65.100 A and V

Christmas Exam 1996

ANSWERS

1a. 35.67 mg H₂O = 0.03567g / 18.0 g mol^-1 = 0.00198 mol H₂O = 0.00396 mol H

152.2 mg CO₂ = 0.1522g / 44.0 g mol^-1 = 0.00346 mol CO₂ = 0.00346 mol C

Both of these are divisible by 0.0005 (approximately). The empirical formula is thus C₇H₈

1b. C₇H₈ would have a molecular weight of 7 x 12 + 8 x 1 = 92 g/mol. Since the actual molecular weight is 92 g/mol, the molecular formula must be C₇H₈.

2. The unbalanced half reactions are:

As₂O₃ ® H₃AsO₄

and NO₃^- ® HNO₂

The balanced half reactions are:

As₂O₃ + 5 H₂O ® 2 H₃AsO₄ + 4 H⁺ + 4 e^-

and NO₃^- + 3 H⁺ + 2 e^- ® HNO₂ + H₂O

Multiplying the second reaction to cancel the electrons results in:

As₂O₃ + 5 H₂O ® 2 H₃AsO₄ + 4 H⁺ + 4 e^-

and 2 NO₃^- + 6 H⁺ + 4 e^- ® 2 HNO₂ + 2 H₂O

Adding gives the final answer:

As₂O₃ + 2 NO₃^- +3 H₂O + 2 H⁺ ® 2 H₃AsO₄ + 2 HNO₂

3. For isoelectronic species (those with the same electronic configurations), the atomic radius decreases with increasing atomic number, since electrons are being added to the same orbitals, and thus might be expected to be about the same distance from the nucleus. However, the nuclear charge increases going from S to Ca to Sc to Ti, so these extra electrons are drawn inward by increasing electrostatic attraction to the nucleus.

Thus, in increasing order of diameter are Ti⁺⁴, Sc⁺³, and Ca⁺², and S^-2

4. Xenon can form compounds because it can make use of unfilled d-orbitals. Since these orbitals lie at approximately the same energy as its 5p orbitals, there is no energy barrier to using them.

5. NO₂⁺

# of valence electrons = 5 + (2 x 6) - 1 = 16

Lewis structure is therefore:

We see that around the central atom there are two bonding groups and no lone pairs. The VSEPR notation is therefore AX₂ and the shape is therefore linear.

NO₂^-

# of valence electrons = 5 + (2 x 6) + 1 = 18

Lewis structure is therefore:

We see that around the central atom there are two bonding groups and one lone pair. The VSEPR notation is therefore AX₂E and the shape is therefore bent.

XeF₃⁺

# of valence electrons = 8 + (3 x 7) - 1 = 28

Lewis structure is therefore:

We see that around the central atom there are three bonding groups and two lone pairs. The VSEPR notation is therefore AX₃E₂ and the shape is therefore t-shaped.

6. The reverse of the first reaction is:

2 NO₂ ® N₂O₄ DH^o = -57.2 kJ (note sign change!!)

Twice the second reaction is:

2 NO + O₂ ® 2 NO₂ DH = -114 kJ (note this is twice the value for the reaction as given!!)

Adding these two reactions gives the desired reaction, and so adding the DH^o values gives the DH^o of the sum reaction:

2 NO + O₂ ® N₂O₄ DH^o = -57.2 - 114 = -171.2 kJ

7a. DH^o = 3 DH^o_f (S_(s)) + 2 DH^o_f (H₂O_(g)) - 2 DH^o_f (H₂S_(g)) - DH^o_f (SO_2(g))

= 3 (0) + 2 (-241.8) - 2 (-20.6) - (-296.8)

= -145.6 kJ/mol

DS^o = 3 S^o (S_(s)) + 2 S^o (H₂O_(g)) - 2 S^o (H₂S_(g)) - S^o (SO_2(g))

= 3 (31.8) + 2 (188.7) - 2 (205.7) - (248.1)

= -186.7 J K^-1 mol^-1

But DG^o = DH^o - T DS^o

= -145600 J/mol - 298 K (-186.7 J K^-1 mol^-1)

= -90,000 J/mol

= -90 kJ/mol

7b. At 25^oC, DG<0, therefore the reaction is spontaneous.

