Extra Credit 48

Last updated
Save as PDF

Page ID: 83445

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Q19.25C

Find the values of delta G for the following reactions in the voltaic cells

a. 2V³⁺(aq)+Ni(s)→2V²⁺(aq)+Ni²⁺(aq)

b. 2Al(s)+3Br₂(l)→2Al³⁺(aq)+6Br^-

c. 2Cu²⁺(aq)+Sn²⁺(aq)→2Cu+Sn⁴⁺

General How To Guide

Let's see what relationships we know between what we are given and what we need to find.

We know that we can relate delta G and E_cell values through the following equation (where F is faraday's constant, and n is the amount of electrons transferred): deltaG=-n(F)(E_Cell)

We can find E_Cell through the following equation: E_Cell=E_{reduction half reaction}-E_{oxidation half reaction}

If we knew the E for each half reaction, then we can solve the problem.

Step 1) Half Reactions

We see that we have to separate each equation into it's two half reactions. Remember that oxidation is loss of electrons, so the one that ends up with less electrons is being oxidized and is therefore the anode. Similarly, reduction is the gain of electrons, so the one that ends up with more electrons is being reduced and is therefore the cathode.

Step 2) Find amount of electrons

By looking at our half reactions and combining them, we should be able to easily tell how many electrons are being transferred. Don't forget to balance the overall equation

Step 3) Find the corresponding E for each half reaction

We can find the standard reduction potential(E°) for each half reaction by simply looking at a chart that is provided here.

Note that because this is a chart of reduction potentials, the sign is important. The more positive it is, the more likely it is to be reduced.

Step 4) Find Delta G

We now know all that we need to know to plug into the equation given. Simply calculate the value and you have your answer. F stands for Faraday's constant which is 96485 Coulombs/(mole of electrons).

Specific Problem Solutions Q19.25C

a) step 1)

First lets look at Vanadium: V³⁺→V²⁺

We can see that the charges are not balanced, so we have to add an electron(with a -1 charge) to the reactant side.

e^-1+ V³⁺→V²⁺

This atom is gaining an electron, so it is being reduced.

Now let look at Nickel: Ni→Ni²⁺

We can see that the charges are not balanced, so we add two electrons to the product side.

Ni→Ni²⁺+ 2e^-1

This atom is losing electrons, so it is being oxidized.

step 2)

We see that Nickel needs twice as many electrons as Vanadium, so our overall equation is going to have two Vanadium each gaining an electron. Multiply each species in the reduction reaction(Vanadium) by two then combine the two reactions.

The overall reaction is 2e^-1+ 2V³⁺+Ni→2V²⁺+Ni²⁺+ 2e^-1

Double check that all of the charges and atoms balance and that it matches the original equation (NOTE: these electrons should cancel out; there should be the same amount(2) of electrons on the product and reactant sides so that they cancel out and we have not altered the equation in any way. This is a good check point to make sure that you have correctly balanced the half reactions).

2 electrons are being transferred; we will use this as N in our equation

step 3)

By looking at the chart, we see that the Standard Reduction Potential for the Vanadium half reaction is -.255V, and Nickel is -.257V. Remeber from step 1 which is oxidation and which is reduction.

E_Cell=E_{reduction half reaction}-E_{oxidation half reaction}

E_Cell=-.255V-(-.257V)

E_Cell=.002V

step 4)

We now know all we need to find delta G. "n" comes from the electrons transfered in step 2. F is Faraday's constant, which can be looked up and should be given. E_Cellis taken from step 3.

deltaG=-n(F)(E_Cell)

deltaG=-(2 moles of electrons) x (96485 C/(mole of electrons)) x (.002V)

deltaG= -385.94 J

b) step 1)

First lets look at Aluminum: Al → Al³⁺

We can see that the charges are not balanced, so we have to add three electrons(with a -1 charge) to the product side.

Al → Al³⁺+ 3e^-1

This atom is losing electrons, so it is being oxidized.

Now let look at Bromine: Br₂→ 2Br^-

We can see that the charges are not balanced, so we add an electron to the product side.

2e^-1+ Br₂→ 2Br^-

This atom is gaining an electron, so it is being reduced.

step 2)

We see that in order to balance out the electrons, we need to multiply the Bromine half reaction by 3 and the Aluminum by 2 to get a total of 6 electrons being transferred.

