Extra Credit 41

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Q17.6.1

Which member of each pair of metals is more likely to corrode (oxidize)?

Mg or Ca
Au or Hg
Fe or Zn
Ag or Pt

The first step in this type of problem is to reference the activity series and find the positions of all eight of these metals. The activity series lists metals in order from most reactive to least reactive, the most reactive in this case being the metals most likely to oxidize, like the alkali metals.

Here is a mini activity series with all eight of these metals on it:

K highest

Pt lowest

The metal that is more likely to corrode is the one higher up on the activity series.

1. Mg vs. Ca: As seen on the activity series, Ca is higher than Mg, meaning Ca is more likely to be oxidized.

2. Au vs. Hg: Although these are both more unreactive metals, Hg is above Au on the activity series, so Hg is more likely to corrode.

3. Fe vs. Zn: Zn is more likely to be oxidized. This is why metals like zinc and aluminum are attached to the sides of iron ships; the zinc will corrode before the iron does, protecting the ship from rust.

4. Ag vs. Pt: Ag is more likely to be oxidized. Platinum is the lowest metal on the activity series, meaning it is very unreactive.

In general the alkali metals are the most likely to be oxidized, followed by the alkali earth metals; both of these groups can be oxidized with water.

Q12.3.4

How much and in what direction will each of the following affect the rate of the reaction: $$\ce{CO}(g)+\ce{NO2}(g)⟶\ce{CO2}(g)+\ce{NO}(g)$$ if the rate law for the reaction is the following equation? $$\ce{rate}=k[\ce{NO2}]^2$$

Decreasing the pressure of NO₂ from 0.50 atm to 0.250 atm.
Increasing the concentration of CO from 0.01 M to 0.03 M.

The rate law is already given, so this problem should be a simple matter of plugging numbers into the rate law equation and comparing them.

1. When we plug in .5 atm and .25 atm into the rate law: $$k[.5]^2=.25k$$ $$k[.25]^2=.0625k$$

Then we divide the final value by the initial value to compute how many times larger or smaller the final value is from the initial: $$.0625k/.25k=1/4$$

This shows that halving the pressure of NO₂ (half of .5 is .25) results in a rate that is one quarter of the size of the original rate. Thus the rate of the reaction decreases to 1/4 of the original rate.

2. The rate law is the only way that the concentrations of the reactants can affect the rate of the reaction. Thus, looking at the rate law of this equation again: $$\ce{rate}=k[\ce{NO2}]^2$$ The concentration of CO is not present in this rate law, meaning that [CO] has no effect on the rate of the reaction whatsoever.

Q12.5.13

Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H₂, and iodine, I₂. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:

Temperature (K)	k (M⁻¹ s⁻¹)
555	6.23 × 10⁻⁷
575	2.42 × 10⁻⁶
645	1.44 × 10⁻⁴
700	2.01 × 10⁻³

What is the value of the activation energy (in kJ/mol) for this reaction?

The only solvable equation we have correlating temperature and k to activation energy is this one: $$\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$$

Thus we solve for the activation energy by choosing two temperatures and k values and inputting them into our equation. For my solution my variables will be:

T₁=645 K

T₂=700 K

k₁=1.44 × 10⁻⁴ M^-1 s^-1

k₂=2.01 × 10⁻³ M^-1 s^-1

R=.008314 kJ/K•mol (make sure R is in terms of kJ because the answer wants E_a in kJ)

The math is relatively simple:

$$\ln \left(\dfrac{2.01\times10^{-3}}{1.44\times10^{-4}}\right)=\dfrac{E_{\textrm a}}{.008314}\left(\dfrac{1}{645}-\dfrac{1}{700}\right)$$

$$\ln \left(13.958\right)=\dfrac{E_{\textrm a}}{.008314}\left(\dfrac{55}{451500}\right)$$

$$2.6361=.014652E_{\textrm a}$$

$$E_{\textrm a}=180{\textrm {kJ}}/{\textrm {mol}}$$

Q21.4.8

The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.

$$\ce{^{34}_{15}P}$$
$$\ce{^{239}_{92}U}$$
$$\ce{^{38}_{20}Ca}$$
$$\ce{^{3}_{1}H}$$
$$\ce{^{245}_{94}Pu}$$

Here is the belt of stability:

None of these five nuclei lie on the black line, where the nucleus would be stable and would not decay. This particular belt of stability graph shows what type of decay will occur in each region that is not on the black line, the official Belt of Stability. However, if the graph was not labeled, the type of decay that a nuclei undergoes depends on which decay moves it closer to the belt of stability on the graph. This generally means that...

