Extra Credit 1

Q17.1.1

If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?

S17.1.1

Given: Current (in Amps) and Time

Asked For: Charge (in Coulombs)

Strategy:

A. Apply the definition of Amps and use a series of stoichiometric conversions.

1 Amp = 1 Coulomb / Second , so 2.5 Amps = 2.5 C/s

\[35 minutes \times \frac{60 seconds}{1 minute} \times \frac{2.5 C}{1 second} = 5.3 \times 10^3 C\]

Q17.7.4

A current of 2.345 A passes through the cell shown in Figure for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?)

S17.7.4

Given: Current (in Amps), Time, Total Pressure, Temperature, and cell equation.

Asked For: Volume of hydrogen gas produced

Strategy:

A. Start by finding the number of moles of hydrogen gas produced using stoichiometric conversions and the balanced equation for the cell.

B. Use PV=nRT to relate the moles found in part A and the pressure to the volume of hydrogen gas.

\[45 minutes \times \frac{60 seconds}{1 minute} \times \frac{2.345 C}{1 second} \times \frac{1 mole e^-}{96 500C} \times \frac{2 mole H_2}{4 mole e^-} = 0.0328 \quad moleH_2\]

\[PV = nRT\]

\[= 1(V) = (.0328)(.08206)(298.15)\]

\[= V = 0.79L\]

Q.19.2.1

1. [Pt(H₂O)₂Br₂]
2. [Pt(NH₃)(py)(Cl)(Br)] (py = pyridine, C₅H₅N)
3. [Zn(NH₃)₂Cl₂]
4. [Zn(NH₃)(py)(Cl)(Br)]
5. [Ni(H₂O)₄Cl₂]
6. [Fe(en)₂(CN)₂]⁺ (en = ethylenediamine, C₂H₈N₂)

S.19.2.1

Given: Formula of complex

Asked For: Coordination number

Strategy:

A. Define coordination number.

B. Identify polydentates if any.

Coordination number - the number of sites on the central atom that surrounding ligands are bound to

1. Since aquo- and bromo- ligands are mondentate, there are a total of 4 ligands bound to the central atom. Coordination number: 4
2. Since all of these ligands are monodentate, there are 4 ligand binding sites on the central atom. Coordination number: 4
3. Since neither the amine nor the chloro groups are polydentate, there are a total of 4 ligands bound to the central atom. Coordination number: 4
4. As in part 2, these ligands are all monodentate. The identity of the transition metal does not effect the coordination number, hence the coordination number is still 4.
5. Since both aquo- and chloro- ligands are monodentate, there are a total of 6 ligand binding sites. Coordination number: 6
6. Since enthyline diammine ligand is bidentate and cyano- ligand is monodentate there are a total of 6 ligand binding sites. Coordination number: 6

\[(2 \times 2) + 2 = 6\]

Q12.3.13

Nitrosyl chloride, NOCl, decomposes to NO and Cl₂.

2NOCl(g)⟶2NO(g)+Cl2(g)

Determine the rate equation, the rate constant, and the overall order for this reaction from the following data:

S12.3.13

Given: Experimentally determined rates with corresponding concentrations.

Asked For: Rate Law

Strategy:

A. Determine the reaction order with respect to the single reactant, NOCl, using 2 data points.

B. Determine the rate law constant, k, using one of the data points and the information found in part A.

Using the first and second data points we get: \[8.0 \times 10^-10 = k(.10)^x\] and \[3.2 \times 10^-9 - k(.20)^x\] Dividing the two equations we get: \[.25 = (0.5)^x\] \[x = 2\]

So overall, the general rate law is: \[k[NOCl]^2\] Plugging in the first data point to this equation: \[8.0 \times 10^-10 = k(.10)^2\] \[k = 8.0 \times 10^-8 M^-1 s^-1\]

S12.6.5

What is the rate equation for the elementary termolecular reaction A+2B⟶productsA+2B⟶products? For 3A⟶products3A⟶products?

S12.6.5

Given: Elementary reactions

Asked For: General Rate Law

Strategy:

A. Recognize the relationship between stoichiometric coefficients and the reaction order for elementary reactions.

Since these are elementary reactions, the stoichiometric coefficients are equivalent to the reaction orders for each reactant.

For \(A+2B⟶\ce{products}\), the stoichiometric coefficient for A is 1 while that of B is 2. Therefore, k[A]¹[B]² = k[A][B]².

For \(3A⟶\ce{products}\), the stoichiometric coefficient for A is 3. Therefore, k[A]³.

Q21.4.17

If 1.000 g of \[_{88}^{226}Sr\] produces 0.0001 mL of the gas \[_{86}^{222}Sr\] at STP (standard temperature and pressure) in 24 h, what is the half-life of ²²⁶Ra in years?

