Extra Credit 32

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Jonathan Nguyen B08

Q17.4.5

Use the data in Table P1 to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.

AgCl(s)⇌Ag⁺(aq)+Cl⁻(aq)
CdS(s)⇌Cd²⁺(aq)+S²⁻(aq) at 377 K
Hg²⁺(aq)+4Br⁻(aq)⇌[HgBr4]²⁻(aq)
H2O(l)⇌H⁺(aq)+OH⁻(aq) at 25 °C

All of these problems follow the same procedure. We need to find E°_cellby the equation

E°_cell=E°_cathode- E°_anode

The cathode is where reduction happens and the anode is where oxidation happens.

Then, we need to find the equilibrium constant by the equation

K=e^{nFE°_cell/RT}

E°_anode: Ag⁺(aq) + e^-⇌ Ag (s) E˚ = .7996

E°_cathode: AgCl (s) + e^- ⇌ Ag (s) + Cl^- E˚ = .2223

E°_cell= -.5773

K=e^{(1)(96485)(-.5773)/(8.314)(298)}=1.72 x 10^-10

ANSWER: 1.72 x 10^-10

E°_anode: Cd²⁺(aq) + 2e^- ⇌ Cd(s) E˚ = -.4030

E°_cathode: S(s) + 2e^-⇌ S^2- (aq) E˚ = -.407

E°_cell= -.004

K=e^{(2)(96485)(-.004)/(8.314)(377)}= .7817

ANSWER: .7817

E°_anode: Hg(l) + 4Br^-(aq) ⇌ [HgBr₄]^2-(aq) + 2e^- E˚ = -.4030

E°_cathode: Hg²⁺(aq) + 2e^- ⇌ Hg(l) E˚ = .7973

E°_cell= 1.2003

K=e^{(2)(96485)(1.2003)/(8.314)(298)}= 3.99 x 10⁴⁰

ANSWER: 3.99 x 10⁴⁰

4. This is not a redox reaction.

ANSWER: 1 x 10^-14; This is the equilibrium constant for water.

Q12.1.5

A study of the rate of the reaction represented as 2A⟶B gave the following data:

Time (s)	0.0	5.0	10.0	15.0	20.0	25.0	35.0
[A] (M)	1.00	0.952	0.625	0.465	0.370	0.308	0.230

Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.
Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate?
Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.

1. The average rate of disappearance of A can be found using the equation

$$average \; rate=\frac{Δ[concentration]}{Δt}$$

For 0s to 10s, we do

$$average \; rate=\frac{.625-1}{10-0}$$

= .0375 M/s

For 10s to 20s, we do

$$average \; rate=\frac{.370-.625}{20-10}$$

= .0255 M/s

All you have to do is essentially find the slope between the two points.

ANSWER: .0375 M/s and .0255 M/s

2. To solve this problem, graph the points.

Screen Shot 2017-06-08 at 9.42.05 PM.png

ANSWER: By looking at the graph, estimated around .05 M/s

3. From the equation given in the problem, the rate of reaction is

$$rate=\frac{(-1)(Δ[A])}{(2)(Δt)}=\frac{Δ[B]}{Δt}$$

We use the value in #1 for Δ[A].

(-.0375 M/s) * (-1/2) = .01875 M/s

ANSWER: .0188 M/s

We follow the same procedure to find the instantaneous rate of formation, but we use the answer in #2.

(.05 M/s) * (1/2) = .025 M/s

ANSWER: .025 M/s

Q12.5.3

What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?

ANSWER: Activation energy is the minimum amount of energy the reacting species must posses in order for the chemical reaction to take place. In other words, this energy is required in order for the reaction "to go." The activation energy of a chemical reaction is the difference between the energy of the activated complex and the energy of the reactants.

Q21.3.7

The mass of the atom F919F919 is 18.99840 amu.

Calculate its binding energy per atom in millions of electron volts.
Calculate its binding energy per nucleon.

First, we must find the mass of fluorine. There are 9 protons, 10 neutrons, and 9 electrons.