1. Elements with proton numbers less than 83 that are below the black line will emit positrons with beta decay. This is because they have too many protons to be stable; this type of beta decay turns a proton into a neutron, moving the element upwards on the graph.

2. Elements with proton numbers less than 83 that are above the black line will emit electrons with beta decay. This is because these elements have too many neutrons; this type of beta decay turns a neutron into a proton, moving the element downwards on the graph.

3. Elements with proton numbers 83 or larger are far above the black line in numbers of protons and neutrons, so they undergo alpha decay to eject both protons and neutrons and lower themselves on the graph.

Thus to solve this problem, we need to determine the number of protons and neutrons in the given elements, see where they fall on the graph, and predict the type of decay they undergo depending on whether they are above or below the belt.

1. The P nuclei has 15 protons (because the proton number is equal to the atomic number), and its neutron number is equal to the mass number minus the atomic number: 34-15=19 neutrons. When we plot this on the graph, the point (15, 19) falls above the belt of stability in the blue area. This means that this phosphorous nuclei will undergo electron beta decay.

2. The U nuclei has 92 protons and 239-92=147 neutrons. Uranium has an atomic number higher than 83, so it will undergo alpha decay. This can also be seen on the graph, where the point (92, 147) lies beyond the reach of the black line in the mostly yellow area.

3. The Ca nuclei has 20 protons and 38-20=18 neutrons. When we plot this on the graph, the point (20, 18), with the protons as the x-axis and the neutrons as the y-axis, lies beneath the belt of stability in the orange area. It may be difficult to tell where this point is, but the straight line passing beneath the belt of stability is the line where the number of protons equals the number of neutrons, meaning that (20, 18) lies below this line. This all means that this calcium nuclei will undergo positron beta decay.

4. The H nuclei has 1 proton and 3-1=2 neutrons. When we plot this on the graph, it is difficult to tell where it falls, but one can see that the straight line signifying the number of protons equaling the number of neutrons merges with the belt of stability when the number of protons is small. This means that, for an element with fewer than 14 protons, we can assume that any element with more neutrons than protons undergoes electron beta decay, while any element with more protons than neutrons undergoes positron beta decay. Since this hydrogen nuclei has more neutrons than protons, it lies above the belt of stability and undergoes electron beta decay.

5. The Pu nuclei has 94 protons and 245-94=151 neutrons. Like the uranium nuclei, plutonium has more than 83 protons, meaning that it undergoes alpha decay.

Q20.2.12

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

A few drops of NiBr₂ are dropped onto a piece of iron.
A strip of zinc is placed into a solution of HCl.
Copper is dipped into a solution of ZnCl₂.
A solution of silver nitrate is dropped onto an aluminum plate.

Once again we need to reference the activity series to determine if these reactions take place. If the lone metal is higher up on the activity series than the cation (metal) attached to the anion, then the metal replaces the cation, causing a reaction.

Here is a mini activity series with the metals of importance on it:

K highest

Pt lowest

1. Fe, the lone metal, is higher up on the activity series than Ni, so this reaction will happen.

2. Zn is higher up on the activity series than H, so this reaction will happen.

3. Cu is lower on the activity series than Zn, the cation, so nothing happens. (We can ignore #3 for the rest of the problem.)

4. Al is higher up on the activity series than Ag, so this reaction will happen.

Because one lone metal is reacting with a molecule, all three of the occurring reactions are single displacement reactions. The lone metal will replace the cation.

The basic, unbalanced equations for these reactions are:

1. $$\ce{NiBr_2}(aq)+\ce{Fe}(s)⟶\ce{FeBr_3}(aq)+\ce{Ni}(s)$$

2. $$\ce{HCl}(aq)+\ce{Zn}(s)⟶\ce{ZnCl_2}(aq)+\ce{H_2}(g)$$

4. $$\ce{AgNO_3}(aq)+\ce{Al}(s)⟶\ce{Al(NO_3)_3}(aq)+\ce{Ag}(s)$$

You need to refer to the solubility rules to determine the states of FeBr₃, ZnCl₂, and Al(NO₃)₃. The solubility rules say that almost all compounds with Cl^-, Br^-, and (NO₃)^- in them are soluble, so all three of our ionic compounds should be labeled as aqueous. It should also be noted that these specific compounds are created because Al is +3 as an ion and Zn is +2. Iron can be either +2 or +3, but according to the Internet the common bromide compound is FeBr₃.