The overall reaction is 6e^-1+ 3Br₂+ 2Al →6Br^-+2Al³⁺+ 6e^-1

Double check that all of the charges and atoms balance and that it matches the original equation (NOTE: these electrons should cancel out; there should be the same amount(6) of electrons on the product and reactant sides so that they cancel and we have not altered the equation in any way. This is a good check point to make sure that you have correctly balanced the half reactions). This equation should match the original, but also include electrons.

6 electrons are being transferred; we will use this as N in our equation

step 3)

By looking at the chart, we see that the Standard Reduction Potential for the Aluminum half reaction is -1.676V, and Bromine is 1.087V. Remember from step 1 which is oxidation and which is reduction.

E_Cell=E_{reduction half reaction}-E_{oxidation half reaction}

E_Cell=1.087V-(-1.676V)

E_Cell=2.763V

step 4)

We now know all we need to find delta G. "n" comes from the electrons transferred in step 2. F is Faraday's constant, which can be looked up and should be given. E_Cellis taken from step 3.

deltaG=-n(F)(E_Cell)

deltaG=-(6 moles of electrons) x (96485 C/(mole of electrons)) x (2.763V)

deltaG= -1.60x10⁶ J

c) step 1)

First lets look at Copper: Cu²⁺→ Cu

We can see that the charges are not balanced, so we have to add two electrons(with a -1 charge) to the reactant side.

Cu²⁺ + 2e^-1→ Cu

This atom is gaining an electron, so it is being reduced.

Now let look at Bromine: Sn²⁺→ Sn⁴⁺

We can see that the charges are not balanced, so we add an electron to the product side.

Sn²⁺→ Sn⁴⁺ + 2e^-1

This atom is losing electrons, so it is being oxidized.

step 2)

We see that the number of electrons between the two reactions is already balanced; 2 electrons are being transferred.

The overall reaction is Cu²⁺ + 2e^-1+ Sn²⁺ → Cu + Sn⁴⁺ + 2e^-1

Double check that all of the charges and atoms balance and that it matches the original equation (NOTE: these electrons should cancel out; there should be the same amount(2) of electrons on the product and reactant sides so that they cancel and we have not altered the equation in any way. This is a good check point to make sure that you have correctly balanced the half reactions). This equation should match the original, but also include electrons.

2 electrons are being transferred; we will use this as N in our equation

step 3)

By looking at the chart, we see that the Standard Reduction Potential for the Copper half reaction is .3419V, and Tin is .154V. Remember from step 1 which is oxidation and which is reduction.

E_Cell=E_{reduction half reaction}-E_{oxidation half reaction}

E_Cell=.3419V-(.154V)

E_Cell=.1879V

step 4)

We now know all we need to find delta G. "n" comes from the electrons transferred in step 2. F is Faraday's constant, which can be looked up and should be given. E_Cellis taken from step 3.

deltaG=-n(F)(E_Cell)

deltaG=-(2 moles of electrons) x (96485 C/(mole of electrons)) x (.1879V)

deltaG= -3.63x10⁴ J

Q20.11A

Write the balanced reaction for the thermal decomposition of ZnCO₃(s).

General How To Guide

Step 1) Know your components

Because this is decomposition reaction, we know that there will be only one reactant. That means that when predicting our products, we have very few elements to work with.

Step 2) Predict products

Using descriptive chemistry, we can predict what the compound will break down in to. Nonmetal oxides often break down into an element and a compound. Carbonates decompose into carbon dioxide and an oxide. This website has some helpful information regarding what breaks down in to what.

Step 3) Balance

Now that you know what it will break down into, simply balance the equation.

Specific Problem Solutions Q20.11A

Step 1)

Our reactant, ZnCO₃ is a carbonate of a transition metal. It consists of only three elements.

Step 2)

We recognized that this is a carbonate of a transition metal. Descriptive chemistry tells us that this will likely break down in to carbon dioxide and the oxide of the metal.

If we balance oxidation states and charges, we see that the resulting products will be ZnO(s) and CO₂(g)

Step 3)

When we balance this equation, we find that we don't need coefficients and it balances quite nicely.

