½=.693/k
ln(k2/k1) = (Ea/R)[(1/T1) – (1/T2)] k = Ae-Ea/RT
R = .082058L.atm.mol-1K-1 = 8.314J mol-1K-1
1. The rate law for a reaction is found to be rate = k[A]2[B]. The reaction rate at 25°C
was found to be .00442M/s when [A] = 0.10M and [B] = 0.20M. What would the rate
be if [A] = 0.20M and [B] = 0.80M?
2. Write the rate law for the reaction. Initial concentrations and rates are given.
Run #
[M] (mol/L)
[N] (mol/L)
Rate (M/sec)
1
.002
.003
.0030
2
.002
.012
.012
3
.004
.012
.048
3. For a first order reaction, k = .0021/s. How long will it take for the reactant
concentration to drop to one eighth (1/8) its original concentration? If you do not have
enough information to calculate an answer, what information would you need?
4. If it takes 3.44 minutes for the concentration of the reactant in a second order reaction
to drop from 1.50M to 0.500M, how long would it take for the concentration to drop
from .50M to 0.25M?
5. If it takes 30.0 minutes for the concentration of the reactant in a first order reaction to
drop from 0.80M to 0.40M, what would the concentration be after 45.0min.?
6. For the reaction A↔B:
a) What does it mean if a plot of 1/[A] vs t is linear? (t=time)
b) What does it mean if a plot of ln[A] vs. t is linear?
7. If the temperature changes, which of the following will change and which will remain
constant? For each, check the appropriate box.
Will change
Will not change
a) the reaction rate
b) the rate constant, k
c) the equilibrium constant, K
d) the activation energy, Ea
8. What is the activation energy for a reaction that runs 3.2 times faster at 40.ºC than it
does at 25ºC?
9. For the reaction of the ammonium ion with nitrous acid, the net reaction is
NH4+ + HNO2 
 N2 + 2H2O + H+
The rate law for the formation of H+ is
a. R = k[NH4+][HNO2].

b. R = k
[NH 4 ][HNO2 ]
[N 2 ][H 2 O]2 [H  ]
.
c. R = k[H+].
d. R = k
[HNO2 ]
[H  ]
.
e. We cannot tell from the data given.
10. A reaction is thought to occur via the following steps. What is the overall chemical
equation for the reaction?
step 1 O3  O2 + O
(slow)
step 2 O3 + Cl  ClO + O2
step 3 ClO + O  Cl + O2
11. Why do you have to “fix” the rate law when the first step is not the slow step? What
is wrong with a rate law written from/for a later step?
A(aq) + 2B(aq) ↔ AB2(aq). 2.0mol of A and 5.0mol of B are allowed to reach
equilibrium in a 10.0L vessel. At equilibrium, the vessel contained 1.5 mol of AB2.
(Next two questions)
12. How many mols of A would be present at equilibrium?
13. What is the value of the equilibrium constant?
For the reaction CaC2O4(s) ↔ Ca+2(aq) + C2O4-2(aq), K = 2.3 x 10-9. (next two
questions)
14. What is the equilibrium expression for this reaction?
15. If someone mixed 1.00 mole of calcium ions with 0.500 mole of oxalate ions
(1.00mol Ca+2 and 0.500mol C2O4-2) in a 5.00L vessel, what would be the
predominate species present at equilibrium? Which reactants and/or products would
be present in appreciable amounts?
16. What general temperature and pressure conditions would cause a maximum yield of
water? 2H2(g) + O2(g) ↔ 2H2O(g) ΔH < 0.
17. If PbCl2(s) is placed into water and allowed to reach equilibrium (a saturated
PbCl2(aq) solution), [Pb+2] = .0159M, calculate K for the equilibrium
PbCl2(s) ↔ Pb+2(aq) + 2Cl-(aq).
18. In the equilibrium 2A(aq) + B(aq) ↔ A2B(aq) it was found (at equilibrium) that the
concentration of A was 2.5x10-5M and the concentration of B was 1.5x10-5M. The
value of K was 2.13x109. What was the concentration of A2B at equilibrium?
19. For the reaction HAc(aq) ↔ H+(aq) + Ac-(aq), K = 1.7x10-5. What would be the
equilibrium concentrations of all three species in a solution that is initially (preequilibrium) 0.20M HAc?
5 points extra credit For the reaction 2A(aq) + B(aq) ↔ A2B(aq), the rate law was found
to be rate = k[A][B]. When the concentration of both A and B were 0.10M at 25⁰C,
the rate was measured to be 0.113M/min. What would be the rate if the reaction was
run at 35⁰C using concentrations of both A and B equal to .015M? The activation
energy for the reaction is 55kJ/mol.