Now we need to balance the equations:

1. $$\ce{3NiBr_2}(aq)+\ce{2Fe}(s)⟶\ce{2FeBr_3}(aq)+\ce{3Ni}(s)$$

2. $$\ce{2HCl}(aq)+\ce{Zn}(s)⟶\ce{ZnCl_2}(aq)+\ce{H_2}(g)$$

4. $$\ce{3AgNO_3}(aq)+\ce{Al}(s)⟶\ce{Al(NO_3)_3}(aq)+\ce{3Ag}(s)$$

This problem asks for the net ionic equation and the complete ionic equation. It is easier to do the complete ionic equation first; this involves turning each aqueous molecule into ions, because when they are in aqueous solution they dissociate. This is what this looks like:

1. $$\ce{3Ni^2+}(aq)+\ce{6Br^-}(aq)+\ce{2Fe}(s)⟶\ce{2Fe^3+}(aq)+\ce{6Br^-}(aq)+\ce{3Ni}(s)$$

2. $$\ce{2H^+}(aq)+\ce{2Cl^-}(aq)+\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{2Cl^-}(aq)+\ce{H_2}(g)$$

4. $$\ce{3Ag^+}(aq)+\ce{3NO_3^-}(aq)+\ce{Al}(s)⟶\ce{Al^3+}(aq)+\ce{3NO_3^-}(aq)+\ce{3Ag}(s)$$

Notice how some of the ions are on both sides of the equations, like the six bromine ions? The net ionic equation is almost exactly like the complete ionic equation except that you leave out any ions that are on both sides of the equation. Note that if an ion changes states, like solid Zn becoming aqueous, the ion remains in the equation.

1. $$\ce{3Ni^2+}(aq)+\ce{2Fe}(s)⟶\ce{2Fe^3+}(aq)+\ce{3Ni}(s)$$

2. $$\ce{2H^+}(aq)+\ce{Zn}(s)⟶\ce{Zn^2+}(aq)+\ce{H_2}(g)$$

4. $$\ce{3Ag^+}(aq)+\ce{Al}(s)⟶\ce{Al^3+}(aq)+\ce{3Ag}(s)$$

Note that these are all redox reactions and could all be used to make galvanic cells.

Q20.5.7

Occasionally, you will find high-quality electronic equipment that has its electronic components plated in gold. What is the advantage of this?

One can discover the answer to this question by referencing the activity series of metals. Gold is at the very bottom of the series, meaning that gold will not displace other metals in solution and is incredibly unreactive. This is important for electronic equipment because it protects the wires from corrosion; gold also does not corrode well, meaning it can protect the equipment without needing to be replaced.

You might notice that Pt is the lower than Au on the activity series; it is the most unreactive metal, but it would be enormously expensive to plate electronics with platinum, since it is more expensive than gold.

Q24.6.3

Will the value of Δ_o increase or decrease if I⁻ ligands are replaced by NO₂⁻ ligands? Why?

We need to refer to the spectrochemical series to solve this problem. This series ranks ligands based on the magnitude of Δ_o their metal complexes exhibit.

$$\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}$$

The ligands on the left side of the spectrochemical series are strong-field ligands. They have larger values of Δ_o for various reasons, including their smaller sizes and lone pairs.

The ligands on the right side are weak-field ligands with smaller values of Δ_o. Many of these ligands are large halides, like Cl^-.

With this information in mind, the value of Δ_o should increase if I⁻ is replaced by NO₂⁻ because I^- is a weak-field ligand and NO₂⁻is a strong-field ligand. The NO₂⁻ will cause the complex to have a higher Δ_o and will most likely create a low-spin configuration as well.

Q12.1

The following data were obtained for the reaction of methane with oxygen: $$\ce{CH_4}(g)+\ce{2O_2}(g)⟶\ce{CO_2}(g)+\ce{2H_2O}(l)$$

time (min)	[CH₄] (mol/L)	[CO₂] (mol/L)
0	0.050	0
10	0.030	0.020
20	0.020	?
30	0.015	?

How many moles of CO₂ are produced for each mole of CH₄ that is used up?
What concentration of CH₄ is used up after 10 minutes?
What is the concentration of carbon dioxide produced after 20 minutes?
Write an equation for reaction rate in terms of Δ[CO₂] over a time interval.
What is the reaction rate for the formation of carbon dioxide between 10 and 20 minutes?
What is the average reaction rate between 0 and 30 minutes?
Write an expression for reaction rate relating Δ[O₂] to Δ[CO₂].
At what rate is O₂ used up between 10 and 20 minutes?

This problem involves differential rate laws, or Δ[A]/Δt.

1. When we look at the balanced equation for this reaction, one can see that both CH₄ and CO₂ have a coefficient of 1, meaning that the moles of CH₄ lost will exactly equal the moles of CO₂ gained. One mole of CO₂ is produced for every mole of CH₄ used up.

