Name: ________________________
CHM101, Section 5, Fall
2000
Second hourly exam, Thursday 19 October 2000
Chapters 4–6
Definitions
(2 points each; 20 points total)
1. Oxidation number—The actual or
hypothetical charge on an atom in an ion, compounds, or in an elemental state.
2. Strong electrolyte—A substance
that ionizes totally in aqueous solution and conducts electricity well.
3. Spectator ion—An ion that does not
participate in a reaction.
4. Ideal gas—A gas with no
intermolecular forces and whose volume is much larger than the volume of its
molecules, i.e., a gas that obeys the ideal-gas law.
5. Molar volume of a gas—The volume
occupied at STP by one mole of a gas (= 22.4 L).
6. Charles’s law—V/T = constant; or
V is proportional to T, where T is the absolute temperature.
7. Endothermic reaction—A reaction
that requires external energy for it to go.
8. Joule—The unit of energy in the
mks system; equals 1 N-m.
9. Hess’s law—The change of
enthalpy for a reaction is independent of the path.
10. Specific heat—The heat capacity
per gram-degree C.
Short answers and problems (4 points each; 20 points total)
1 cal = 4.184 J
1. What are the oxidation numbers of each of the elements in the
following chemical species?
(a) Al2O3
+3, -2 (b) NaMnO4
+1, +7, -2. (c) ClO-
+1, -2. (d) HSbF6
+1, +5, -1.
2. Which of the following solutions has the highest and which the lowest
concentration of NO3-?
(a)
0.10 M KNO3 0.10 M; intermediate.
(b)
0.040 M Ca(NO3)2 0.08 M; the lowest.
(c)
0.040 M Al(NO3)3 0.12 M; the highest.
(d)
0.050 M Mg(NO3)2 0.10 M; intermediate.
3. According to the kinetic-molecular theory of gases:
(a) What creates the pressure of a gas on the walls of its container? The
collisions of the molecules with the walls.
(b) The absolute temperature of a gas is directly related to what
quantity? The average kinetic energy of its molecules.
(c) What happens to the kinetic energy of two molecules when they
collide? It is redistributed between the molecules, but the total remains the
same.
(d) What kind of motion do the molecules of the gas follow? Random,
linear paths.
4. A 10.25-g sample of a metal alloy is heated to 99.10 ºC and then dropped into 20.0 g water in a calorimeter. The water temperature rises from 18.51 ºC to 22.03 ºC. What is the specific heat of the alloy?
Heat
lost by alloy = heat gained by water
malloy(Sp. ht.alloy)ΔTalloy = mw(Sp.
ht.w)ΔTw
(Sp. ht.alloy) = (mw/malloy)(ΔTw/ΔTalloy)(Sp.
ht.w) =
(20.0 g/10.25 g)[(22.03 ºC– 18.51 ºC)/(99.10 ºC – 22.03 ºC)94.184 J g-1)
=
0.373 J g-1 ºC-1
5. Given that ΔU = q + w, where U = internal energy, q = heat, and w = work, derive the expression for enthalpy H = U + PV in the way that we did in class.
ΔU
= q + w = qp – PΔV (at constant pressure)
qp = ΔU + PΔV = Δ(U +PV)
Define the new variable H = U + PV
Then qp = ΔH
Problems (10 points each; 60 points total)
R = 0.082058 L atm K-1 mol-1; 1 atm = 760 mmHg = 760 Torr = 101.325 kPa = 1.01325 bar
1. An aqueous solution is 0.0554 M NaCl and 0.0145 M Na2SO4.
What are [Na+], [Cl-], and [SO42-]
in this solution?
[Na+] = 0.0554 M + 2(0.0145 M)
= 0.0844 M
[Cl-] = 0.0554 M
[SO42-] = 0.0145 M
2. Balance this redox equation and show your work:
Zn + Cr2O72- + H+ → Zn2+ + Cr3+ + H2O
Step 1: Balance oxidation and reduction. Since Zn goes from 0 to +2 and Cr goes from +6 to +3, 3 Zn’s are needs for 2 Cr’s. This gives:
3 Zn + Cr2O72- + H+ → 3 Zn2+ + 2 Cr3+ + H2O
Step 2: Balance the O’s.
3 Zn + Cr2O72- + H+ → 3 Zn2+ + 2 Cr3+ + 7 H2O
Step 3: Balance the H’s.
3 Zn + Cr2O72- + 14 H+ → 3 Zn2+ + 2 Cr3+ + 7 H2O
Step 4: Check to see if the charges balance.
-2 + 14 = 6 + 6
Thus the balanced redox equation is:
3 Zn + Cr2O72- + 14 H+ → 3 Zn2+ + 2 Cr3+ + 7 H2O
3. If the gas in 4.65 L at STP is changed to a temperature of 15 ºC and a pressure of 756 Torr, what will its new volume be?
Use
the combined gas law: P1V1/T1 = P2V2/T2
Solve for V2 = (P1/P2)(T2/T1)V1
=
[1 atm/(756 Torr/760 Torr atm-1)][288 K/273 K] 4.65 L = 4.93 L
4. At what temperature will O2(g) at 0.982 atm pressure have a density of 1.05 g/L?
Use
the ideal-gas law in the form that includes density:
PV = nRT = (m/W)RT
T = (PW/R)(V/m) = PW/RD = [(0.982 atm)(32 g mol-1)]/[(0.082 L atm mol-1
K-1)(1.05 g L-1)]
T = 365 K
5. Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) as follows:
CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(s) ΔH = -128.0 kJ
How many kilojoules of heat are evolved when 3.50 kg of CaC2 reacts with 1.25 L of H2O?
This problem has two steps. First, find which reactant is limiting. Second, use that reactant to calculate the amount of heat evolved when all of it reacts.
(1) Limiting reactant.
3500
g CaC2 = 3500 g/64 g mol-1 = 54.7 mol CaC2
1250 g H2O = 1250 g/18 g mol-1 = 69.4 mol H2O
Since 1 mol CaC2 needs 2 of H2O, 54.7 mol CaC2
needs 109.4 mol H2O.
(Equivalent to the proportion ½ = 54.7/x; x = 109.4)
Since there are only 69.4 mol H2O, it is the limiting reactant.
(2) The amount of heat evolved.
Set up the proportion (2 mol H2O/69.4 mol H2O) = (-128.0 kJ/x). x = -4440 kJ
6. Use the following equations
N2H4(l)
+ O2(g) → N2(g) + 2 H2O(l)
ΔH = -622.2 kJ
H2(g) + ½ O2(g) → H2O(l)
ΔH = -285.8 kJ
H2(g) + O2(g) → H2O2(l)
ΔH = -187.8 kJ
to calculate the change in enthalpy for the reaction
N2H4(l) + 2 H2O2(l) → N2(g) + 4 H2O(l) ΔH = ??
Show your work.
The first equation fixes the N2H4 and the N2 in the final equation. It needs a coefficient of 1:
N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ΔH = -622.2 kJ
The third equation fixes the H2O2 in the final equation. It needs a coefficient of –2:
2 H2O2(l)→ 2 H2(g) + 2 O2(g) ΔH = 375.6 kJ
The second equation can be used to cancel the H2 and O2 from the second equation, for they do not appear in the final equation. It needs a coefficient of 2:
2H2(g) + O2(g) → 2 H2O(l) ΔH = -571.6 kJ
The sum of the three adjusted equations and their enthalpies is the desired equation and its enthalpy:
N2H4(l) + 2 H2O2(l) → N2(g) + 4 H2O(l) ΔH = -818.2 kJ