Extra Credit 45

Q17.6.4

Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?

A17.6.4

Since B corrodes when with A, B oxidizes in comparison to A, or B is the donor of electrons (anode) and A is the cathode. When A and C are together, C is the cathode and A is the anode since C is the metal that corrodes. This means that A can donate its electrons to C, but accepts electrons from B.

Hence, since donors are higher up (or have positive E values) on the standard reduction potentials chart, B would be highest since it donates to A. A would be next on the chart (or between metals B and C) since it donates electrons to C, causing itself (metal A) to oxidize/corrode.

On the standard reduction potentials table, B would have the greatest reduction potential > A > C.

When B and C come into contact, since B is higher up (with larger E value) than C, it would be the donor of electrons, or would lose electrons and become oxidized. This oxidation would lead to metal B becoming corroded, so metal B corrodes when in contact with metal C.

Q12.3.8

The rate constant for the radioactive decay of 14C is 1.21 × 10−4 year−1. The products of the decay are nitrogen atoms and electrons (beta particles):

C146⟶614N+e−C146146Ne

C146⟶614N+e−

rate=k[C146]
ratekC146

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 × 10−9 M?

A12.3.8

Based on the given information, you can plug that into the given rate formula and solve for the instantaneous rate of production of N atoms.

rate= k[¹⁴₆C]

k= 1.21 * 10^-4 year^-1

C= 6.5 * 10^-9 M

rate = (1.21 * 10^-4 year^-1 ) * (6.5 * 10^-9 M)

= (7.87 * 10^-13) M/year

Q12.5.17

Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first A+BC⟶AB+CABCABC reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?

A12.5.17

C bounces off of B and then goes back to being bounded to B when A is shot as a straight shot. When C leaves B, A and B are bonded, but then when C comes back to B, A leaves (shoots back down, away from the B-C molecule) and C and B are bonded together again and form a molecule.

At an angled shot, A binds to B and they become one molecule together, C bounces off and becomes a single atom. C and the A-B molecule move around (they have energy from their collision). The angled shot enables A to bind to B stronger because C gets hit off in another direction, and cannot come back to the A-B molecule. Even when it eventually does find it and hits it, it does not have enough energy to break the A-B bond and bind to B instead of A.

Q21.4.12

rite a nuclear reaction for each step in the formation of ²⁰⁸₈₂Pb from ²²⁸₉₀Th, which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.

A21.4.12

²²⁸₉₀Th → ²²⁴₈₈Ra + ⁴₂He

²²⁴₈₈Ra → ²²⁰₈₆Rn + ⁴₂He

²²⁰₈₆Rn → ²¹⁶₈₄Po + ⁴₂He

²¹⁶₈₄Po → ²¹²₈₂Pb + ⁴₂He

²¹²₈₂Pb → ²¹²₈₃Bi + ⁰_-1e

²¹²₈₃Bi → ²¹²₈₄Po + ⁰_-1e

²¹²₈₄Po → ²⁰⁸₈₂ Pb + ⁴₂He

Q20.2.16

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

1. Zn(s) + HCl(aq) →
2. 3HNO3(aq) + AlCl3(aq) →
3. K2CrO4(aq) + Ba(NO3)2(aq) →
4. Zn(s) + Ni2+(aq) →

A20.2.16

a) This is a precipitation reaction because Zinc Chloride (which would be a solid since zinc is insoluble according to solubility rules) is produced and Hydrogen gas.

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2 (s)

b) There is no reaction because there is not an acid and base, so it is not an acid-base reaction, no precipitate is formed (HCl is aq and AlNO3 is always soluble- again, this is due to solubility rules) so it's not a precipitation reaction, and the oxidation numbers of the elements do not change in the equation, so it is not a redox reaction.

3HNO3(aq) + AlCl3(aq) → Al(NO3)3 (aq) + 3HCl (aq)

c) This is a precipitation reaction because chromates like BaCrO4 are insoluble. Again, not dealing with acids and bases and not with changing oxidation numbers.

