Extra Credit 45

Q17.6.4

Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?

ANSWER

Corrosion is the process of deterioration/oxidation of a metal as a result of its chemical reactions with its environment, which turns the metal to its natural state as an ore. All metals can corrode, but some corrode faster than others due to their reduction potential. Metals with a smaller reduction potential are more prone to corrosion. When two metals are together, it is a fact that the one with the smaller reduction potential will corrode/ oxidize first. This metal is known as the sacrificial anode, and protects the other metal from corrosion. For this problem, since we know that B corrodes and A doesn't when they are together, we can assume that the reduction potential \((E^{o})\) of B is less than that of A: \(E^{o}_{B} < E^{o}_{A}\). Since A corrodes when in contact with C, we can assume that \(E^{o}_{A} < E^{o}_{C}\). We now know that overall \(E^{o}_{B} < E^{o}_{A} < E^{o}_{C}\). So, when B comes in contact with C, we know it will corrode first and protect C because it has the lowest reduction potential. B - corrodes, C - does not corrode.

Q12.3.8

The rate constant for the radioactive decay of ¹⁴C is 1.21 × 10⁻⁴ year⁻¹. The products of the decay are nitrogen atoms and electrons (beta particles):

\[^{6}_{14}C → ^{6}_{14}N + e^{-}\]

\[\text{rate}=k[\ce{^6_{14}C}]\]

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 × 10⁻⁹ M?

ANSWER

In this question, we are asked to find the rate of production of nitrogen atoms from the decay of a carbon-14 sample that has a concentration of 6.5 x 10^-9 M. The rate of decay of carbon-14 is equal to the rate of production of the nitrogen atoms. Our equation to solve this problem is given to us, but the general form of the equation is this:

\[\text{rate} = \text{k}[\text{A}]^{m}\]

k = rate constant \((1.21x10^{-4} year^{-1})\)

\( [^{14}_{6}C] \)= the concentration of C-14 (6.5 × 10^-9 M)

rate = what we are solving for

To solve, we must plug in all our values in to get our answer!

\(rate=(1.21x10^{-4} year^{-1})(6.5 x 10^{-9}M)\)

rate = 7.87 x 10^-13 M • year^-1 ≈ 7.9 x 10^-13 M/year (only have 2 significant figures because the least precise measurement has 2 significant figures)

Q12.5.17

Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first A+BC⟶AB+C reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?

ANSWER

The A atom has enough energy to react with BC; however, the different angles at which it bounces off of BC without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.

Q21.4.12

Write a nuclear reaction for each step in the formation of \(^{208}_{82}Pb\) from \(^{228}_{90}Th\) , which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order.

ANSWER

To answer this question one must understand what a nuclear symbol means. The top number indicates the mass number, which is the sum of protons and neutrons in a given isotope, while the bottom number indicates the atomic number, which is the number of protons in an element (which never changes). From the atomic number, the element can be determined. For example, an isotope \(^{23}_{12}X\), where X is some unknown nuclide, has 12 protons which, by definition, makes it the element magnesium. Since the mass number is the number of neutrons and protons, if we subtract the 12 protons from the isotope's mass number (23), we find that there are 11 neutrons. This isotope can be written as \(^{23}_{12}Mg\). There are many types of ways that a nucleus can decay. This can be done by α, β, or γ decay.

An alpha decay is when the nucleus emits an alpha particle, which is denoted as \(^4_2He\) . It has a charge of 2 and a mass number of 4. To get the new nucleus, you subtract 4 from the mass number and subtract 2 from the protons of the original nucleus. If the magnesium from above underwent alpha decay, the new nucleus would be \(^{19}_{10}Ne\).

A Beta emission is when the nucleus emits a beta particle, which is denoted as \(^0_{-1}β\) . It has no mass and a charge of -1. To get the new nucleus, you add 1 to the atomic number. If the magnesium above under went beta emission, the new nucleus would be \(^{23}_{13}Al\) .

A gamma ray has no charge or weight so nothing changes when it is emitted, and its symbol is denoted as . The new nucleus looks exactly the same as the original.

