Answers, Hourly 2

CHM101, Section 5, Fall 2001
Answers to Second hourly exam, 18 October 2001
Chapters 4–6

Definitions (3 points each; 30 points total)

      1. Oxidation number—The actual or hypothetical charge on an atom that represents the number of electrons it has gained, lost, or shared through bonding.
      2. Weak acid—An acid only some of whose molecules ionize in aqueous solution.
      3. Reduction—Gain in electrons by an atom.
      4. Boyle’s law—PV = constant for an ideal gas.
      5. Molar volume of a gas—The volume occupied by one mole of a gas at STP (0 ºC and 1 atm).
      6. Ideal gas—A gas that strictly obeys the simple gas laws and has a molar volume of 22.4141 at STP. Alt.: A gas whose molecules are not attracted to one another.
      7. Enthalpy—The thermodynamic function H = U + PV.
      8. Calorimeter—A device for measuring flows of heat within a system.
      9. Heat—Energy transferred in response to differences in temperature.
      10. Heat capacity—The amount of energy required to change the temperature of a body or system by 1 ºC.

Short answers and short problems (5 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) Fe₂O₃ +3, -2
            (b) S₈ 0
            (c) Na₂O₂ +1, -1
            (d) KClO₄ +1, +7, -2
            (e) S₄O₆^2- +2.5, -2

      2. A solution is prepared by mixing 100.0 mL 0.438 M NaCl, 100.0 mL 0.0512 M MgCl₂, and 300.0 mL of water. What are [Na⁺], [Mg²⁺], and [Cl^-]in the resulting solution?
      [Na⁺] = 0.438 M/5 = 0.0876 M
      [Mg²⁺] = 0.0512 M/5 = 0.0102 M
      [Cl^-] = [0.438 M + (2 x 0.0512 M)}/5 = 0.1081 M

3. A piece of stainless steel (sp. heat = 0.50 J g^-1 ºC^-1) is taken from an oven at 178 ºC and immersed in 225 mL of water at 25.9 ºC. The temperature of the water rises to 42.4 ºC. What is the mass of the piece of steel? (Be sure to express the answer to the correct number of significant figures.)

q_ss + q_w = 0
m_sssp.ht._ssΔT_ss = -m_wsp.ht._wΔT_w
m_ss = -m_w(sp.ht._w/sp.ht._ss)(ΔT_w/ΔT_ss)
= -225 g(4.184/0.50)[(42.4 – 25.9)/(42.4 – 178)]
= 230 g

4. Given that ΔU = q + w, where U = internal energy, q = heat, and w = work, derive the expression for enthalpy H = U + PV.

ΔU = q + w = q - PΔV
q_p = ΔU + PΔV = Δ(U + PV)
H = U + PV

Problems (10 points each; 50 points total)

R = 0.082058 L atm K^-1 mol^-1; 1 atm = 760 mmHg = 760 Torr = 101.325 kPa = 1.01325 bar

1. Balance this redox equation and show your work:

3 Mn²⁺+ ClO₃^- + 3 H₂O → 3 MnO₂(s) + Cl^- + 6 H⁺

2. Use the ideal gas law to convert the kinetic-molecular expression for pressure, P = (1/3)(N/V)mu², to the relation between kinetic energy and temperature e_k = (3/2)(RT/N_A), where m is molecular mass, u is molecular velocity, N is the number of molecules, and N_A is Avogadro’s number.

P = (1/3)(N/V)mu²
PV = (1/3)(2/1)N(0.5mu²) = nRT
= 0.5mu² = (3/2)(n/N)(RT)
= (3/2)(RT/N_A) = e_k

3. At what pressure will gaseous nitrogen have a density of 0.985 g/L at 25 ºC?

PV = nRT
P = (m/V)(RT/M)
= (0.985 g/L)(0.082 L atm mol^-1 K^-1)(298 K)/(28 g mol^-1)
= 0.86 atm

4. The reaction of sodium peroxide and water is a source of gaseous oxygen:

2 Na₂O₂ + 2 H₂O → 4 NaOH(aq) + O₂(g) ΔH = -287 kJ

How many kilojoules of heat are evolved when 150.0 g of Na₂O₂ reacts with 50 mL of H₂O?

First check to see which reactant is limiting. If you start with the 150.0 g Na₂O₂, it will require 35 g of H₂O (156/150 = 36/x), which makes H₂O in excess and Na₂O₂ limiting.

Second, use the 150 g Na₂O₂ and the given ΔH to solve for the new ΔH: (156/150 = -287/x); x = -275 kJ.

5. Use the following equations

Br₂(l) → Br₂(g)   ΔH = 30.91 kJ
Br₂(g) → 2 Br(g)   ΔH = 192.9 kJ
Cl₂(g) → 2 Cl (g)   ΔH = 243.4 kJ
Br₂(l) + Cl₂(g) → 2 BrCl(g)   ΔH = 29.2 kJ

to calculate the change in enthalpy for the reaction

BrCl(g) → Br(g) + Cl(g) ΔH = ??

Show your work.

To create the final equation, you need to multiply the given equations by ½, ½, ½, and -½, respectively. That creates the new set of equations (in different order):

BrCl(g) → ½ Br₂(l) + ½ Cl₂(g)
½ Br₂(l) → ½ Br₂(g)
½ Br₂(g) → Br(g)
½ Cl₂(g) → Cl(g)

that sum to BrCl(g) → Br(g) + Cl(g). To get the final ΔH, just multiply the original values by the new multipliers and add:

ΔH = ½(30.91) + ½(192.9) + ½(243.4) – ½ (29.2) = 219.0 kJ.

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