Extra Credit 28

Q17.4.1

For the standard cell potentials given here, determine the ΔG° for the cell in kJ.

1. 0.000 V, n = 2
2. +0.434 V, n = 2
3. −2.439 V, n = 1

S17.4.1

Before doing any of these problems, we must analyze the elements given to us first. We are given an \(E^o_{cell}\) ,which is the value in Volts (V), and the number of electrons transferred in the reaction (from balanced reaction), symbolized as "n." They ask us to find the \(\Delta G^o\) with the given information. Now, we can go to a list of equations that deal with these thermodynamic values. One equation that will be useful is the variation of the Nernst Equation: \[\Delta G^o=-nFE^o_{cell}\] From this equation, we see that we have the values of "n," \(E^o_{cell}\) , and "F," which is the Faraday constant valued at 96.485 \({kJ\over V mol}\) . These three values correspond to those in the equation, and can be plugged in to find \(\Delta G^o\). These steps will be performed in numbers 1-3.

1. 0.000V, n=2

\(\Delta G^o= -nFE^o_{cell}\)

\(\Delta G^o\)= -(2 mol)(96.485 \({kJ\over V mol}\))(0.000 V)

\(\Delta G^o\)= 0.000 kJ

2. +0.434 V, n = 2

\(\Delta G^o\)= \(-nFE^o_{cell}\)

\(\Delta G^o\)= -(2 mol)(96.485 \({kJ\over V mol}\))(0.434 V)

\(\Delta G^o\)= -83.7 kJ

3. −2.439 V, n = 1

\(\Delta G^o\)= \(-nFE^o_{cell}\)

\(\Delta G^o\)=-(1 mol)((96.485 \({kJ\over V mol}\))(-2.439 V)

\(\Delta G^o\)= 235.33 kJ

Tara's Phase II Edit: All of my solutions matched yours! However instead of -83.8 kJ I got -83.7 kJ so I changed that.

Q12.1.1

What is the difference between average rate, initial rate, and instantaneous rate?

S12.1.1

First, a general reaction rate must be defined to know what any variation of a rate is. The reaction rate is defined as the measure of the change in concentration of the reactants or products per unit time. The rate of a chemical reaction is not a constant and rather changes continuously, and can be influenced by temperature. Rate of a reaction can be defined as the disappearance of any reactant or appearance of any product. Thus, an average rate is the average reaction rate over a given period of time in the reaction, the instantaneous rate is the reaction rate at a specific given moment during the reaction, and the initial rate is the instantaneous rate at the very start of the reaction (when the product begins to form).

The instantaneous rate of a reaction can be denoted as \[ \lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t} \]

Q12.4.19

Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:

S12.4.19

First we need to understand what the question is asking for: the average rate constant. The average rate constant is the variable "k" when discussing kinetics and it can be defined as the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances. Knowing that we need to find K in this first order reaction, we can look to formulas that include "k," initial and final concentrations \([A]_o and [A]_t\), and half life time "t." Since this is a first order reaction, we can look to the first order equations, and doing that we find one that includes the variables given in the question: \[\ln[A]_t=-kt+\ln[A]_o\]

For the first reaction, we have an initial concentration of 4.88 M, and a percentage decomposed. To find the final concentration, we must multiply the initial concentration by the percentage decomposed to know how much decomposed, and subtract that from the original to find out how much is left: 4.88M x 0.52= 2.54 M and 4.88M-2.54M=2.34M

Now, we have the variables we need, and we plug it into the equation above:

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[2.34M]=-k(300s)+\ln[4.88M]\)

k=\({-(\ln[2.34M]-\ln[4.88M])}\over 300\)

\(k=2.45x10^{-3}\)

Since it asks for the rate constant of each experiment, we now must do the same procedure for each data set to find the rate constant:

Second experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[1.66M]=-k(300s)+\ln[3.52M]\)

k=\({-(\ln[1.66M]-\ln[3.52M])}\over 300\)

\(k=2.51x10^{-3}\)

Third experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[1.07M]=-k(300s)+\ln[2.29M]\)

k=\({-(\ln[1.07M]-\ln[2.29M])}\over 300\)

\(k=2.54x10^{-3}\)