ZnCO_3(s)→ ZnO_(s) + CO_2(g)

Q21.19B

If the ion Ni²⁺ is linked with strong-field ligands to produce an octahedral complex, the complex has one unpaired electron, but if linked with weak-field ligands, the compound has three unpaired electrons. How does one account for this?

General How To Guide

Valence Electrons Crash Course

This is a transition metal complex. Start by finding the number of electrons in the d orbital. There is a fairly simple pattern to this. D orbitals start at element 21, which has 1 electron in group three. Each element you go over to the right(up until Zn which has a full shell), the element gets another electron. For example, Vanadium has 3 valence electrons. Then you have to adjust this for charge. If the metal ion has a positive charge, for example +3, that means it has three less electrons. be careful with this, because elements with d orbitals tend to lose their s orbital electrons first.

Step 1) find the number of valence electrons

Orbitals Crash Course

We know that electron orbitals exist as areas where an electron is most likely to be found. For d orbital electrons, these are known as the xy, yz, xz, x²y², and z²orbitals(for a visualization of these, click here and look at figure 21.6.1 b). They are named off of what axis they are based on. In octahedral complexes, the x²y² and z²orbitals are of a higher energy than the xy, yz, due to repulsion between the orbitals, creating two distinct energy levels. This is known as Crystal Field Splitting, and the difference between the two levels is known as the Crystal Field Splitting Energy, or delta o. The xy, yz, and xz orbitals go down in energy by 2/5 of delta o and are referred to as the t_2g energy level, while the x²y² and z²orbitals go up by 3/5 of delta o and are referred to as the e_g orbital.

Electron Pairing Rules Crash Course

Electrons tend towards whatever requires the least amount of energy. When they are filling orbitals, they will fill the lowest energy orbital first; this is known as the Aufbau Principle. This means that in octahedral complexes that undergo Crystal Field Splitting, they will fill the lowest(t_tg) orbital first before jumping up to the higher(e_g) orbital.

Two electrons can share an orbital if necessary, but because they have the same negative charge, they repel and would prefer to stay in different orbitals. This aversion to having two electrons in the same orbital before each one at that particular energy level is filled is known as Hund's rule.

These two electron tendencies contradict each other. They simultaneously prefer to not share orbitals, and not jump to the next one. What happens when an each slot in an orbital is already occupied by one electron? Does the next electron pair up with another to avoid entering a higher energy level(Aufbau), or jump to the next energy level to avoid pairing up and repulsive charges? the answer lies in the statement: electrons tend toward whatever requires the least amount of energy.

If the energy required to jump up a level(Crystal Field Splitting Energy or delta o) is less than the energy required to pair with another electron, it will jump. If the energy required to share an orbital is less than the Crystal Field Splitting Energy, it will share. Well how do we know which one is bigger? That depends on the ligand attached to it.

Ligands Crash Course

Weak field ligands create small Crystal Field Splitting Energies, so delta o will be smaller than the pairing energy and the electrons will jump. Each orbital will have one electron before any start to pair. This is known as a high spin configuration.

Strong field ligands create large Crystal Field Splitting Energies, so delta o will be larger relative to the pairing energy and electrons will share orbitals. The lowest orbitals will all have two electrons before any go up to the next level. This is known as a low spin configuration.

If you look at figure 21.6.2(click here), you can see examples of diagrams of the electrons in the orbitals.

To determine if a ligand is weak or stron, check what side of the Spectrochemical Series it is on. They are ranked by how strong they are, so those on the lowest end are the weakest ligands and those on the larger end are the strongest.

The strength of the ligand attached to the metal determines the Crystal Field Splitting Energy, which determines how electrons pair.

Step 2) determine whether the ligand is weak or strong field by looking at the Spectrochemical Series

D Orbital Electron Configuration Charts

If you look at figure 21.6.2(click here), you can see examples of diagrams of the electrons in the orbitals.

Each arrow represents an electron. Each dash represents and orbital and can hold two electrons. If the configuration is high spin(see ligands crash course), fill each dash with a single electron before putting pairing any. If the configuration is low spin, fill the lowest level with paired electrons until it's filled.

