Extra Credit 48

Q12.3.11

Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:

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

S12.3.11

First, identify the rate equation:

rate = k[C₂H₅OH]^x

As shown by the data table, regardless of how the concentration of acetaldehyde changes, the rate remains constant. From this one can infer that the rate constant, k, is equal to the rate of the reaction in each trial and that the order with respect to the concentration of acetaldehyde is zero.

rate = k[C₂H₅OH]⁰

2.0 x 10^-2 mol/(L x h) = k(1)

k = 2.0 x 10^-2 mol/(L x h)

Lastly, to determine reaction order, the respective orders of all of the products (in this case, just acetaldehyde) are added together.

Reaction Order: 0.

Good job, I agree!

Q12.6.3

Phosgene, COCl₂, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by:

1. Write the overall reaction equation.
2. Identify any reaction intermediates.
3. Identify any catalysts (the question asked again for intermediates).

S12.6.3

First, we write the overall reaction equation by adding the two equations together, removing any compound that appears on both the product and the reactant side.

Cl + CO --> COCl - Reaction 1

COCl + Cl₂ --> COCl₂ + Cl - Reaction 2

Cl + CO + COCl + Cl₂ --> COCl + COCl₂ + Cl

CO + Cl₂ --> COCl₂

Reaction intermediates are compounds that are produced within the reaction but then reacts further to give the observed products of a chemical reaction. They typically provide a more favorable pathway to create the end product. In order to determine what the intermediates are, we look for what product is created in the reaction but then is used up later (and in this case, crossed out).

Intermediate: COCl

The reason why Cl however, is not an intermediate is that it is present in the beginning of the reaction 1. The catalyst, on the other hand, is present throughout the entire reaction, being Cl₂. This compound speeds up the reaction to make it more favorable but is not consumed in the end.

Catalyst: Cl₂

Q14.4.7

Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:

What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?

S14.4.7

In order to find the reaction order with respect to benzoyl peroxide, we compare the concentrations of two reactions with the rate of the two reactions. In this case, lets look at experiment 1 and 3. Experiment 1 has double the concentration of experiment 3, from this, we can set up the equation:

( [Exp 1 BP] / [Exp 3 BP] )^x = (Exp 1 rate) / (Exp 3 Rate)

Where x = the respective order and [BP] is the concentration of Benzoyl Peroxide. From this we can elaborate:

( [1.00M BP] / [0.5 BP] )^x = (2.22 x 10^-4 M/S) / (1.12 x 10^-4 M/S)

2^x = 1.98 (approximately 2)

2^x = 2

x = 1

Therefore, the reaction order with respect to benzoyl peroxide is 1.

Now, we find the rate law by writing the rate equation and elaborating the equation for an experiment trial.

rate = k[BP]¹

@ Exp 1

2.22 x 10^-4 M/S = k[1]

k = 2.22 x 10^-4 M/S

Good job!

Q17.7.2

What mass of each product is produced in each of the electrolytic cells of Q17.7.1 if a total charge of 3.33 × 10⁵ C passes through each cell? Assume the voltage is sufficient to perform the reduction.

S17.7.2

The products in Q17.7.1 are

1. CaCl₂
2. LiH
3. AlCl₃
4. CrBr₃

In order to identify the mass of each product produced, the following stoichiometry can be used:

3.33 x 10⁵C x (1 mole of electrons / 96485 C) x ( 1 mole of element in question / moles of electrons transferred ) x ( molar mass of element in question / 1 mol of element in question) = y grams of element in question.

Lets find the mass of Ca produced, for example. The reduction of Ca²⁺ is:

Ca²⁺(aq) + 2e^- --> Ca(s)

Plugging in the respective values into the equation above we get:

3.33 x 10⁵C x (1 mole of electrons / 96485 C) x ( 1 mole of Ca²⁺ / 2 moles of electrons transferred ) x ( 40.078g Ca²⁺ / 1 mol of element in question) = 69.1 grams of Ca.

With this, the grams of each element produced are:

1. mass of Ca=69.1g mass of Cl2=122g
2. mass of Li=23.9g mass of H2=3.48g
3. mass of Al=31.0g mass of Cl2=122g
4. mass of Cr=59.8g mass of Br2=276g

Q20.3.3

What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?

S20.3.3

A galvanic cell transforms the energy released through a spontaneous redox reaction into electrical energy that can be used to perform work. A galvanic cell converts chemical energy into electrical energy. An electrolytic cell converts, on the other hand, electrical energy into chemical energy through a nonspontaneous redox reaction. If you were to generate electricity, you would use a galvanic cell.

Q20.5.14

Calculate the pH of this cell constructed with the following half reactions when the potential is 0 at 25 °C

MnO^-4 + 8H⁺ +5e^-→Mn²⁺+4H₂O

Au³⁺ + 3e^-→Au(s)

Under this condition, the concentrations of other species in the cell are:

• 0.36 M: MnO−4
• 0.004 M: Au3+
• 0.001 M: Mn2+

S20.5.14

First, we add the two equations together and balance them by multiplying each equation to equal the same amount of electrons and flip the first equation due to Manganese's higher oxidation potential (and lower reduction potential). Doing so will yield the following equation:

3Mn²⁺+ 12H₂O + 5Au^3+-→ 5Au(s) + 3MnO^-4 + 24H⁺

(15 electrons transferred)

Now that we have the balanced equation, we can then apply the Nerst equation. With the information given we get:

E_cell = (RT/nF)lnK

0 ={ [ (8.314 J / mol x k ) (298k) ] / [ (15 moles of electrons)(96485 J) ] } lnK

.001712 = lnK

K = 1.0017

Now that we have the equilibrium constant, we can then elaborate the [H⁺] in this reaction.

K = ( [H⁺]²⁴[MnO^-4]³ ) / ( [Mn²⁺]³[Au³⁺]⁵ )

Plugging in the values given we get:

1.0017 = ( [H⁺]²⁴[.36M]³ ) / ( [.001M]³[.004M]⁵ )

Which simplifies to:

[H⁺]²⁴ = 2.199 x 10^-20M

[H⁺] = 0.15168

Then to get pH of the cell, we take the negative log of the concentration of hydrogen (or hydronium) ions.

-log[0.15168] = 0.8191

pH = 0.8191

Correct Formula: \(E_{cell}=E^\circ_{cell}-\frac{0.0592V}{n}logQ\)

Correct Solution: pH=0.71

Q21.4.15

²³⁹Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the ²³⁹Pu present today will be present in 1000 y?

S21.4.15

In order to find out the fraction of the ²³⁹Pu present, we first need to find out how much of the half-life ²³⁹Pu went through in 1000 years. To do this we simply divide the time given by the half-life of ²³⁹Pu.

1000 years / 24, 0000 years = .004 of ²³⁹Pu's half life

Then, we use that value as the amount of "time" passed in our half-life equation.

(1/2)^.004 = .973 x 100% = 97.3% of ²³⁹Pu present.

Good job!

Q24.6.9

The ionic radii of V²⁺, Fe²⁺, and Zn²⁺ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.

S24.6.9

The ionic radii of the elements given are so similar due to the fact that the valence electrons in each element are shielded from attraction to the nucleolus. Since the positive attraction from the nucleolus is inhibited, when electrons are added to the inner 3d orbital (an orbital closer to the positive attraction), it pushes the outer 4s orbital away from the nucleolus with relatively the same attractive force the nucleolus exerts onto the 4s orbital. As a result, the atomic and ionic radii of the transition metals listed are almost constant.