Fourth experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[0.834M]=-k(300s)+\ln[1.81M]\)

k=\({-(\ln[0.834M]-\ln[1.81M])}\over 300\)

\(k=2.58x10^{-3}\)

Fifth Experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[3.49M]=-k(180s)+\ln[5.33M]\)

k=\({-(\ln[3.49M]-\ln[5.33M])}\over 180\)

\(k=2.35x10^{-3}\)

Sixth Experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[2.60M]=-k(180s)+\ln[4.05M]\)

k=\({-(\ln[2.60M]-\ln[4.05M])}\over 180\)

\(k=2.46x10^{-3}\)

Seventh Experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[1.89M]=-k(180s)+\ln[2.95M]\)

k=\({-(\ln[1.89M]-\ln[2.95M])}\over 180\)

\(k=2.47x10^{-3}\)

Eighth experiment

\(\ln[A]_t=-kt+\ln[A]_o\)

\(\ln[1.11M]=-k(180s)+\ln[1.72M]\)

k=\({-(\ln[1.11M]-\ln[1.72M])}\over 180\)

\(k=2.43x10^{-3}\)

Tara's Phase II Edit: I fixed some of the mathjax equations because the "k=" was sitting on top of the fraction which looked a little odd. In addition, I got 2.46 x 10^-3 for k instead of 2.44 x 10^-3 for the 6th experiment.

Q21.3.3

Complete each of the following equations by adding the missing species:

1. \(^{27}_{13}Al+^4_2He\rightarrow ?+ ^1_0n\)
2. \(^{239}_{94}Pu+?\rightarrow ^{242}_{96}Cm+ ^1_0n\)
3. \(^{14}_{7}N+^4_2He\rightarrow ?+ ^1_1H\)
4. \(^{235}_{92}U+\rightarrow ?+ ^{135}_{55}Cs+4^1_0n\)

S21.3.1

This problem has to do with nuclear decay and solving for the missing nuclear species in its appropriate place. To do this, we must first understand what is happening here. We start with a radioactive element that has a nuclear particle emitted onto it, causing it to decay. The three nuclear particles are alpha, beta, and gamma particles. Alpha is a helium nucleus, \(^4_2He\), Beta which can be \(^1_0e\) or \(^{-1}_0e\), and gamma which is \(^0_0\gamma\). The number in the numerator is the mass number of the particle, and the denominator is the number of protons. Even though this is not relevant to this problem, this notation can also provide information about the number neutrons in a nuclear element because the mass number equal the amount of protons plus neutrons. So, when approaching this type of problem we can think of balancing both sides to have equal numbers for mass and protons. We can solve these problems using simple algebra. Once we discover what numbers go in the numerator and denominator of the fractions, we can use the number of protons to know then what element or nuclear particle fulfills the reaction.

1. \(^{27}_{13}Al+^4_2He\rightarrow ?+ ^1_0n\)

We need to balance the mass and protons.

27+4=x+1

Solving for x results in

31=x+1

x=30

Now we know the mass, so lets do protons following the same algebra

13+2=x+0

x=15

Now we know the mass and protons, and using the protons we can look to the periodic table to identify the missing element. 15 protons corresponds to the atomic number, which is phosphorous, "P."

The complete equation is

\(^{27}_{13}Al+^4_2He\rightarrow ^{30}_{15}P + ^1_0n\)

We can follow the same procedure for the rest of the problems

2. \(^{239}_{94}Pu+?\rightarrow ^{242}_{96}Cm+ ^1_0n\)

239+x=242+1

x=4

94+x=96+0

x=2

Proton count of 2 corresponds to helium, and knowing this particle is realizing this is an alpha particle- helium nucleus

\(^{239}_{94}Pu+^4_2He\rightarrow ^{242}_{96}Cm+ ^1_0n\)

3. \(^{14}_{7}N+^4_2He\rightarrow ?+ ^1_1H\)

14+4=x+1

x=17

7+2=x+1

x=8

Proton count of 8 corresponds to element of oxygen

\(^{14}_{7}N+^4_2He\rightarrow ^{17}_8O+ ^1_1H\)

4. \(^{235}_{92}U+\rightarrow ?+ ^{135}_{55}Cs+4^1_0n\)

When we have a coefficient in front of an element or particle, we must multiply the mass and proton count by that coefficient in order to maintain balance

235=x+135+4(1)

235=x+135+4

x=96

92=x+55+4(0)

92=x+55+0

x=37

Proton count of 37 corresponds to element of rubidium

\(^{235}_{92}U+\rightarrow ^{96}_{37}Rb+ ^{135}_{55}Cs+4^1_0n\)

Q21.7.5

Given specimens neon-24 \(t_{1\over2}=3.38min\) and bismuth-211 \(t_{1\over2}=2.14min\) of equal mass, which one would have greater activity and why?

