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

Solution:

1. Given the current 1.234 A and 1 mole of Al³⁺. Using the equations C=A x t, and moles of e^-= (C)/Faraday's constant. Since we have one mole of Al³⁺, it takes 3 moles of electrons to reduce Al³⁺. Using this value and plugging in A=1.234, we get 3=(1.234 x time)/96,486. Solving for time, we get t= 2.35 x 10⁵ seconds.

2. Given current 22.2 A and 1 mole of Ca²⁺. Using the equations C= A x t, and moles of elections = (C)/Faraday's constant. Since we have one mole of Ca²⁺, it takes 2 moles of electrons to reduce Ca²⁺. Using this value and plugging in A=22.2, we get 2=(22.2 x time)/96,486. Solving for time, we get t= 8.7 x 10³ seconds.

3. Given current 37.45 A, and 1 mole of Cr⁵⁺. Using equations C=A x t, and moles of electrons = (C)/Faraday's constant. Since we have one mole of Cr⁵⁺, it takes 5 moles of electrons to reduce Cr⁵⁺. Using this value and plugging in A=37.45, we get 5=(37.45 x time)/96,486. Solving for time, we get t=1.29 x 10⁴ seconds.

4. Given current 3.57 A, and 1 mole Au³⁺. Using equations C= A x t, and moles of electrons = (C)/Faraday's constant. Since we have one mole of Au³⁺, it takes 3 moles of electrons to reduce Au³⁺. Using this value and plugging in A=3.57, we get 3=(3.57 x time)/96,486. Solving for time, we get t=8.1 x 10⁴ seconds.

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.

Solution: Based on the given concentrations of [NH₃], the rate value stays the same regardless of the different NH₃ concentrations. That means that the overall order of the rate equation will be zero. The rate equation will end up to be Rate = k[NH₃]⁰. After plugging in the rate value and the NH₃ concentration values, we get k=1.5 x 10^-6 Ms^-1.

Q12.6.4

Define these terms:

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

S12.6.4

1. unimolecular reaction: A unimolecular reaction occurs when a molecule rearranges itself to produce one or more products. An example of this is radioactive decay, in which particles are emitted from an atom. Unimolecular reactions are often first-order reactions. The rate=k[A].

2. bimolecular reaction: A bimolecular reaction involves the collision of two particles. Bimolecular reactions are common in organic reactions such as nucleophilic substitution. The rate of reaction depends on the product of the concentrations of both species involved, which makes bimolecular reactions second-order reactions.

The rate=k[A][B].

3. elementary reaction: An elementary reaction is a single step reaction with a single transition state and no intermediates. A multi-step reaction is composed of multiple elementary reactions.

4. overall reaction: The combination of all the steps in a reaction. The overall reaction does not include the intermediates in the equation.

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?

Solution:

1. Tl-208 undergoes a β decay, which means that Tl gain a proton and become Pb. However, the total atomic mass does not change. Therefore Tl-208 becomes Pb-208 after a β decay.

2. From the question we know that t_1/2=3.1 min. And since all the decay is first order reaction, from the kinetics chapter, we learned that t_1/2=0.0693/k, which gives us k=0.02235. After finding out that k=0.02235, we plug that value into the first order rate equation, [A]_t=[A]₀e^-kt. Plugging in [A]_t=0.01 for 99% decay, and [A]₀=1, we solve for t. We find that it takes 206 mins for 99% of Tl to decay.

3. For this part, we are given t=60 mins for the first order rate equation. After plugging in [A]₀=1, k=0.02235, and t= 60 mins. We find [A]_t=0.26. From this, we say that 26% of 208Tl remains un-decayed after 1 hour.

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?

S20.3.4

Half cells are connected with a slat bridge to prevent charge buildup by allowing ions to exchange between the bridge and the half cell. A salt bridge is always necessary for a galvanic cell or else the charge build-up will result in the reaction ending and no more voltage will be created.

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

Solutions:

1. For half reaction 1, I don't agree with the proposed potential for the Cu²⁺(aq) + 2e^-→ Cu(s). I found that the potential of that reaction was E^o = -2.53 V.

2. For half reaction 2, I don't agree with the proposed potential for Ce⁴⁺(aq)+4e^-→Ce(s). I found that using the potentials of Ce⁴⁺+e^-→Ce³⁺ (E^o=1.44) and Ce³⁺+3e^-→Ce(s) (E^o=-2.336). We get the E^o=(1.44-2.336= -0.896 V).

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?

S14.1.1

Through studying of chemical kinetics, we can understand the reaction rate change based on time. However, with a balanced chemical equation, we cannot predict how fast a reactant is consumed to produce a product, we can only determine the rate of the reaction by conducting an experiment to empirically determine the rate.

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?

Solution:

Given: The order for both reactants are first order, thus the rate law=k[1-bromopropane][S₂O₃^2-].

[Initial Reaction Rate]:By plugging in a concentration of 0.4 M for both reactants, and rate constant of 8.05 x 10^-4 M^-1s^-1, we get the rate=1.29 x 10^-4 M/s.

[Reaction Rate for decreased concentration]: By plugging in a concentration of 0.2 M for both reactants and a rate constant of 8.05 x 10^-4 M^-1s^-1, we get the rate=3.22 x 10^-5 M/s.