2. The table tells us how the amount of CH₄ at zero minutes and at ten minutes, so this subproblem is simply the subtraction of the two numbers. $$.05-.03=.02$$ So .02 M of CH₄ is used up. It should also be noted that the amount of CH₄ lost equals the amount of CO₂ gained (according to question #1); the table shows that .02 M of CO₂ is made, so .02 M of CH₄ must have been used.

3. This question uses the logic that we discovered in the last two questions: the amount of CH₄ lost is the amount of CO₂ made. Thus we can subtract the amount of CH₄ at 20 minutes from the amount of CH₄ at zero minutes, and that will give us the amount of CH₄lost as well as the amount of CO₂. $$.05-.02=.03$$ So .03 M CO₂ is produced after 20 minutes.

4. The basic form of the reaction rate for an equation is Δ[A]/Δt, where Δ[A] is the change in the concentration of some molecule in the equation and Δt is the change in time. This rate equation is then modified depending on whether the molecule has a coefficient in the reaction equation or if the molecule is a reactant or a product.

The basic, unmodified reaction rate for CO₂ is thus Δ[CO₂]/Δt. Plugging values into this equation will always give you the rate of gain or loss of CO₂, but it may not represent the differential rate law for the entire reaction. Looking at the reaction equation, CO₂ has a coefficient of 1, meaning that nothing needs to be multiplied by the rate law to give the proper equation. CO₂ is a product of this reaction as well, so the equation will be positive. (If CO₂ was a reactant, the rate law would need a negative sign in front of it to guarantee that the reaction rate is positive; calculating rate for a reactant is naturally a negative number, because it is losing molarity.)

So the overall reaction rate for the equation in terms of CO₂ is Δ[CO₂]/Δt.

5. This problem can be solved by using the rate equation created in question 4. We learned in question 3 that .03 M CO₂ is created once 20 minutes have passed, so we subtract the molarity of CO₂ at ten minutes from this number and divide it by the change in time. $$\left (\dfrac{\Delta[\mathrm{CO_2}]}{\Delta t} \right)=\left (\dfrac{.03 M-.02M}{20 \textrm{min}-10 \textrm{min}}\right)=\dfrac{.01}{10}=.001$$ So the reaction rate during this interval of time is .001 M/min.

6. This problem is the same type of problem as question 5 expect that we need to calculate how much CO₂ is made when 30 minutes have passed. Again, this is simple subtraction of the amounts of CH₄ at zero minutes and thirty minutes. $$.05-.015=.035$$ So there is .035 M of CO₂ at thirty minutes. Now to plug everything into the rate equation: $$\left (\dfrac{\Delta[\mathrm{CO_2}]}{\Delta t} \right)=\left (\dfrac{.035 M-0M}{30 \textrm{min}-0 \textrm{min}}\right)=\dfrac{.035}{30}=.0012$$ So the average rate of this reaction during this time is .0012 M/min.

7. This problem will involve some of the ideas used in question 4. Δ[CO₂]/Δt is the overall rate for the reaction; how do we make Δ[O₂]/Δt output the same answer as that equation? The first step as stated before is to look at the reaction equation. $$\ce{CH_4}(g)+\ce{2O_2}(g)⟶\ce{CO_2}(g)+\ce{2H_2O}(l)$$ Oxygen has a coefficient of 2, meaning that its rate is twice as fast as carbon dioxide's rate. Two moles of oxygen are used up for every mole CO₂ produced. Thus oxygen's rate needs to be halved, making it Δ[O₂]/2Δt. Furthermore, O₂ is a reactant, meaning its rate will naturally be negative because it is losing molarity. Thus we need to add a negative sign to the rate of O₂ so that it equals the rate of CO₂ being produced. $$\left (\dfrac{\Delta[\mathrm{CO_2}]}{\Delta t} \right)=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{O_2}]}{\Delta t} \right)$$

8. The easiest way to do this problem is to manipulate the equality above. We calculated the rate of CO₂ creation from 10 to 20 minutes in question 5, so we can simply multiply both sides of the above equation by -2 and plug in the rate of CO₂, .001 M/min, to get our answer. $$(-2)\left (\dfrac{\Delta[\mathrm{CO_2}]}{\Delta t} \right)=(-2)\left(-\dfrac{1}{2}\right)\left (\dfrac{\Delta[\mathrm{O_2}]}{\Delta t} \right)$$ $$-2\left (\dfrac{\Delta[\mathrm{CO_2}]}{\Delta t} \right)=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}$$ $$-2(.001)=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}$$ $$\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=-.002$$

So the rate that O₂ is used up is -.002 M/min, with the negative sign signifying the loss of O₂.

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