Extra Credit 36

Question 12.2.3

In the PhET Reactions & Rates interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature?

Answer 12.2.3

First you must go to the following link to run the simulation in order to visualize what the questions is asking: PhET: Reactions & Rates

The above image shows the reaction speed at a constant temperature and an increased concentration than the initial concentration by a factor of 2. The rate of reaction is increased drastically compared to the initial concentration, which you can see for yourself if you time the rate of the reaction from the start and you document how quickly the bar graphs change of each respective collision. This happens because, according to the collision theory, when the concentration of reactants increases then there is a greater number of total collisions, which means there is a greater number of collisions that effectively produce products. Therefore, as the number of reactants goes up, so does the reaction rate (remember the reaction rate is dependent on the frequency of effective collisions). Note that the reverse is also true - as you decrease the concentration of reactants, then the reaction rate decreases as well, which represents a directly proportional relationship between concentration and reaction rate.

The above image represents the rate of reaction for just one effective collision when the temperature is drastically reduced. The reaction rate is decreased with a decrease in temperature because the molecules would have a lower speed, or a kinetic energy lower than the activation energy required for the reaction to occur, which means a lower proportion of molecules have enough kinetic energy to participate in the reaction.

This image shows that the reaction reaches an equilibrium rate faster than all of the other conditions; this condition is an increase in total average energy, which is shown as the green bar on the bottom right-hand graph. The rate of the reaction was in fact too fast to record using the stopwatch, which shows that effective collisions are occurring very often, as a higher proportion of molecules have enough speed or kinetic energy to overcome the activation energy barrier.

Question 12.5.8

In an experiment, a sample of NaClO₃ was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher?

Answer 12.5.8

First off, it is important to recognize that this decomposition reaction is a first-order reaction, which can be written as follows: \(\mathrm2NaClO_3\to2NaCl + 3O_2\)

Understanding this, it is important to be able to then be able to recognize which equation would be most useful given the initial conditions presented by the question. Since we are dealing with time, percentage of material left, and temperature, the only viable equation that could relate all of this would be the Arrhenius Equation, which is written as follows: \(\mathrm \ln(\frac{k_2}{k_1}) = \frac {Ea}{R}({\frac1{t_1}}-{\frac{1}{t_{2}}})\)

However, this problem does not give us enough information such as what the activation energy is or the initial temperature in order to mathematically solve this problem. Additionally, the problem tells us to approximate how long the decomposition would take, which means we are asked to answer this question conceptually based on our knowledge of thermodynamics and reaction rates. As a general rule of thumb, we know that for every 10˚C rise in temperature the rate of reaction doubles. Since the question tells us that there is a 20˚C rise in temperature we can deduce that the reaction rate doubles twice, as per the general rule mentioned before. This means the overall reaction rate for this decomposition would quadruple, or would be 4 times faster than the reaction rate at the initial temperature.

We can gut check this answer by recalling how an increase in the average kinetic energy (temperature) decreases the time it takes for the reaction to take place and increase the reaction rate. Thus, if we increase the temperature we should have a faster reaction rate.

Question 14.6.9

The following reactions are given:

\(A+B\underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}} C+D\)

\(D+E\underset{}{\stackrel{k_{2}}{\rightarrow}} F\)

A. What is the relationship between the relative magnitudes of \(k_{−1}\) and \(k_{2}\) if these reactions have the following rate law?

\(\frac{∆F}{∆t}=k\frac{[A][B][E]}{[C]}\)

B. How does the magnitude of \(k_{1}\) compare to that of \(k_{2}\)?

C. Under what conditions would you expect the rate law to be

\(\frac{∆F}{∆t}=k'[A][B]\)

Assume that the rates of the forward and reverse reactions in the first equation are equal.

Answer 14.6.9

a) The first thing you should notice about this reaction mechanism is that the rate of the forward reaction equals the rate of the reverse reaction in the first reaction. In addition, there is an intermediate D - remember an intermediate is produced and then consumed before the end of the reaction. As such, we can already determine that D will allow us to relate the magnitudes of \(k_{-1}\) and \(k_{2}\). In order to determine this relationship we must relate the rate laws of reaction 1 an 2 together and set the combined rate law equal to the \(\frac{∆F}{∆t}\) substitution that is already given in the problem. Also, because the intermediate D is on the reactant side of the second step of the overall reaction we can surmise that this reaction is most likely the rate determining step.

