1. B
Let us calculate work first, then we will find out the change in internal energy by subtracting
work performed by the system from the heat transferred to the system.
w = ∆V×Pexternal = 3.00L×2.00 atm = 6.00 L×atm = 608 J
∆U = q – w t. = 2.00 kJ – 0.61 kJ = 1.39 kJ
2. A
The reaction is: Ag+ + NO3- + Na+ + Cl- Æ AgCl↓ + Na+ + ClThe formula weight of AgNO3 is 169.9 g/mol, so 1.691 g of AgNO3 is 9.95 mmol. The enthalpy
of the actual event that took place is - 216 × 3.03 = - 654 J. The standard enthalpy of our
reaction is -654 J/9.95 mmol = -65.7 kJ/mol
3. A
∆S = nR×ln(V2/V1) = 2×8.314×ln(0.25) = -23.1 J/(mol×K)
4. E
The equilibrium constant will be equal to 1 when ∆Gr = ∆Hr - T∆Sr = 0. ∆Hr will be equal to
T∆Sr at 1042 K and so that is the temperature at which ∆Gr = 0. Above it, Keq will be greater
than 1 (∆G < 0), below it, Keq < 1 and ∆G > 0.
5. C
The vapor pressure of a substance does not depend on the volume of the container. The vapor
pressure will remain 23.8 Torr.
6. E
Four phases: Gas, liquid, solid rhombic, solid monoclinic
Three triple points: two indicated by arrows and another one at which gas, liquid, and solid
monoclinic coexist.
7. C
The second reaction is the first reaction in reverse with coefficients divided by two. So, K2 =
(K1)-1/2 = 0.60
8. A
Reaction A is not because the number of molecules of gas in it does not change.
9. B
NH2- + H+ Æ NH3
10.
pKb increases from the strongest base… E < A < D < F < C < B… to the weakest base
11.
12. B
If the molar solubility of M2A3 is s = 7.0×10-6, then
[M3+] = 2s = 1.4×10-5
[A2-] = 3s = 2.1×10-5
Ksp = [M3+]2[A2-]3 = 4s2×27s3 = 1.8×10-24
13.
The formula weights of tartaric acid and its monopotassium salts are 150 and 188 g/mol,
respectively.
If we have a mixture of a weak acid and its salt, then we have a buffer. From that, we can figure
out the molar ratio of acid to base.
pH = pKa – log([HA]/[A-])
For Gewürz:
3.0 = 3.2 – log([HA]/[A-]);
log([HA]/[A-]) = +0.2;
[HA]/[A-] = 1.58
For Mama Rosa:
3.5 = 3.2 – log([HA]/[A-]);
log([HA]/[A-]) = -0.3;
[HA]/[A-] = 0.50
For Gewürz, the molar ratio of acid/base is 1.58:1.00.
The mass ratio is 1.58×150 : 1.00×188 = 1.26 or 56% : 44% by mass. Therefore in 8.3 g there is
approximately 4.7 g of acid and 3.6 g of base. So, 3.6 g/L of potassium hydrogen tartrate.
For Mama Rosa, the molar ratio of acid/base is 2.0:1.0.
The mass ratio is 0.5×150 : 1.0×188 = 0.40 or 29% : 71% by mass. Therefore in 7.7 g in one
liter there is approximately 2.2 g of acid and 5.5 g of base. So, 5.5 g/L of potassium hydrogen
tartrate.
KOH would convert any acid into its potassium salt. To match the pH of Mama Rosa, we will
need to match the acid/base ratio in it. In Gewürz, there is a 61:39 molar ratio of acid/base, so in
one liter, there is approximately 30 mmol of acid and 19 mmol of base. We need to add enough
KOH to convert enough acid into base so that the molar acid/base ratio becomes 0.50. If ‘x’ is
the number of mmol of KOH added, then:
(30-x)/(19+x) = 0.50
x ≈ 13.7 mmol of KOH which is (FWKOH = 56) approximately 0.8 g. That means we would need
to add 0.8 g/L of KOH.
14.
You have 80 mg Na2SO3 in a liter of your solution. Given the formula weight of Na2SO3 of 126
g/mol, the concentration of sulfite in this solution is 6.3×10-4 mol/L.
Ksp(CaSO3) = [Ca2+][SO32-] = 6.8×10-8
You will observe precipitation of CaSO3 when you add enough Ca2+ (by way of calcium nitrate)
so the quotient Q = [Ca2+][SO32-] of actual concentrations becomes equal to Ksp. Since we have
just calculated [SO32-], then the required
[Ca2+] = Ksp/([SO32-]) = 6.8×10-8/(6.3×10-4) = 1.1×10-4 mol/L. So, to see the onset of
precipitation, you will need to add 1.1×10-4 mol of Ca(NO3)2 to your 1000 mL solution. That’s
0.018 g of calcium nitrate.
The osmotic pressure depends only on the concentration (in mol/L) and not the identity of the
dissolved solutes. All we need to do is calculate the total concentration of all dissolved species.
What we will have is 6.3×10-4 mol/L of Na2SO3 and 1.1×10-4 mol/L of Ca(NO3)2. Each of these
salts dissociates fully and produces three ions, so the total concentration of dissolved species will
be:
c = 3×6.3×10-4 + 3×1.1×10-4 = 2.22×10-3 mol/L
Π = RTc = 0.082 × 298 × 2.22×10-3 = 0.054 atm
Note that we do not need ‘i’ here (or, rather i = 1 for each ion).
15.
Three of the structures have the empirical formula C5H8XY and one is C5H8F2. The three
C5H8XY will all have the same number of isomers. C5H8F2 will have fewer.
Br
Cl
F
Cl
CF2
F
2-Bromo-3-fluoro-2-pentene
Br
3-Bromo-5-chloro-1-pentene
1,1-Difluoro-3-methyl-1-butene
1-Chloro-1-fluorocyclopentane
BONUS question.
The two layers are perfluorohexane (bottom layer) and a mixture of isooctane and toluene (top
layer). A crystal of I2 was added to color the layer. Note that I2 dissolves much better in the
hydrocarbon layer.
The two layers separate at lower T but mix at higher T. Note that the layers separate again upon
cooling. Apparently, at a temperature slightly above RT, ∆G for the mixing changes sign (to
negative at higher T). The entropy of mixing is typically positive (favorable) and in this case
T∆S overcomes the unfavorable positive ∆H at a temperature slightly above RT.
Note that it is not a chemical reaction.