Step 3) fill in the electron configuration chart and see how many electrons are unpaired

Specific Problem Solutions Q21.19B

Step 1)

We see that Nickel, straight off the periodic table, would have 8 d orbital electrons. We also see that it has a 2+ charge, so it must be missing two electrons. one comes from the s orbital, so our d orbital only loses one.

There are 7 d orbital electrons

Step 2)

This question asks about both weak and strong field ligands and the unpaired electrons in both, so we don't really need to determine what the ligand is as we are creating an orbital diagram for both.

Step 3)

(note that the space between the energy levels (delto o) is larger on the strong field)

Strong Field Ligands Weak Field Ligands

(electrons will pair within the same energy level first) (Electrons occupy all orbitals before pairing)

e_g ___^___ _______ e_g ___^___ ___^___

t_2g __^_^__ __^__^__ __^__ __

t_2g __^_^__ __^__^__ __^__^__

We see that the strong field ligand complex has 1 unpaired electron and the weak field ligand complex has 3 due to different electron pairing patterns due to different Crystal Field Splitting Energies.

Q24.27B

Which sets of data correspond to a:

a) Zero order reaction

b) First order reaction

c) Second order reaction

I		II		III
Time(s)	[A], M	Time(s)	[B], M	Time(s)	[C], M
25	1	0	5	0	2.23
50	.85	25	2.5	25	1.82
75	.70	50	1.67	50	1.49
100	.55	75	1.25	75	1.21
125	.40	100	1	-	-
150	.25	-	-	-	-
200	.10	-	-	-	-

General How To Guide

What are orders of reaction?

Order of reaction are used to describe the rates of reaction, specifically in relation to concentration. Each order has its own set of equations to be used.

Zero order reactions follow the following set of equations:

Rate = k

Half Life = [A]/(2k)

[A] -[A_o] = kt

In zero order reactions, the rate is independent of concentration. A straight line can be achieved by plotting [A] vs t.

First order reactions follow the following set of equations:

Rate = k[A]

Half Life = ln(2)/k

log([A_o]/[A]) = kt/2.303

In first order reactions, the rate is dependent on concentration. A straight line can be achieved by plotting log([A]) vs t.

Second order reactions follow the following set of equations:

Rate = k[A]²

Half Life = 1/(k[A])

1/[A] = 1/[A_o]+ kt

In second order reactions, the rate is dependent on concentration. A straight line can be achieved by plotting 1/[A] vs t.

How to determine the order of reaction?

See which rate law works to describe the data set. This can be done by finding the half life, or by graphing the data and seeing which manipulation(none, log([A]), or 1/[A]) created a straight line.

Specific Problem Solutions Q24.27B

Zero order reactions are probably one of the easiest to find. Because the rate is independent of concentration, we see that it should remain the same regardless of what concentration we look at. To find the rate, find how much the concentration changes and divide it by how much the time changes. For every two data points in reaction A, we get that the change in concentration over change in time is always equal to .15/25. Because it doesn't matter what point we look at, the rate will always be the same, this is a zero order reaction.

A is a zero order reaction

First order reactions can be found by two methods. We can use a lot of math to find which of the rate equations describes the data, or we can simply graph it. First order reactions will form a straight line when manipulated to be log([concentration]) vs t. The only data that shows a straight line when graphed this way is data set C.

C is a first order reaction

Second order reactions can also be found using graphs. When the relationship between 1/[concentration] vs t is linear, then the reaction is second order. When we plot 1/[B] vs t. we get a straight line. Therfor, B is a second order reaction.

B is a second order

In case you don't have a graphing tool, here is the math method: if the k value for two different data points are the same using this equation, it is second order.

1/[A]−1/[A_o]=kt

First test point:

1/1.67-1/2.5=25k

k=.008

Second test point:

1/1.25-1/1=kt

k=.008

Using second order manipulation on the data makes the k value constant. That means this, if graphed, would be linear and the reaction is second order.

