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

In this question we must find standard gibbs free energy (ΔG°) when being given voltage and number of electrons transferred. We can solve these problems using an equation relating the two values:

\[\Delta G^{\circ}-n\cdot F\cdot E^{\circ}\]

For each problem we are given E° (cell potential), n (number of moles of electrons transferred), and F (faraday's constant = 96.485 kJ/V*mol). Thus, we can plug in our known values into the equation to solve for ΔG°.

1. First begin by listing the given values:

E°= 0.000 V, n = 2, F = 96.485 Kj/V*mol

Now plug into the known equation:

\[\Delta G^{\circ}=-(2 mol)\cdot (96.485 \frac{kj}{mol\cdot V})\cdot (0.00V)\]

Since our E° is 0, it is easy to see that the product of our given values will also be 0. Thus, we know that our ΔG° will also be 0.00 kJ

2. Once again, we begin by listing the given values.

E° = 0.434 V, n = 2, F = 96.485 kJ/V*mol

We can plug into the known equation, making sure that the units of every value will cancel out to leave us with kJ:

\[\Delta G^{\circ}=-(2 mol)\cdot (96.485 \frac{kj}{mol\cdot V})\cdot (0.0434V)\]

We can see that the mol of electrons will cancel with the mols from faraday's constant, and the volts from our cell potential will

also cancel out with the volts from faraday's constant, leaving us with kJ in the final answer

ΔG° = -83.75 kJ Do not forget the negative sign!

3. Once again, we can solve the same way, but this time our E° is negative:

E°= -2.439 V, n = 1, F = 96.485 Kj/V*mol

\[\Delta G^{\circ}=-(1 mol)\cdot (96.485 \frac{kj}{mol\cdot V})\cdot (-2.439V)\]

ΔG° = 235.33 kJ The negative signs from the equation and from our E° cancel out to generate a positive ΔG°

Q12.1.1

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

S12.1.1

A rate is a measurement of how something changes over time. In the context of chemistry, this can relate to concentration, voltage, enthalpy, etc. Understanding rate leads to an understanding of how chemistry can change over time. First, the instantaneous rate is the specific rate of change at a certain point in time. This point in time can be at the beginning, end, or anywhere in between the start and end of a functions rate. The initial rate is this very instantaneous rate at the beginning of a function; hence it is the "initial". Finally, average rate is not the rate at a single point, but rather the collection of all rates over a certain period of time. It is the mean value of all of the rates of a certain function over a specified period of time.

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

We are asked to find rate constants for experiments in which we are given initial concentrations, time, percent decomposition, and the fact that this is a first order reaction. Using this information, we are able to use the rate law equations for first order reactions in order to determine rate constants.

The rate law for First Order reactions is:

\[ln[A]_{f}=-kt+ln[A]_{0}\]

[A]_f = final concentration

[A]0 = initial concentration

t = time

k = rate constant

Looking at the equation, the only unknowns are final concentration and the rate constant. However, since we are given initial concentrations and percent decomposition, we can find the final concentrations as the percent decomposition determines how much of the initial concentration will be remaining. Thus, for the first experiment we can find the final concentration:

\[[A]_{i}\cdot percentdecomposed=(4.88M)(0.52)=2.54M\]

We know that 2.54 M of our initial concentration was decomposed, so we can subtract it from the initial concentration to find the remaining concentration:

4.88 M - 2.54 M = 2.34 M

Now we know that the initial concentration is 4.88 M, final concentration is 2.54 M, and time is 300 s. The only unknown is the rate constant, so we can manipulate the rate law reaction to solve for the rate constant:

ln[A]_f = -kt + ln[A]₀

ln[A]_f - ln[A]₀ = -kt We can subtract ln[A]₀ from both sides of the equation

-[(ln[A]_f - ln[A]₀ )/t] = k The final step is to divide both sides by -t, and now we have isolated k and can plug in our known values.