If DG = 0, then DH^o = T DS^o. Thus, T = DH^o / DS^o

= -145000 J mol^-1 / -186.7 J K^-1mol^-1

= 780 K

= 507^oC

8a. The Pauli Exclusion Principle states that no two electrons in an atom may have identical quantum numbers.

8b. Hund’s rule states that electrons in degenerate orbitals (i.e. those with the same energies) must have the maximum un-pairing possible. i.e. electrons are placed into empty degenerate orbitals first, and only then are subsequent electrons paired up.

8c. The solution to Schrodinger’s wave equation are the four quantum numbers: (n, l, m_l and m_s). These are the principle quantum number, the angular momentum (or shape) number, the magnetic quantum number and the spin number.

8d. Fe⁺³ has 23 electrons (since Fe has 26). The electronic structure is thus:

1s²2s²2p⁶3s²3p⁶4s²3d³

9a. The expression for K_p for this reaction is:

K_p = (p_NO)⁴ / [(p_N2O)²(p_O2)]

9b. To solve this, we need a value for K_p:

K_p = exp (-DG^o/RT)

= exp (-137900 J mol^-1 / (8.314 J K^-1mol^-1 x 298 K)

= exp(-55.7)

= 6.7 x 10^-25

rearranging the expression in part a, we find (p_NO)⁴ = K_p [(p_N2O)²(p_O2)]

thus, (p_NO)⁴ = 6.7 x 10^-25 (0.1)²(0.2)

= 1.3 x 10^-27

thus, p_NO = 1.9 x 10^-7 atm

10a Adding more NO_(g) causes the equilibrium to shift RIGHT (according to Le Chatelier’s principle)

10b. Removing NOBr_(g) causes the equilibrium to shift RIGHT (according to Le Chatelier’s principle)

10c. Compressing the mixture the equilibrium to shift RIGHT since there are a smaller number of moles on the right hand side of this equation

10d. Increasing the temperature causes the equilibrium to shift LEFT, since this in an exothermic reaction as written.

11a. According to the ideal gas law, P = nRT/V

Thus, P = 1 mol (0.082 L atm K^-1mol^-1)(298 K) / 1 L

= 24.4 L

11b. According to the van der Waals’ equation, P = nRT/(V-nb) - a(N²/V²)

Thus, P = 1 mol (0.082 L atm K^-1mol^-1)(298 K) / (1 L - 1 mol (0.032 L mol^-1)

- 1.35 L² atm mol^-2 (1 mol)²/(1 L)²

= 23.9 atm

11c. The pressure calculated using the ideal gas law assumes that (a) molecules occupy space themselves and (b) molecules in the gas phase do not interact. Both of these assumptions are false. The van der Waals’ equation takes both phenomena into consideration.

11d. As above, but for ammonia,

Thus, P = 1 mol (0.082 L atm K^-1mol^-1)(298 K) / (1 L - 1 mol (0.037 L mol^-1)

- 4.17 L² atm mol^-2 (1 mol)²/(1 L)²

= 21.2 atm

11e. The calculated pressure for ammonia is lower than for Argon because ammonia is a very polar molecule. Thus, the intermolecular attractions are greater, the molecules are "drawn together" and the pressure decreases.

12a. A sigma bond is formed whenever there is positive "end-on" overlap of orbitals.

12b. A Pi bond is formed by sideways overlap of p-orbitals

12c. The hybridization in CH₄ is sp³. In NH₃ it is sp³. (Both have four charge clouds - remember that NH₃ has one lone pair.)

12d. There are three resonance forms of the CO₃^-2 ion, differing in the position of the double bond.

12e. Antibonding molecular orbitals have a NODE (i.e. zero electron density) between the two atoms, whereas bonding molecular orbitals have positive electron density between the two atoms.