Q25.10D

Complete the following nuclear equations:

a) ${_{19}^{37}}K \rightarrow ? + {_{-1}^{0}}\beta$

b) ${_{94}^{239}}Pu \rightarrow ? + {_{2}^{4}}He^{2+}$

c) ${_{92}^{235}}U +{_{0}^{1}}n\rightarrow ? + 2{_{0}^{1}}n+{_{52}^{137}}Te$

d) ${_{95}^{241}}Am+{_{2}^{4}}He^{2+}\rightarrow ?+2{_{0}^{1}}n$

e) ${_{4}^{9}}Be +{_{1}^{1}}H\rightarrow ? + {_{2}^{4}}He^{2+}$

General How To Guide

While nuclear reactions may seem complicated at first glance, if we just remember that they follow basically the same rules as normal chemical reactions they seem much less daunting.

First let's break down the components of what we are looking at.

` ${_{p}^{m}}Sy$

This is the general notation for nuclear chemistry species. The m stands for the total number of protons and neutrons combined r mass number. Sy is the Symbol of the element, which we can use to find the proton number by looking at the periodic table. P stands for the number of protons in the atom or the atomic number. We can find the amount of neutrons by subtracting number of protons(p) from the total number of things in the nucleus(m).

Note: p doesnt always stand for the amount of protons, but can show the charge. For example, ${_{-1}^{0}}e$ is an electron with no mass(0) and a negative charge.

Note: because p can be redundant when given the symbol, p is sometimes left off

Step 1) Conserve Mass

Make sure the total of 'm' on both sides adds up to the same number. Even though this is nuclear chemistry and matter can be converted to energy, this should still balance.

Step 2) Conserve Charge

Find the charge that will balance both sides of the equation using p(the atomic number of the elements).

Step 3) Find the Corresponding Symbol

If this is a large element this part is fairly simply. Just find the element that corresponds to the atomic number you found for p.

If the unknown is some sort of radioactive particle or something small, this can get more difficult and requires some memorization. There are some common particles that we work with. Look at Table 5.3.1 for what they are. Note that the alpha particle(Helium nucleus) can also be written as a ${_{2}^{4}}He$ with or without a 2+ charge written in, and the Beta particle(a electron or positron) can also be written as ${_{-1}^{0}}e$ for an electron or with a +1 charge for a positron.

Specific Problem Solutions Q25.10.D

a) ${_{19}^{37}}K \rightarrow ? + {_{-1}^{0}}\beta$

Step 1)

The total mass on the product side should be the same as on the reactant side.

On the reactant side, the total mass number is 37.

On the product side, the total mass number is m+0.

Together, these create 37=0+m.

It's clear to see that m is 37.

Step 2)

The total charge on the product side should be the same as on the reactant side.

On the reactant side, the charge is 19.

On the product side, the charge is p-1.

Together, these create 19=p-1.

It's clear to see that p is 20.

Step 3)

p is 20, which means its symbol is simply going to be an element.

The element with the atomic number 20 is Ca

Our final answer is

${_{20}^{37}}Ca$

b) ${_{94}^{239}}Pu \rightarrow ? + {_{2}^{4}}He^{2+}$

Step 1)

The total mass on the product side should be the same as on the reactant side.

On the reactant side, the total mass number is 239.

On the product side, the total mass number is m+4.

Together, these create 239=4+m.

It's clear to see that m is 235.

Step 2)

The total charge on the product side should be the same as on the reactant side.

On the reactant side, the charge is 94.

On the product side, the charge is p+2.

Together, these create 94=p+2.

It's clear to see that p is 92.

Step 3)

p is 92, which means its symbol is simply going to be an element.

The element with the atomic number 92 is U

Our final answer is

${_{235}^{92}}U$

c) ${_{92}^{235}}U +{_{0}^{1}}n\rightarrow ? + 2{_{0}^{1}}n+{_{52}^{137}}Te$

Step 1)

The total mass on the product side should be the same as on the reactant side.

On the reactant side, the total mass number is 235+1.

On the product side, the total mass number is m+2(1)+137.

Together, these create 236=m+139.

It's clear to see that m is 97.

Step 2)

The total charge on the product side should be the same as on the reactant side.

On the reactant side, the charge is 92.

On the product side, the charge is p+52.

Together, these create 92=p+52.

It's clear to see that p is 40.

Step 3)

p is 40, which means its symbol is simply going to be an element.