Rate Law for Equation 1: \(k_{1}[A][B]=k_{-1}[C][D]\) and isolating [D] we get \([D] = k_{1}\frac{[A][B]}{k_{-1}[C]}\)

Rate Law for Equation 2: \(k_{2}[D][E]\)

Combined Rate Laws Substituting [D]: \(k_{2}k_{1}\frac{[A][B][E]}{k_{-1}[C]}\)

Since \(\frac{∆F}{∆t}=\frac{k[A][B][E]}{[C]}\) , then the relationship between relative magnitudes of \(k_{-1}\) and \(k_{2 }\) is: \[k_{2} = \frac{k}{k_{1}}(k_{-1})\]This indicates that there is a direct relationship between the aforementioned magnitudes and that \(k_1\)>>\(k_2\), as \(k_2\) would be the rate determining or slow step given that k = \(k_2\) from the relationship we just determined, as \(k_{-1}\) and \(k_2\) are equal and would therefore cancel each other out in the equation.

b) We can compare the magnitudes of \(k_1\) and \(k_2\) by using the relationship we determined regarding the relative magnitudes from part a. All we have to do is rewrite the equation in terms of \(k_2\), which gives us: \(k_2=(k)(k_{-1})\frac{1}{k1}\). From this we can tell that there is an inverse relationship between \(k_1\) and \(k_2\). In addition, since we concluded that the \(k_2\) may be the slow step we can assert that \(k_1\)>>\(k_2\).

c) The overall rate law given to us in part c is the same rate law for the first reaction going in the forward reaction. The only time the rate of the forward reaction would be the overall reaction rate is if the rate of the forward \(k_1\) reaction is the slow step, as the overall rate is only as fast as its slowest step.

Question 20.2.7

Which of these metals produces H₂ in acidic solution?

1. Ag
2. Cd
3. Ca
4. Cu

Answer 20.2.7

In order to answer this problem you have to write out the equation that is described by the question. An acidic solution indicates the presence of H^+ and H2 is produced, so we would use this information to find an equation within the table of standard reduction potentials. The equation is as follows: \(2H^{+} + 2e^{-}\to H_{2}(g)\) with \(E˚= 0 V\). Recall that a reaction occurs when \(E˚_{cathode} - E˚_{anode} > 0\). It is important to mention that reduction occurs at the cathode and in an acidic solution the \(H^{+}\) would gain electrons and would be the reduction reaction that occurs at the cathode. Since \(E˚_{cathode}\) equals 0, any of the metals listed that have a standard reduction potential that is negative would result in a net E˚ that is greater than 0 and therefore would proceed as written and produce \(H_2\) in an acidic solution. Because the reactions of both \(Cd^{2+}\) and \(Ca^{2+}\) have standard reduction potentials of -0.403V and -2.84V respectively, they would produce \(H_2\) in an acidic solution. In addition, you can also find the answer to this problem by analyzing a table with activity series values for oxidation reactions of the metals listed below. Only the metals above \(H_2\) would be able to be oxidized when combined with \(H^{+}\) to produce \(H_2\) in an acidic solution. Therefore, we can confirm that Cd and Ca are able to produce \(H_2\) in an acidic solution.

Question 20.5.2

A. What is the relationship between the measured cell potential and the total charge that passes through a cell?

B. Which of these is dependent on concentration?

C. Which is dependent on the identity of the oxidant or the reductant?

D. Which is dependent on the number of electrons transferred?

Answer 20.5.2

a) The relationship between the measured cell potential (E) and the total charge that passes through a cell given by the number of electrons transferred n throughout the system is determined by an equation involving Faraday's constant (96,485 \(\frac{Coulombs}{mol of e^{-}}\)). This equation is: \(∆G=-nFE_{cell}\). The equation measures the maximum amount of work that can be performed by the cell, as the maximum work is equal to the total charge that passes through the cell multiplied by the measured cell potential of the cell.

b) Only \(E_{cell}\) is dependent on concentration, as the Nernst Equation, which gives the value of \(E_{cell}\) in which \(E_{cell} = E˚- (\frac{0.0592V}{n})logQ\). Q is the reaction quotient that depends on the concentrations of the products as well as the reactants in any given equation, as Q can be written in a fraction form with the products on the numerator to their stoichiometric power and reactants on the denominator to their respective stoichiometric power as well.

c) Given that \(E_{cell}\) is dependent on the value of E˚, and E˚ is calculated by the equation \(E˚_{cathode} - E˚_{anode}\), then \(E_{cell}\) would correspondingly be dependent on the identity of the oxidant or reductant, as the oxidant corresponds to the \(E˚_{cathode}\) value and the reductant corresponds to the \(E˚_{anode}\) value.

d) The variable n is dependent on the number of electrons transferred between the cathode and the anode, which is given by the number of electrons present in the balanced redox equation.

Question 20.9.11

Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.