A nucleus can decay by different ways, multiple times. In this question we will be writing 7 step equations starting with \(^{228}_{90}Th\) until we get to our final product of \(^{208}_{82}Pb\) . The first step is an alpha decay so our first equation will look like this:

\[\ce{_{90}^{228}Th \rightarrow _{88}^{224}Ra + _{2}^{4}He}\]

Our next step is also an alpha decay so the second equation will look like this. Notice how the new nucleus from the first equation is used to perform the 2nd alpha decay. \[\ce{_{88}^{224}Ra\rightarrow _{86}^{220}Rn + _{2}^{4}He}\]

The next step is also an alpha decay so the third equation will look like this. Again, we use the new nucleus from the previous equation to perform the 3rd alpha decay. \[\ce{_{86}^{220}Rn\rightarrow _{84}^{216}Po + _{2}^{4}He}\]

The next step is also and alpha decay so the 4th equation will look like this. Again, we use the new nucleus from the previous equation to perform the 4th alpha decay. \[\ce{_{84}^{216}Po\rightarrow _{82}^{212}Pb + _{2}^{4}He}\]

The next step is a beta emission so the 5th equation will look like this. Again, we use the new nucleus from the previous equation to perform the 1st beta emission.\[\ce{_{82}^{212}Pb\rightarrow_{83}^{212}Bi + _{-1}^{0}β}\]

The next step is also a beta emission so the 6th equation looks like this. Again, we use the new nucleus from the previous equation to perform the 2nd beta emission.\[_{83}^{212}Bi → _{84}^{212}Po+_{-1}^{0}β\]

Our final step is another alpha decay so the 7th equation looks like this. Again, we use the new nucleus from the previous equation to perform the final alpha decay.\[_{84}^{212}Po→ _{82}^{208}Pb+_{2}^{4}He\]

Now we have completed the nuclear decay and our final product is \(^{208}_{82}Pb\).

Q20.2.16

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

1. \(Zn(s)+HCl(aq)\rightarrow\)
2. \(3HNO_3(aq)+AlCl_3(aq)\rightarrow\)
3. \(K_2CrO_4(aq)+Ba(NO_3)_2(aq)\rightarrow\)
4. \(Zn(s)+Ni^{2+}(aq)\rightarrow\)

ANSWERS

1. \(Zn(s)+2HCl(aq)\rightarrow H_2(g)+ZnCl_2(aq)\) This reaction is a redox reaction because the Zn is losing 2 electrons (oxidized) and hydrogen is is gaining 2 electrons.
2. \(3HNO_3(aq)+AlCl_3(aq)\rightarrow 3HCl(aq)+ Al(NO_3)_3(aq)\) This reaction is none of the above listed; this is a double replacement reaction, as HNO₃ is a strong acid while AlCl₃ is a Lewis acid (accepts an electron pair).
3. \(K_2CrO_4(aq)+Ba(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+BaCrO_4(s)\) This is a precipitation reaction because the product of BaCrO₄ is insoluble.
4. \(Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+ Ni(s)\) This is a redox reaction because there are 2 electrons being transferred from going to Zn(s) to Zn²⁺(aq) and Ni²⁺(aq) to Ni(s). However, this would only occur in an electrochemical cell; if Zn was just next to Ni²⁺ no reaction would occur.

Q20.5.11

The chemical equation for the combustion of butane is as follows:

This reaction has ΔH° = −2877 kJ/mol. Calculate E°_cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?

ANSWER

To solve this problem we need 3 equations : \(\Delta G^o= \Delta H^o-T\Delta S^o\), \(\Delta G^o= -nFE^o_{cell}\) and the equation \(\Delta G^{o}_{f}=\Delta G^{o}_{products}-\Delta G^{o}_{reactants}\)

To get \(\Delta G^o\) , we must use a \(\Delta G^o\) table and the equation \(\Delta G^{o}_{f}=\Delta G^{o}_{products}-\Delta G^{o}_{reactants}\) to calculate the overall Gibbs Free Energy. From the table we can find that C₄H₁₀ has \(\Delta G^o\) = -134.3 , O₂ has \(\Delta G^o\)=0 , CO₂ has \(\Delta G^o\) = -394.4 and H₂O has \(\Delta G^o\)=-228.6. When we plug them into the equation, we must multiply by them by their coefficients from the reaction.