The element with the atomic number 40 is Zr

Our final answer is

${_{40}^{97}}Zr$

d) ${_{95}^{241}}Am+{_{2}^{4}}He^{2+}\rightarrow ?+2{_{0}^{1}}n$

Step 1)

The total mass on the product side should be the same as on the reactant side.

On the reactant side, the total mass number is 241+4.

On the product side, the total mass number is m+2(1).

Together, these create 245=m+2

It's clear to see that m is 243.

Step 2)

The total charge on the product side should be the same as on the reactant side.

On the reactant side, the charge is 95+2.

On the product side, the charge is p+2(0).

Together, these create 97=p+0.

It's clear to see that p is 97.

Step 3)

p is 97, which means its symbol is simply going to be an element.

The element with the atomic number 97 is Bk

Our final answer is

${_{97}^{234}}Bk$

e) ${_{4}^{9}}Be +{_{1}^{1}}H\rightarrow ? + {_{2}^{4}}He^{2+}$

Step 1)

The total mass on the product side should be the same as on the reactant side.

On the reactant side, the total mass number is 9+1.

On the product side, the total mass number is m+4.

Together, these create 10=m+4

It's clear to see that m is 6.

Step 2)

The total charge on the product side should be the same as on the reactant side.

On the reactant side, the charge is 4+1.

On the product side, the charge is p+2.

Together, these create 5=p+2.

It's clear to see that p is 3.

Step 3)

p is 3, which means its symbol is simply going to be an element.

The element with the atomic number 3 is Li

Our final answer is

${_{3}^{6}}Li$

Q25.41F

Of the following pairs of nuclides, which do you expect to be more abundant in natural resources? Why?

a) $_{12}^{24}Mg$ or $_{12}^{26}Mg$

b) $_{10}^{21}Ne$ or $_{10}^{22}Ne$

c) $_{5}^{10}B$ or $_{5}^{11}B$

General How To Guide

As one probably knows, radiative materials are not very stable. Substances that aren't stable tend to decompose rapidly, so they are less likely to be abundant in natural resources, or exist in a stable state anywhere. What this question is really asking is which of these isotopes are more stable or less likely to be radioactive. There are a few rules that can tell us which ones be less radioactive.

Rule 1) Isotopes with an even number of neutrons are likely to be more stable than those with an odd number.

Recall: we find the number of neutrons by subtracting the proton and mass numbers

Rule 2) In isotopes with less than 20 protons, a one to one ratio of protons to neutrons is going to be the most stable.

Note: Rule 1 is more important than Rule 2

Belt of Stability

The belt of stability can be seen here. It is simply a graph of the number or neutrons and protons in stable isotopes. Isotopes that lie on or near the bet of stability will be more stable than those that are farther.

Specific Problem Solutions Q25.41F

Both of these isotopes have an even number of nutrons, so rule one isnt going to be informative here. Instead, rule two tells us that those with a 1:1 ration of protons to neutrons will be more stable. $_{12}^{26}Mg$ has 12 protons and 14 nuetrons. This is not a 1:1 ratio. $_{12}^{24}Mg$ has 12 protons and 12 nuetrons; this is a 1:1 ratio, so $_{12}^{24}Mg$ is more stable and common.

$_{12}^{24}Mg$ will be more abundant.

$_{10}^{21}Ne$ had 11 neutrons whereas $_{10}^{22}Ne$ has 12. Isotopes with an even number of protons are more stable. 12 is an even number, 11 is not, so the one with 12 will be more stable.

$_{10}^{22}Ne$ will be more abundant.

$_{5}^{10}B$ has 5 neutrons(odd) and a 1:1 ratio. $_{5}^{11}B$ has 6 neutrons(even) and a 6:5 ratio.

According to rule 1, $_{5}^{10}B$ should be less stable; according to rule 2, it should more stable. Remember that rule 1(neutron number) is more significant than rule 2(ratio), so isootope that rule 1 idicates is more stable will actually be more stable.

$_{5}^{11}B$ has an even number of neutrons, so it will be more stable.

$_{5}^{11}B$ will be more abundant.

