Extra Credit 49

SOLUTION

Using the equation n=It/F we can determine the time taken to reduce 1 mole of each ion, where n is the number of moles of electrons, I is the current in Amps, t is the time in seconds, and F is Faraday's constant (~ 96500 Coulombs/mol). For convenience, 1 Coulomb = Amp x sec. Because we are trying to find the time is takes to reduce the ions, we can rewrite the equation as follows:

$$n_{e}=\frac{It}{F}$$

$$t=\frac{n_{e}F}{I}$$

1. 1 mol Al³⁺ x (3 mol e^-/1 mol Al³⁺) = 3 mol e^-

$$t=\frac{(3mole^{-})(96,485Cmol^{-1}e^{-1})}{1.234A}$$

$$t=\frac{(289600 A s)}{1.234A}$$

$$t=\frac{(234602.92 sec)(1 min)}{60 sec} = 3910.05 min$$

It will take 3910.05 minutes to reduce 1 mole of Al³⁺ at 1.234 Amps.

2. 1 mol Ca²⁺ x (2 mol e^- / 1 mol Ca²⁺) = 2 mol e^-

$$t=\frac{(2mole^{-})(96,485Cmol^{-1}e^{-1})}{22.2A}$$

$$t=\frac{(193000 A s)}{22.2 A}$$

$$t=\frac{(8693.69 sec)(1 min)}{60 sec} = 3144.89 min$$

It will take 144.89 minutes to reduce 1 mole of Ca²⁺ at 22.2 Amps.

3. 1 mol Cr⁵⁺ x (5 mol e^- / 1 mol Cr⁵⁺) = 5 mol e^-

^$$t=\frac{(5mole^{-})(96,485Cmol^{-1}e^{-1})}{37.45A} $$

$$t=\frac{(482500 A s)}{37.45 A}$$

$$t=\frac{(2883.84 sec)(1 min)}{60 sec} = 214.73 min$$

It would take 214.73 minutes to reduce 1 mole of Cr⁵⁺ at 37.45 Amps.

4. 1 mol Au³⁺ x (3 mol e^- / 1 mol Au³⁺) = 3 mol e^-

t=\frac{(3mole^{-})(96,485Cmol^{-1}e^{-1})}{3.57A}

$$t=\frac{(289500 A s)}{3.57 A}$$

$$t=\frac{(81092.44 sec)(1 min)}{60 sec} = 1351.54 min$$

It would take 1352.54 minutes to reduce 1 mole of Au³⁺at 3.57 Amps.

Payal: All solutions are correct and now in mathjax form.

SOLUTION

rate = k[NH₃], From the Table, we can see that the rate is constant despite the concentration of ammonia changing, so the rate must be independent of concentration, which means that it is a first order reaction.

k, the rate constant, can be found by plugging in some of the values from the table. Let's use the first concentration and the first rate to find the rate constant, k.

$$1.5x10^{-6}= k[1.0x10^{-3}]^{0} $$

$$1.5x10^{-6}\frac{M}{s}=k$$

$k=1.5x10^{-6}\frac{M}{s}​​​​​​​$, therefore we can write the rate equation as follows: $rate= 1.5x10^{-6}[NH_{3}]^{^{0}}A$

SOLUTION

1. unimolecular reaction - a reaction where one molecule exists on the reactant side of the chemical reaction equation
2. bimolecular reaction - a reaction where two molecules exist on the reactant side of the chemical reaction equation
3. elementary reaction - a reaction where there is only one step and does not have any intermediates
4. overall reaction - the complete reaction after all steps have been completed

SOLUTION

1. ²⁰⁸Tl → β + ²⁰⁸Pb because a β particle has an atomic number of -1 and a mass number of zero, so the ²⁰⁸Tl which has an atomic number of 81 changes to 82, which is Pb. The mass number does not change because beta particles do not add mass.

2. We can use the equations t_1/2=ln(2)/λ and N=N_oe^-λt to solve for how long it will take for 99.0% of a sample of pure ²⁰⁸Tl to decay.

