Extra Credit 49

Q17.7.3

How long would it take to reduce 1 mole of each of the following ions using the current indicated? Assume the voltage is sufficient to perform the reduction.

1. Al³⁺, 1.234 A
2. Ca²⁺, 22.2 A
3. Cr⁵⁺, 37.45 A
4. Au³⁺, 3.57 A

In order to solve for t, time, we must know the relation between current and moles of electrons in order to use the equation

where n is the moles of electrons, F is the faraday's constant and A is the current in amps which is Coulomb per second.

Because we need to the number of moles of electrons to complete the equation, although not necessary, but it's suggested to write out balanced equations.

and from here on, we just plug in numbers into the equation.

Balanced equations	n, moles of electrons	F, Faraday's constant 96,485	A in Coulombs	t in seconds
	3		1.234	234,566 s
	2		22.2	8692.34 s
	5		37.45	12881.8 s
	3		3.57	81079.8 s

Q12.3.12

Under certain conditions the decomposition of ammonia on a metal surface gives the following data:

Determine the rate equation, the rate constant, and the overall order for this reaction.

Since the concentration of the ammonia does not affect the rate, this equation is a zero order

or alternatively, we can use the rate equation to determine this

Rate = k[NH₃]ⁿ

rate 1: 1.5x10^-6=k[1x10^-3]ⁿ

rate 2: 1.5x10^-6=k[2.0x10^-3]ⁿ

^{Solve for k from one of the equations, in this case we will use rate 2}

^{Subing this to the rate 1 equation will grant us the following}

^{dividing the constants will give us the solution}

and this is only possible if n = o.

rate equation	rate constant	overall order
Rate=k[NH3] or Rate=k	Rate=k,	overall order = o

Q12.6.4

Define these terms:

1. unimolecular reaction
2. bimolecular reaction
3. elementary reaction
4. overall reaction

Q21.4.16

The isotope ²⁰⁸Tl undergoes β decay with a half-life of 3.1 min.

1. What isotope is produced by the decay?
2. How long will it take for 99.0% of a sample of pure ²⁰⁸Tl to decay?
3. What percentage of a sample of pure ²⁰⁸Tl remains un-decayed after 1.0 h?

1. ²⁰⁸Tl + ⁰ß —> ²⁰⁸Pb

the beta decay of Tl with 81 protons will result in Pb with 82 protons.

2. To calculate the the time for 99% of pure Tl to decay, we must fine the k constant using the integrated form of the 1st order reaction. This is a 1st order because it's a unimolecular reaction. Don't forget to convert the units of half life into seconds.

k= .6931/(3.1*60)

k = .003727 s^-

with this equation, we can then use the equation to figure out the time needed to decay 99% of Tl which means 1 percent left.

n, convert from percent to decimal	, the initial or 100%	k	solution
1%=0.01	100% = 1	.003727	1235.62 seconds

We can then use the same equation to figure out the percentage left after 1h. This requires raising both sides of the equations to be the power of e. After some algebra, this should be the general solution.

, the initial or 100%	k	t,time in seconds	n
100% = 1	.003727	1h = 3600 seconds	.00000149 or .000149 percent left.

Q20.3.4

What is the purpose of a salt bridge in a galvanic cell? Is it always necessary to use a salt bridge in a galvanic cell?

The purpose of the salt bridge is to balance the charges between the cells. If one cell is being oxidized, the excess electrons will halt the reaction if there are no more materials to accept the electrons. The salt bridges are always necessary in a galvanic cell.

Q20.5.15

Based on Table 19.2 and Table P2, do you agree with the proposed potentials for the following half-reactions? Why or why not?

1. Cu²⁺(aq) + 2e⁻ → Cu(s), E° = 0.68 V
2. Ce⁴⁺(aq) + 4e⁻ → Ce(s), E° = −0.62 V

Table 19.2 suggests that

which means the 1st assumption is false. However, the second assumption is correct.

Because Ce³⁺ is being oxidized, we can reverse the equation

to find that the oxidation reaction of the molecule Ce³⁺ to Ce has an E^o of 2.336

Using the equation E^o= E(Cathod) - E(anode), we get E^o=1.72-2.336 which is equal to -0.62.

Q14.1.1

What information can you obtain by studying the chemical kinetics of a reaction? Does a balanced chemical equation provide the same information? Why or why not?

By studying the kinetics of a reaction, we know how fast a reaction can go. By doing so allows us to expect the outcome of a reaction. Things such as half life and rate of reaction is derived from the studies of kinetics.

Balanced chemical equation does not provide the same information. A balanced chemical equation can tell us the concentration of the reactants and products and what will form, what will be lost. The equation however does not contain information that allows chemists to derive the rate of reaction such as the orientation of the molecules, the frequency they will collide and much more.

Q14.4.8

1-Bromopropane is a colorless liquid that reacts with S₂O₃²⁻ according to the following reaction:

C3H7Br+S2O2−3→C3H7S2O−3+Br−C3H7Br+S2O32−→C3H7S2O3−+Br−

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?

To determine the initial reaction rate. We must recall the rate law equation and substitute the reactants into the equation.

Rate=k[C₃H₇Br]¹[S₂O₃^2-]¹

8.05x10^-4M⁻¹·s⁻¹=k[.4M][.4M]

k=.005031 M·s^-1

if the concentration was lowered to 20 mmol/100ml

8.05x10^-4M⁻¹·s⁻¹=k[.2M][.2M]

k= .020125 M·s^-1