Mass = (mass of proton * 9) + (mass of neutron * 10) + (mass of electron * 9)

Mass = (9 * 1.00728) + (10 * 1.008) + (9 * .00055)

= 19.15 amu

Then, subtract this mass by the mass given in the problem.

19.15 amu - 18.99840 amu = .1516 amu

Convert from amu to kg. 1 amu = 1.6605 x 10^-²⁷ kg

Then, find the energy by using the equation

E = mc²

where m is mass and c is the speed of light.

E = (2.517 x 10^-²⁸) * (2.9979 x 10⁸ m/s)²= 2.262 x 10^-11 J

Convert from J to MeV. 1 MeV = 1.602 x 10^-13 J

1. ANSWER: 141.22 MeV/atom

To calculate binding energy per nucleon, divide by number of nucleons, which was 19.

2. ANSWER: 7.432 MeV/nucleon

Q20.2.3

In each redox reaction, determine which species is oxidized and which is reduced:

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)
BrO₃⁻(aq) + 2MnO₂(s) + H₂O(l) → Br⁻(aq) + 2MnO₄⁻(aq) + 2H⁺(aq)

In order to determine which species is oxidized and which is reduced, we have to get the oxidation number of each species in the reactants and products and compare them. Oxidation is the loss of electrons, while reduction is the gain of electrons. (OIL RIG: Oxidation Is Loss and Reduction Is Gain)

In the first problem, the charge of Zn goes from 0 to +2. The charge of H goes from +1 to 0.

ANSWER: Zn is oxidized while H is reduced.

In the second problem, the charge of Cu goes from 0 to 2. The charge of N goes from +5 to +4.

ANSWER: Cu is oxidized while N is reduced.

In the third problem, the charge of Br goes from +5 to -1. The charge of Mn goes from +4 to +7.

ANSWER: Mn is oxidized while Br is reduced.

Q20.4.22

Calculate E°_cell and ΔG° for the redox reaction represented by the cell diagram Pt(s)∣Cl₂(g, 1 atm)∥ZnCl₂(aq, 1 M)∣Zn(s). Will this reaction occur spontaneously?

From the cell diagram, we get the half reactions:

Zn²⁺(aq) + 2e^- ⇌ Zn (s) E˚= -.7618

Cl2(g) + 2e− ⇌ 2Cl− E˚= 1.396

The first half reaction is the cathode reaction because of the placement in the cell diagram.

The equation for E°_cell is:

E°_cell=E°_cathode- E°_anode

E°_cell=(-.7618)-(1.396) = -2.1578

ANSWER: E°_cell=-2.1578

To convert to ΔG°, we use the equation

ΔG°=-nFE°_cell= (-2)(96485 J/V*mol)(-2.1578 V) = 416390.7 J/mol

ANSWER: ΔG°=416390.7

This reaction will not occur spontaneously. In order for the reaction to occur spontaneously, ΔG° must be negative and E°_cellmust be positive.

Q20.9.7

What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl₂ is electrolyzed using a current of 12.4 A for 1.0 h?

We must first find the number of moles of electrons. We start with the equation

$$n=\frac{It}{F}$$

where I=current (amps), t=time (secs), F=faraday's constant (96500 C/mole).

$$n=\frac{(12.4 \; A)(3600 \; secs)}{(96500 \; C/mole)}$$

n=.46259

Then find the moles of Cl₂. There are 2 moles of electrons for every mole of Cl_2.Multiply by the fraction of 1/2.

.46259 * (1/2)

= .23129 moles Cl₂

Then use the equation

$$V=\frac{nRT}{P}$$

to find the volume.

$$V=\frac{(.23129 \; moles)(.08206)(273 \; K)}{(1 \; atm)}$$

= 5.18 L

ANSWER: 5.18 L

Q14.6.5

Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health?

ANSWER: Having free radicals in the body can be very dangerous. They can cause instability, which can lead to many diseases and hurt James Bond's health. They can attack DNA and cause unwanted reactions to take place.

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