Exam #1
Chemistry 102
General Chemistry
Tuesday September 28, 2004
Name:
KEY
.
The exam is worth a total of 100 points; there are six pages and
ten questions. Please show all work to receive full credit for an
answer.
By putting your name on this exam, you agree to abide by
California State University, Northridge policies of academic
honesty and integrity
Calculators are allowed for this exam. On the back pages you
will find a list of useful equations and a periodic table.
Good Luck!
Circle one correct answer for problems 1-4. No work need be shown to receive full
credit.
1. The decomposition of N2O(g) to nitrogen and oxygen
N2O(g)  N2(g) + 1/2O2 (g)
has the rate law expression:
Rate=k[N2O]2
Where k=1.1 x10-3 M-1s-1 at 565°C. If the initial concentration of a sample of N2O
is 0.50M, what is the concentration after one hour at 565°C? (3 pts)
(a)
(b)
(c)
(d)
(e)
9.5 x 10-3M
2.2 x10-3 M
0.029 M
0.17M √ use: 1/[A]t = akt + 1/[A]o, where a=1 and [A]o=0.50M
0.46 M
2. If a catalyst is added to a chemical reaction, the equilibrium yield of a product will
be the same, and the time taken to come to equilibrium will be less . than before.
(2 pts)
(a) higher; less
(b) lower; the same
(c) higher; the same
(d) lower; less
(e) the same; less √
3. Consider the following gas-phase reaction:
A(g)  B(g) + C(g) ΔH>0
This reaction will be most product-favored under conditions of
and high . temperature (2 pts)
(a)
(b)
(c)
(d)
(e)
4
low . pressure
high; high
low; high √
high; moderate;
high; low
low; low
Order the following aqueous solutions in terms of their freezing points, from
highest to lowest (assume all salts fully dissociate in water) (3 pts)
0.25 m naphthalene (C10H8), 0.15m CaCl2, 0.20m NH4NO3, 0.1 m NaCl
effective molality:
0.25
0.45
0.4m
0.2m
(a)
NaCl>C10H8> NH4NO3 > CaCl2√
(b)
C10H8>NaCl> NH4NO3> CaCl2
(c)
NH4NO3> CaCl2> C10H8> NaCl
(d)
CaCl2>NaCl> NH4NO3> C10H8
(e)
NaCl> NH4NO3> CaCl2> C10H8
5. The data below were obtained at 25°C for the reaction (15 pts):
NO2 (g) + O3 (g)  NO3(g) + O2 (g)
Trial
[NO2]o (M)
[O3]o (M)
Initial Rate
(M/s)
1
0.21
0.70
6.30 x10-3
2
0.63
0.70
1.89 x10-2
3
0.21
0.35
1.57x10-3
4
0.42
0.35
3.15 x10-3
5
0.60
0.20
(a) Write the rate law expression for this reaction
Compare runs 1 and 2 or runs 3 and 4 to determine order wrt [NO2]: first order.
Compare runs 1 and 3 for order wrt [O3] (0.7/0.35)n=(6.30 x10-3/1.57x10-3);
n=2, thus second order wrt [O3]
overall: rate = k [NO2][O3]2
(b) find the value of the rate constant (k) at 25°C.
substituting into any trial gives: 6.30 x10-3/(0.21)(0.7)2 =k=0.06122 1/M2•s
(c) What is the initial rate in trial 5?
Trial 5: rate = 0.06122 1/M2•s (0.60M)(0.20M)2 = 0.0014 M/s
6. For the following overall reaction:
2NO (g) + 2H2(g)  N2 + 2H2O
The following mechanism has been proposed:
(i)
(ii)
(iii)
NO + NO  N2O2
N2O2 + H2  N2O + H2O
N2O + H2  N2 + H2O
(a) write the rate law that would be observed if the first step in this mechanism (i) is
slow (rate-limiting) (5 pts).
Rate = k1 [NO][NO] = k1[NO]2
(b) write the rate law that would be observed if the second step in this mechanism (ii)
is slow (rate-limiting) and the first step is a fast equilibrium (5 pts).
Rate = k2[N2O2] [H2]
but [N2O2] is an intermediate
Assuming step 1 is a fast equilibrium, then k1[NO][NO]=k-1[N2O2]
And [N2O2]= (k1/k-1 )[NO]2
Thus, rate = (k2k1/k-1)[NO]2[H2]
(c) Suppose the experimentally observed rate constant for this reaction at 100°C is
100 times that at 25°C. Calculate the activation energy for this process in kJ/mol.