S21.4.17

Given: Mass of reactant, volume of product, temperature and pressure conditions, time elapsed

Asked For: Half Life

Strategy:

A. Convert the moles of product to grams of product.

B. Figure out the mass of reactant remaining after 24 hours.

C. Calculate k, the decay constant.

Converting to moles: \[(1 atm)(.0000001 L) = n(.08206)(273.15 K)\] \[= n = 4.46 \times 10^-9 moles\] \[4.46 \times 10^-9 moles \times \frac{222g}{1mole} = 9.90 \times 10^-7 g\] Mass reactant remaining: \[1.00 - 9.90 \times 10^-7 = .999g\] Calculating k: since \[k = \frac{-1}{t} ln \frac{A}{A0}\] \[k= \frac{-1}{24 hours} ln \frac{.999}{1} = 4.13 \times 10^-8\]

Calculating Half Life: \[t1/2 = \frac{.693}{k} = \frac{.693}{4.13 \times 10^-8} = 16792834 hours\] Converting hours to years: \[16792834 hours \times \frac{1 day}{24 hours} \times \frac{1 year}{365 days} = 1917 years\]

Q20.3.5

One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K⁺ and Cl⁻ ions do. What would happen if the diffusion rates of the anions and cations differed significantly?

S20.3.5

Ions of similar rates of diffusion are needed to prevent discrepancy in the rate at which the electrodes can be neutralized. If ions with significantly different rates of diffusion were used, a charge will build up on one electrode more than the other. This decreases the amount of electrons that can flow through the wire.

Q20.5.16

For each reaction, calculate E°_cell and then determine ΔG°. Indicate whether each reaction is spontaneous.

1. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
2. K₂S₂O₆(aq) + I₂(s) → 2KI(aq) + 2K₂SO₄(aq)
3. Sn(s) + CuSO₄(aq) → Cu(s) + SnSO₄(aq)

S20.5.16

Given: Overall reaction

Asked For: E°_cell and ΔG°

Strategy:

A. Differentiate between the reduction and oxidation reaction using the standard reduction potential.

B. Plug in values to calculate E°_cell, and use E°_cell to calculate ΔG°.

1. Looking at the two reduction potentials that make up the overall reaction:

Na⁺(aq) + e^- ⇌ Na(s) E°=–2.713 and 2H₂O(l) + 2e^- ⇌ H₂(g) + 2OH^-(aq) E°=-0.828

Since Na(s) is getting oxidized to form Na⁺(aq), it is the anode. Therefore, H₂O(l) is the cathode.

\[E°cell = E°cathode-E°anode = -0.828-(-2.713) =1.885V\]

Calculating ΔG°:

First, balance the redox reaction, paying careful attention to charge.

Oxidation: 2(Na(s) ⇌ Na⁺(aq) + e^-) → 2Na(s) ⇌ 2Na⁺(aq) + 2e^-

Reduction: 2H₂O(l) + 2e^- ⇌ H₂(g) + 2OH^-

Overall: 2Na(s) + 2H₂O(l) ⇌ 2Na⁺(aq) + H₂(g) + 2OH^-

Therefore, a total of 2 electrons are transferred in this process. (n = 2)

Then, apply this and the value calculated above to find the ΔG°.

\[ΔG° = nFE° = (2 \quad moles \quad e^-)(96,486J/(V⋅mol))(1.885V) = 363752.22 J\]

2. Cathode/Oxidation: 2I^-⇌ I₂(s) + 2e^- (aq)

Anode/Reduction: 2SO₄²⁻ + 4H⁺ + 2e⁻ ⇌ S₂O₆²⁻ + 2H₂O(l)

Overall: 2SO₄²⁻ + 4H⁺ + I₂(s) ⇌ S₂O₆²⁻ + 2H₂O(l) + 2I^-(aq)

\[E°cell = E°cathode-E°anode = 0.5355-(-0.25) =0.7855V\]

Since n=2 in this reaction:

\[ΔG° = nFE° = (2 \quad moles \quad e^-)(96,486J/(V⋅mol))(0.7855V) = 151579.506 J\

3. Cathode/Oxidation: Sn(s) ⇌ Sn²⁺ + 2e⁻

Anode/Reduction: 2(Cu²⁺(aq) + e^- ⇌ Cu⁺(aq)) → 2Cu²⁺(aq) + 2e^- ⇌ 2Cu⁺(aq)

Overall: Sn(s) + CuSO₄(aq) → Cu(s) + SnSO₄(aq)

\[E°cell = E°cathode-E°anode = -0.14-(0.3419) =-0.4819V\]