K2CrO4(aq) + Ba(NO3)2(aq) → BaCrO4 (s) + 2KNO3 (aq)

d) This is a redox reaction because there is a change in the oxidation number of Zn and Ni.

To find the products, split the reaction into two half reactions:

Zn(s) → Zn2+(aq) + 2e-

Ni2+(aq) + 2e- → Ni(s)

Since there are 2e- in the reactants side of the second half reaction and the same number of e- at the products side of the first half reaction, they cancel out and you do not have to multiply the reactions by anything. So, you get:

Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)

Q20.5.11

The chemical equation for the combustion of butane is as follows:

C4H10(g)+132O2(g)→4CO2(g)+5H2O(g)C4H10g132O2g4CO2g5H2Og

This reaction has ΔH° = −2877 kJ/mol. Calculate E°cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?

A20.5.11

Since there is a circle next to delta H, E cell, etc., it means "standard," in that you have to find these values at standard conditions.

K (which is used to find ΔG° with the equation ΔG°=-RT*ln Keq) is 1 at standard conditions (since gas pressures = 1 bar and K= concentrations/gas pressures of products/ concentrations/gas pressures of reactants).

Plugging in K=1 into the equation ΔG°=-RT*ln Keq, we find that ΔG°=0. This means that the reaction is at equilibrium and is neither spontaneous nor non-spontaneous.

To calculate E°cell, use the equation: ΔG°=-nFE°cell, where n is the number of electrons and F is Faraday's constant. Since ΔG°=0, E°cell=0.

To find the change in entropy, or ΔS°, use the equation: ΔG°= ΔH° - TΔS°, where T is temperature in kelvin.

So, the equation (after plugging in given ΔH°, ΔG°, and T values) would be 0 = −2877 kJ/mol - 298*ΔS°. Solving for ΔS°, we get that the change in entropy= -9.65 kJ/mol, which is equivalent to about -9650 J/mol (technically -9654 but sig figs). ΔS° is usually written as J/mol rather than kJ/mol.

Q24.6.7

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

a. [TiCl6]3−

b. [CoCl4]2−

A24.6.7

a. There are 6 ligands bonded to the central atom, so the complex is octahedral.

Since the complex has a negative 3 charge and the oxidation number of Cl is -1, Cl6 gives the complex a -6 charge and Ti must be +3 for the complex to be -3 in total.

Ti is the second transition metal in the first row of transition metals (atomic number 22). Ti3+ has the electron configuration of [Ar] d1 which means that there is one electron in the d orbital. It it unpaired (since there are no other electrons).

Since Cl is pretty low in the Spectrochemical Series, it is a high energy ligand that would cause the complex to be high spin. However, since there is only one electron, whether it is high or low spin does not affect the number of paired/unpaired electrons.

To see how the electron configuration would be for octahedral complexes, go to http://wps.prenhall.com/wps/media/ob...1/blb2405.html and scroll down to Electron Configurations in Octahedral Complexes

b. There are 4 ligand bonded to the central atom, so the complex is either tetrahedral or square planar. In this case, it is tetrahedral. Since the overall charge is -2, and 4 Cl's give a -4 charge, Co's charge must be +2.

Co is the 7th transition metal in the first row of transition metals. Co2+ has an electron configuration of [Ar] d7 because transition metals lose electrons in the s orbital first, before they begin to lose in the d orbital.

As explained above, because of Cl this would be a high spin complex. In a high spin complex, all shells are half filled (not where the bottom ones are completely filled first before moving onto the top shells). Since there are 7 electrons and 5 shells, only two shells will be completely filled and there will be 3 unpaired electrons.

The image below is NOT how the electrons should look, but shows the shells. The only thing is that one more of the tetrahedral upper shell should have an unpaired electron (there are 8 electrons in the image instead of 7). You can ignore the sq planar side of the image since this complex was tetrahedral.

NOTE: I tried to attach an image here but could not do it. If you go to http://wps.prenhall.com/wps/media/ob...1/blb2405.html and scroll down to SAMPLE EXERCISE 24.9, you can see the image.