Q21.2.27

Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for ⁸⁷Sr if the measured mass of ⁸⁷Sr is 86.908877 amu.

a) the calculated mass

b) the mass defect

c) the nuclear binding energy

d) the nuclear binding energy per nucleon

General How To Guide

Lets start with some definitions:

The calculated mass comes from adding together the mass of all the protons and neutrons. While we generally consider them to both have the same mass(1 amu) that doesn't work with small scale nuclear chemistry. Treating them the same is fine and dandy when working with very few sig figs, when 1/1000 of an atomic mass unit wont greatly affect calculations. In nuclear chemistry, those few decimal places are of great importance. The mass of protons and neutrons can be found in Table 5.3.1.

The number of protons comes from the atomic number of the element.

The number of neutrons come from the mass number minus the number of protons.

calculated mass = (number of neutrons)x(mass of a neutron)+(number of protons)x(mass of a proton)

Note: this should be extremely close to the mass number.

The mass defect is simply the difference between the measured mass and the calculated mass. It is the amount of matter that is converted into energy.

mass defect = calculated mass - measured mass

The nuclear binding energy can be found from the mass defect using what is probably the most cited physics equation in popular culture: E=mc² Where c is the speed of light(3.998x10⁸ m/s) and m is the mass defect(generally in kg), or amount of matter converted into energy.

Nuclear Binding Energy = (mass defect)(3.998x10⁸ m/s)²

Note: be careful with units and wether the question is asking for J or kJ

The nuclear binding energy per nucleon is just a different way of measuring the nuclear binding energy. Rather than measuring the energy for the whole reaction, you divide the total energy by the amount of nucleons in the reaction to get the binding energy per nucleon. Also relevant, a nucleon just means either a proton or neutron, or anything in the nucleus. The number of nucleons is the same thing as the mass number given in the nuclear species symbol.

Energy per nucleon = (Nuclear Binding Energy)/(Total number of nucleons)

Specific Problem Solutions Q21.2.27

To find the calculated mass , we need to know the number of protons and neutrons as well as the mass of each. Looking at a reference, we see that the mass of a proton is 1.0007276 amu and the mass of a neutron is 1.0008665 amu.

The number of protons in Sr is 38. This is found by looking at the atomic number of Sr.

The number of neutrons is given by (mass number) - (number of protons)

neutrons = 87-38 = 49

Now we can find the calculated mass

calculated mass = (number of neutrons)x(mass of a neutron)+(number of protons)x(mass of a proton)

calculated mass = (49)x(1.0008665 amu)+(38)x(1.0007276 amu)

calculated mass = 87.070107 amu

The mass defect is simply the calculated mass minus the measured mass.

mass defect = 87.070107 amu - 86.908877 amu

mass defect = .16123 amu

The nuclear binding energy is given by E = mc²

E = (.16123 amu) x (2.998 x 10⁸m/s)²

E = 1.44914 x 10¹⁶megaelectronvolts

The binding energy per nucleon is just the total binding energy divided by the number of nucleons.

E_p_{er nucleon} = (1.44914 x 10¹⁶megaelectronvolts) / 87

E_p_{er nucleon}= 1.66567 x 10¹⁴megaelectronvolts

Q20.5.2

What is the relationship between the measured cell potential and the total charge that passes through a cell? Which of these is dependent on concentration? Which is dependent on the identity of the oxidant or the reductant? Which is dependent on the number of electrons transferred?

General How To Guide

It's helpful to recall all of our equations relating to Cell Potential.

Equation 1) $E^{\circ}_{cell}=E^{\circ}_{cation}-E^{\circ}_{anion}$

This relates the standard cell potential and the reduction potentials of the anion and cation.

Equation 2) $E_{cell}=E^{\circ}_{cell}-\frac{RT}{nF}ln(Q)$

This relates the charge of the cell in non standard conditions, such as different temperature or concentrations.

The relationship between the measured cell potential and the total charge is given by equation 2.

The total charge going through the cell is dependent on concentration(Q in equation 2) whereas the measured cell potential in standard conditions is not.

Both the measured cell potential an the total charge are dependent on the identity of the oxidant and reductant(the measured cell potential(equation 1) is directly related to the identity, whereas the total charge is related to the cell potential which is related to the identities).

The total charge is also dependent on the number of electrons (n) because of equation 2.

Search

Text Color

Text Size

Margin Size

Font Type