S21.7.5

In order to solve this problem we have to understand what a half life tells us about activity. The rate of decay of a radioactive element, which is the number of nuclei in sample that decay in a given time, is also called the activity. The half life tells us how much time it takes for 50% (one half) of the radioactive element to be left as it decays. So, the faster that it decays, meaning a shorter half life, the faster the rate of decay is, and consequently the one that has greater activity. Now that we understand this information, we need to look at which element has has the shorter half life. In doing this we can see that Bismuth-211, with a half life of 2.14 minutes, would be the one with the greater activity because its half life is shorter than Neon-24 at 3.38 minutes

Q20.4.18

You have built a galvanic cell using an iron nail, a solution of FeCl2, and an SHE. When the cell is connected, you notice that the iron nail begins to corrode. What else do you observe? Under standard conditions, what is \(E_{cell}\)?

S20.4.18

First we need to understand what each element is. Iron nail is solid iron Fe(s), \(FeCl_2\) is an aqueous solution \(FeCl_2\)(aq), and SHE is the Standard Hydrogen Electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. SHE needs to be connected to an inert electrode, which plays no role in the reaction but helps it proceed forward. It is used when there is no solid electrode in a reaction of the elements. In SHE, the half reaction is: \[2H^+(aq)+2e^-\rightarrow H_2(g)\]

Also, the half reaction for the iron is \(Fe^{2+}(aq)+2e^-\rightarrow Fe(s)\) because Fe(s) is oxidized from 0 to 2+. Knowing this information, we can understand that the metal being corroded acts as the anode, so the metal is oxidized. The SHE, then, is being reduced. As it's reduced, we observe that it disappears slowly.

The question then asks what is \(E_{cell}\) under standard conditions. We know the half reactions of each species, and we know which one is cathode and anode. Now we employ the formula to find \(E_{cell}\) under standard conditions: \[E_{cell}^o=E_{cathode}^o-E_{anode}^o\]

We can find the values for anode and cathode using a Standard Reduction Potentials chart, and using our respective half reactions. For the iron anode, the value is -0.440V and the SHE cathode is 0.00V. We can plug this into the equation now.

\(E_{cell}^o=E_{cathode}^o-E_{anode}^o\)

\(E_{cell}^o=0--0.440V\)

\(E_{cell}^o\)= 0.440V

Becuase our \(E_{cell}\) value is positive, we know that this reaction occurs spontaneously so there is no energy source attached to this galvanic cell.

Q20.9.3

Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?

S20.9.3

Mixtures of molten salts are used over pure salts during electrolysis because they have good electric conductivity, a higher heat capacity, better thermal conductivity and can act as a solvent. The pure salts have higher melting points alone, but when they are mixed, their combined melting point (known as the eutectic point) is lower. This helps the reaction move forward, making electrolysis easier.

Q20.9.1

Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?

S20.9.1

First, we need to define some terms. An electrochemical reaction that is thermodynamically favored means it is spontaneous, with a negative \(\Delta G\). A spontaneous reaction does not require a prolonged input of energy to move it forward, however it might need activation energy to get it going. An overvoltage is when the voltage required to force a chemical reaction to occur is larger than expected. For this reaction, it would be the first to start in a series of reactions because it is spontaneous. Since it is already in motion, it takes less energy to push it forward than it would for a non-spontaneous reaction. However, since a non-spontaneous reaction has a push to it, it's easier to push it over the activation energy barrier, but for a spontaneous reaction, it doesn't already have a push. This is why an overvoltage occurs more in a spontaneous reaction; the extra energy is needed to push the reaction over the activation energy threshold.