So we find the equation solving for k:

\[k=-\frac{ln[A]_{f}-ln[A]_{0}}{t}\]

We can now plug in our known values, starting with the first trial:

k = -((ln[2.34] - ln[4.88])/300)

\[k=-\frac{ln[2.34]-ln[4.88]}{t}\]

k = 2.45 x 10^-3 s^-1 The units are s^-1 since this is a first order reaction, as M/(s*M) = 1/s

This process can be repeated for each given equation to find the final concentration, and then plugging into the rate law equation that isolates K.

Here is the second experiment following the same steps.

We know initial concentration is 3.52 M, % decomposition is 52.9%, and time is 300 seconds.

(3.52 M)(0.529) = 1.86 M

3.52 M - 1.86 M = 1.66 M This is our final concentration. We can now plug into the equation we solved for in the first part.

\[k=-\frac{ln[1.66]-ln[3.52]}{t}\]

k = 2.51 x 10^-3 s^-1

The process can be repeated for each experiment, and will yield these rate constants:

Q21.3.3

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

1. \[_{13}^{27}\textrm{Al}+_{2}^{4}\textrm{He}\rightarrow ?+_{0}^{1}\textrm{n}\]
2. \[_{94}^{239}\textrm{Pu}+?\rightarrow_{96}^{242}\textrm{Cm}+_{0}^{1}\textrm{n}\]
3. \[_{7}^{14}\textrm{N}+_{2}^{4}\textrm{He}\rightarrow ?+_{1}^{1}\textrm{H}\]
4. \[_{92}^{235}\textrm{U}\rightarrow ?+_{55}^{135}\textrm{Cs}+_{0}^{1}\textrm{n}\]

S21.3.3

In this problem we need to do some "nuclear accounting" by filling in the correct species in each given nuclear reaction. With a good understanding of alpha, beta, and gamma particles, this task will be a lot easier.

1. To solve these problems, it is easiest to find total mass numbers and atomic numbers on both sides of the equations, and then compare to see what is missing.

\[_{13}^{27}\textrm{Al}+_{2}^{4}\textrm{He}\rightarrow ?+_{0}^{1}\textrm{n}\]

Total Mass # on left side: 27+4 = 31

Total Atomic # on left side: 13 + 2 = 15

We know the the mass # on the right, x, plus the mass of the neutron, 1, is equal to the total masses on the left, 31.

So 1 + x = 31 x = 30

Same steps for the atomic # on the right, y, find that

0 + y = 15 y = 15

Thus the missing species has these masses, and from the periodic table using the atomic number we can find that the element is Phosphorus.

\[_{15}^{30}\textrm{P}\]

2. Here we see the missing item is on the left, so we find the sums on the right.

Total mass on right= 242 + 1 = 243

Atomic number on right = 96 + 0 = 96

From the left side we can see that

239 + x = 243 so x = 4 for mass number

94 + y = 96 so y = 2 for atomic number

From the atomic number 2 we look at the periodic table and see that the element is Helium

\[_{2}^{4}\textrm{He}\]

3.

Mass on left = 14 + 4 = 18

Atomic number on left = 7 + 2 = 9

We know that for the mass number (x) on the right

18 = x + 1 x = 17

And for the atomic number (y) on the right

9 = y + 1 y = 8

So the element with atomic number 8 is Oxygen

\[_{8}^{17}\textrm{O}\]

4.

The total mass on the left is 235 and the atomic number is 92.

From the right we know that the mass number = x + 135 + 1 = x + 136

And that the atomic number = y + 55 + 0 = y + 55

235 = x + 136 x = 99 We can equate the left side to the ride side of the equations to solve for the missing mass and atomic numbers, x and y.

92 = y + 55 y = 37

With atomic number 37, this element is Rubidium

\[_{37}^{99}\textrm{Rb}\]

With lambda and the mass numbers, we can solve for activity using the above equation:

Activity= (0.205min^-1)(24) Activity = (0.324min^-1)(211)

Neon-24: 4.92 Bismuth-211: 68.4

Now we have the activities for both, and we can see that Bismuth - 211 has the bigger activity, most likely as a result of its larger mass number.