(10 pts)
100°C=373K; 25°C=298K
equation to use:
ln(k2/k1) = -Ea/R (1/T2 - 1/T1); ln (100) = -Ea/(8.314J/molK)[1/373K – 1/298K]
solving: Ea = 56.7 kJ/mol
7. Suppose H2(g) and I2(g) are sealed in a flask at 600K with initial partial
pressures P°H2=1.98 atm and P°I2=1.71 atm. At this temperature the gases
quickly reach equilibrium:
H2(g) + I2(g)  2 HI (g)
Keq=92.6
The equilibrium constant for the reaction is 92.6 at 600K.
What are the equilibrium partial pressures of H2, I2, and HI? (20 pts)
Keq = 92.6 = (PHI)eq2/ (PH2)eq(PI2)eq
Q=(PHI)i2/ (PH2)i(PI2)i = 0 atm / (1.98atm)(1.71 atm) = 0, thus equilibrium moves
to the right, towards products
H2(g) + I2(g)  2 HI (g)
Keq=92.6
Initial:
1.98
1.71
0
Change:
-x
-x
+2x
Equilibrium:
1.98-x 1.71-x
2x
Substituting into Keq expression:
92.6 = (2x)2/(1.98-x)(1.71-x)
algebra gives: 92.6 (1.98-x)(1.71-x) = 4x2
multiplying through: 313.5 – 341.7 x + 92.6 x2 = 4x2
or : 313.5 – 341.7x + 88.6x2 = 0
quadratic: x=[-b±(b2-4ac)1/2]/2a
x=2.35 or 1.50
2.35 doesn’t make sense because it gives a negative amount of H2 and I2 at equilibrium.
Thus, x=1.5
And (PHI)eq = 2(1.5) = 3.0 atm
P(I2)eq= 1.71 – 1.5 =0.21 atm
P(H2)eq = 1.98 – 1.5 = 0.48 atm
Check: (3.0)2/(0.21)(0.48) = 89 ~ 92
8. When 39.8 grams of a non-dissociating, non-volatile sugar is dissolved in 200
grams of water (Kb = 0.512 °C kg/mol), the boiling point of the water is raised by
0.30°C. Estimate the molar mass of the sugar. (10 pts)
ΔTb= Tb,soln-Tb,pure= Kbmsolute
ΔTb= 0.30°C
m= moles solute/0.2kg
ΔTb= 0.30°C= 0.512 °C kg/mol(moles solute/0.2kg)
moles solute=0.117 mol
M.W.= 39.8 g/ 0.117 moles= 340 g/mol
9. The Henry’s law constant (kH) for CO2 gas dissolving in water is 4.45 x 10-5
M/mmHg. If a carbonated beverage is under a CO2 pressure of 5.0 atm, calculate
the amount, in grams per liter (g/L), of CO2 gas dissolved in water. (10 pts)
Sg = kHPg
kH=4.45 x 10-5 M/mmHg.
Pg=5.0 atm(760 mmHg/atm) = 3800mmHg
Sg = (4.45 x 10-5 M/mmHg)(3800mmHg) = 0.1691 moles/liter
M.W. (CO2) = 44 g/mol
Sg = 7.44 g/liter
10. For the following endothermic dissolution:
NH4NO3 (s) + H2O (l)  NH4+ (aq) + NO3- (aq)
ΔHmix>0
(a) On the axes below graph of the solubility of NH4NO3 vs. temperature.
Clearly indicate the regions of the graph corresponding to Q< Keq, Q=Keq,
Q>Keq (5 pts)
Q=Ksp
Q>Ksp
Solubility of
NH4NO3
(g/100g H2O)
Q<Ksp
Temperature (°C)
For an endothermic dissolution, solubility of the salt increases with increasing temperature
(b) Describe the types of intermolecular forces involved in the dissolution of
NH4NO3 (s). What type of solvent-solvent and solute-solute interactions are
being broken, and what types of solute-solvent interactions are being formed?
To help explain, draw a picture of the ions in aqueous solution. What can you
say about the sign (positive or negative) of ΔSmix? (10 pts)
Ionic bonds broken in NH4+NO3-; hydrogen bonds broken in H2O; ion-dipole forces
established between hydrated NH4+ ions and NO3- ions and water.
hydrated ions
H
H
H
O
H
O
O
H
H
NH4+
O
O
H
H
H
O
O
H
H
H
H
negative end of water
dipole lined up with NH4+
H
NO3-
H
H
H
O
H
O
positive end of water
dipole lined up with NH4+
Since ΔHmix>0, the driving force for dissolution must be entropic:
ΔSmix>0, that is, the disorder of the solution is greater than the disorder of pure solvent or
of the ionic solid.