The half like of ²⁰⁸Tl is 3.1 min. Therefore, 3.1 min = ln(2) / λ.

3.1 min ≈ 0.692 / λ

λ = 0.22 min^-1

We can plug that into the other equation, N=N_oe^-λt, assuming we start with 1 mole of ²⁰⁸Tl and end with 0.01 mol ²⁰⁸Tl. We can rearrange the equation to more easily solve for t (time).

N=N_oe^-λt

N/N_o = e^-λt

ln[N/N_o] = ln(e^-λt)

ln[N/N_o] = -λt x ln(e)

ln[N/N_o] = -λt

(ln[N/N_o]) / -λ = t

Thus,

(ln[0.01/1]) / -0.22 min^-1 = t

t = 20.60 min

It will take 20.60 minutes for ²⁰⁸Tl to decay 99.0%.

3. From our previous equations, we know the decay constant of ²⁰⁸Tl is 0.22 min. To find the percentage of sample remaining after 1.0 hr, we can use N=N_oe^-λt and assume that we start with an initial concentration of 1 mole. We first have to convert either time or the decay constant to the same unit.

t = 1.0 hr x (60 min / 1 hr) = 60 min. Then,

N = 1 mol x e^{-(0.22/min)(60min)}

N = 1 mol x e^-13.42

N = 1.49x10^-6 mol

(1.49x10^-6 mol / 1 mol initial) x 100 = 1.49x10^-4 %

After 1 hour, there is 1.49x10^-4% of ²⁰⁸Tl remaining after decay.

SOLUTION

The purpose of the salt bridge in the galvanic cell is to maintain charge balance when electrons are flowing from one substance to another. The electrons do not move along the salt bridge, rather they move along a wire and the salt bridge moves anions to the anode after having so many positively charged ions at that site. The salt bridge is necessary when one wants to keep the galvanic cell running, otherwise the flow of electrons and ions will stop.

SOLUTION

1. According to the 19.2 Table following the link above, the E° of the equation is 0.3419 V, not 0.68 V, so the proposed standard potential is incorrect.

2. To find the standard reduction potential for this equation, we must look at the values when the overall reaction is split up:

Ce⁴⁺(aq) + e^- ⇌ Ce³⁺(aq), E° = 1.44 V and

Ce³⁺ + 3e⁻⇌ Ce(s), E° = -2.336 V

From here, we can add the two standard reduction potentials because they both contribute to the same half-reaction.

So, E° = 1.44 V + (-2.336 V) = -0.896 V

Therefore, the given standard reduction potential is not correct.

SOLUTION

The information one can gain from studying the chemical kinetics of a reaction is its rate. The balanced equation does NOT provide any information about the mechanics of the reaction because there is no correlation between the rate order and the stoichiometry.

SOLUTION

The reaction is first order in 1-bromopropane and first order in S₂O₃²⁻, with a rate constant of 8.05 × 10⁻⁴ M⁻¹·s⁻¹. If you began a reaction with 40 mmol/100 mL of C₃H₇Br and an equivalent concentration of S₂O₃²⁻, what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be?

rate = k [C₃H₇Br][S₂O₃^2-]

rate = (8.05 × 10⁻⁴ M⁻¹·s⁻¹)[C₃H₇Br][S₂O₃^2-]

rate = (8.05 × 10⁻⁴ M⁻¹·s⁻¹)[40 mmol/100 mL][40 mmol/100 mL], however, 40 mmol/100mL = 0.40 M, so

rate = (8.05 × 10⁻⁴ M⁻¹·s⁻¹)[0.40M][0.40 M]

rate = 1.288x10^-4 M/s or 1.29x10^-4 M/s

and rate = (8.05 × 10⁻⁴ M⁻¹·s⁻¹)[20 mmol/100 mL][20 mmol/100 mL], where 20 mmol/100 mL = 0.20 M, so

rate = (8.05 × 10⁻⁴ M⁻¹·s⁻¹)[0.20 M][0.20 M]

rate = 3.22x10^-5 M/s

Therefore the answers given below are correct.