\(\Delta G^{o}_{f}=Σ\Delta G^{o}_{products}-Σ\Delta G^{o}_{reactants}\)

\(\Delta G^{o}_{f}= [5(-228.6) + 4(-394.4)] - [-134.3 + \frac{13}{2}(0)]\)

\(\Delta G^{o}_{f}= -2586.3\) kJ/mol

To calculate E°_cell, we use \(\Delta G^o= -nFE^o_{cell}\) and plug in our \(\Delta G^o\) value we just calculated and Faraday's constant which is 96.485 kJ/V mole e^-.

-2586.3= -(1)(96.485 kJ/V mol e^-)E°_cell

E°_cell = 26.8 V

Since our E°_cell value is positive and our \(\Delta G^o\) value is negative, that tells us that our reaction is spontaneous.

To calculate \(\Delta S^o\), we must use this equation : \(\Delta G^o= \Delta H^o-T\Delta S^o\) , and now we have all of our values to calculate it.

-2586.3 kJ/mol = −2877 kJ/mol - (298K)\(\Delta S^o\)

\(\Delta S^o\) = -.976 kJ/ (mol•K) = -9760 J/(mol•K)

Since our \(\Delta S^o\) < 0 , that means that entropy decreases, which means that randomness/ disorder is decreasing.

I don't think this is correct because, at standard conditions, you cannot assume that the equilibrium constant is equal to 1. I would find \(\Delta G^{o}\) by using tabulated data of \(\Delta G^{o}_{f}\) values, then subtract the sum of the \(\Delta G^{o}_{reactants}\) values from the sum of \(\Delta G^{o}_{f}\) values. Then I would use \(\Delta G^{o} = -nFE^{o}_{cell} \) to get E^o_cell.

FROM TARA: I just fixed it so it should be good now.

Q24.6.7

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

1. [TiCl₆]³⁻
2. [CoCl₄]²⁻

ANSWER

To complete this problem, we have to understand crystal field theory. This involves all 5 d-orbitals which look like this:

These orbitals help us create an energy level diagram based on the number of ligands present on a complex ion. The ligands then tell us how to fill up the energy level diagram, either high spin or low spin. The 4 types of energy level diagrams are octahedral (6 ligands), tetrahedral (4 ligands), square planar (4 ligands), and linear (2 ligands). Our complex ion in part 1 of the question has 6 chlorine atoms attached to it, which means that it has 6 ligands and therefore corresponds to an octahedral energy level diagram. To determine if we fill up our diagram as a high spin or low spin complex, we must look at the spectrochemical series. The spectrochemical series tells us the magnitude of the splitting energy - ∆. H₂O to I^- are considered to have low ∆, CN^- to NH₃ are considered to have high ∆. From the table provided, we can see if the energy diagram will be filled as high or low spin. High ∆ = low spin , low ∆ = high spin.

HIGH VS. LOW SPIN

Here is an example of an octahedral diagram being filled in high and low spin. In high spin, each orbital is filled once before returning to the lower level. In low spin, the lower level of the diagram is filled completely before reaching the higher level. Now we have all the information to complete the 1st question.

1) We know that our complex is octahedral so our energy level diagram will look like the one above. Our ligand is classified as having a low ∆, which means it's diagram is filled as high spin. Next, we look at how many electrons will fill the diagram. We can set up an equation to figure out the charge of Ti in our complex.

X + -1(6) = -3

X - 6 = -3

X = 3

Since Ti has a +3 charge, we can look at the periodic table and see that the Ti contributes 1 electron. It's energy level diagram will look like this.

So, we now know that [TiCl₆]³⁻ is octahedral, high spin, and has 1 unpaired electron.

2)[CoCl₄]²⁻ has Cl ligands as well, so it will also be high spin due to its low ∆. Let's again setup an equation to figure out the charge of cobalt in our complex.

X + -1(4) = -2

X - 4 = -2

X = 2

Since Cobalt has a +2 charge, we can look at the periodic table and see that Co contributes 7 electrons. We now have to decide if we use tetrahedral or square planar energy level diagrams. Since our ligands are low energy, meaning high spin, we use the tetrahedral diagram. Our diagram with cobalt will look like this.

So we now know that [CoCl₄]^2- is tetrahedral, high spin, and has 3 unpaired electrons.