1. AgNO₃
2. RbI

Answer 20.9.11

In order to answer this question we would need to look at the standard reduction potential values, as the reduction potential values allow us to determine which of the compounds would be oxidized and reduced. The reason we would need this is because the direction in which each half reaction would proceed when coupled together is found by determining which compound is reduced, which means the half reaction would proceed as written in table, and which compound is oxidized, which means the half reaction from the table would need to be written such that electrons are released as products. The table below is an example of how the reactions are written with the most negative (low reduction potential) voltage values listed first and the positive ones listed next. A more expanded table that also includes the Rb half reaction can be found using the following link: http://www.charleston.k12.il.us/chs/...Potentials.pdf

In order to predict the products we can write the half reactions such that they represent an oxidation and reduction equation respectively. Since Ag has a higher reduction potential value than Rb, it would be the reduction half reaction, which is written directly from the table, and the Rb equation would be the oxidation half reaction that would be written in the reverse order in order to reflect an oxidation reaction. The half reactions are as follows:

Reduction: \(Ag^{+}(aq) + e^{-}\to Ag(s)\)

Oxidation: \(Rb(s)\to Rb^{+}(aq) + e^{-}\)

Note the overall reaction would not include the spectator ions present in the initial question, which are \(NO_{3}\) and I respectively. Moreover, the equations are balanced in terms of electrons such that the electrons can be directly canceled out to form the following net equation:

\(Ag^{+}(aq) + Rb(s)\to Ag(s) + Rb^{+}(aq)\)

Thus, we can predict that the products would be Ag(s) and \(Rb^{+} (aq)\) from the above equation.

Question 21.4.3

What is the change in the nucleus that results from the following decay scenarios?

1. emission of a β particle
2. emission of a β⁺ particle
3. capture of an electron

Answer 21.4.3

This question requires us to have a proper understanding of the different types of radioactive decay. Changes in nucleus that result from the aforementioned decay scenarios would affect either the atomic number (and therefore change the actual element), the mass number (making the atom a different isotope of the element), or both the atomic and mass number. The following table gives us a clear idea of how each decay takes place and what to look for in terms of changes to nuclei:

For the first β emission scenario, which is denoted in the second row of the table, the atomic mass remains the same; however, the atomic number increases by one - thus making a new element. This is because the daughter nucleus gains a proton.

The second positron emission no change in atomic mass, but does have an atomic number decrease by one as a result of the emission, which means a new element is also formed, as the nucleus has one less proton because the proton was emitted and replaced with a neutron.

The third electron capture occurs when a proton and electron are transformed into a neutron, which results in a decrease in proton number by 1, which again is a new daughter nucleus or element.

Question 17.5.4

Consider a battery with the overall reaction:

\(Cu(s) + 2Ag^{+}(aq)\to 2Ag(s) + Cu^{2+}(aq)\)

1. What is the reaction at the anode and cathode?
2. A battery is “dead” when it has no cell potential. What is the value of Q when this battery is dead?
3. If a particular dead battery was found to have \([Cu^{2+}]=0.11M\), what was the concentration of silver ion?

Answer 17.5.4

1) To determine which reaction occurs at which particular electrode it may be useful to use the following mnemonic: Red Cat - reduction @ cathode; An Ox - oxidation @ anode. By simply writing out the oxidation number of each species we can see that copper is going from an oxidation number of 0 to that of \(2^{+}\), whereas silver is going from oxidation number of \(1^{+}\) to 0, which means that copper is undergoing an oxidation reaction while silver is undergoing a reduction reaction respectively. After assigning each half reaction as reduction or oxidation, we simply use the aforementioned mnemonic to determine whether each reaction is at the anode or cathode. Therefore, the reaction of copper is at the anode, while the reaction of silver is at the cathode. The half reaction of copper at then anode is:

\[Cu(s)\to Cu^{2+}(aq) + 2e^{-}\]

And the balanced half reaction (in terms of equal electrons) of silver at the cathode is:

\[2Ag^{+}(aq)+2e^{-}\to 2Ag(s)\]

2) It is important to note that when the question says there is no cell potential it is really saying that the non-standard cell potential is equal to 0 - the standard cell potential on the other hand is still constant from the equation: \(E˚_{cathode}- E˚_{anode}\) In order to solve for Q we must use the following equation (where n is the number of electrons transferred among the balanced half reactions) that relates non standard cell potential to the reaction quotient Q: \[E_{cell}=E˚_{cell}- (\frac{0.0592V}{n})log(Q)\]

Using the fact that \(E_{cell}\) is equal to 0 and \(E˚{cell}\) is 0.4577V (obtained from the aforementioned cathode/anode equation and the table of standard reduction potential values) we can rearrange the preceding equation to solve for Q: \[log(Q)={E˚_{cell}(n)\over 0.0592V}\]

When plugging in 2 for the total number of electrons in the circuit and the E˚ from above we obtain a Q value of \(2.0*10^{15}\) as our answer.

3) Since this problem is asking us to calculate the concentration of silver ion given the same conditions as before (dead battery) we can use the Q we calculate above to answer this question, as Q takes into account all relevant concentrations within the context of this problem. Q only takes into account the aqueous elements within the cell, which gives us: \[Q=\frac{[Cu^{2+}]}{[Ag^{+}]^{2}}\]

Plugging in the value of of Q as well as the given value of \(Cu^{2+}\) concentration at 0.11M, we get a silver ion concentration of \(6.16*10^{-9}M\) in the given dead battery.