Extra Credit 34

Question 12.2.1

Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.

Solution 12.2.1

Collision theory states that the rates of chemical reactions depend on three factors: 1. fraction of molecules with require orientation 2. Fraction of collisions with required energy 3. Collisions frequency

Fraction of collisions with required energy is a function of temperature which means as temperature of the solution increases, fraction of collisions with required energy also increases. Also, as temperature increases, kinetic energy of molecule increases as well which means molecules will collide more often increasing collisions frequency. With increased fraction of collisions with required energy and collisions frequency, the rate of a chemical reaction increases.

As the size of the pieces of magnesium decreases, the rate of the chemical reaction increases. This is because as the size of the pieces of magnesium decreases, the surface area of the pieces of magnesium increases. The more surface area means that there is more surface for hydrochloric acid molecules to collide to the pieces of magnesium increasing collision frequency and thus overall rate of the chemical reaction. Meaning smaller pieces of magnesium metal will react more rapidly since more reactive surface exists.

As molarity of hydrochloric acid increases, the rate of the chemical reaction increases. This is because as molarity of hydrochloric acid increases, the number of moles of hydrochloric acid increases while the volume is held constant. The extra amount of hydrochloric acid molecules will collide with the magnesium metal which increases the number of collisions. These molecules have the required energy for the reaction to occur. With increased molecules with required energy, the overall rate of reaction increases.

Question 12.5.6

How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.

Solution 12.5.6

Collision theory states that the rates of chemical reactions depend on three factors: 1. fraction of molecules with require orientation 2. Fraction of collisions with required energy 3. Collisions frequency

Fraction of collisions with required energy is a function of temperature which means as temperature increases, fraction of collisions with required energy also increases. Also, as temperature increases, kinetic energy of reactants increase as well which means molecules will collide more often increasing collisions frequency. Collisions between molecules will be more violent at higher temperatures. The higher temperatures mean higher molecular velocities. You need to increase the number of the very energetic particles- those with energies equal to or greater than the activation energy. The very energetic particles now collide with enough energy to react. The increased number of collisions and the greater violence of collisions results in more effective collisions. With increased fraction of collisions with required energy and collisions frequency, the rate of a chemical reaction increases.

Question 14.6.7

Above approximately 500 K, the reaction between NO₂ and CO to produce CO₂ and NO follows the second-order rate law Δ[CO₂]/Δt = k[NO₂][CO]. At lower temperatures, however, the rate law is Δ[CO₂]/Δt = k′[NO₂]², for which it is known that NO₃ is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest?

Solution 14.6.7

We are given that NO₃ is an intermediate which means NO₃ is a product or one of the products of first reaction and a reactant of second reaction and thus does not show up in the overall reaction. With these and the description found in the question we know that first reaction has to produce NO₃ using NO₂

\[\mathrm{2NO_{2\;(g)} NO_{3(g)}+\rightleftharpoons NO_{(g)}\;\;\;\;\;\;\;(1) equalibrium}\]

This step has rate law of \[\mathrm{Rate\;Law=k[NO_{2}]^{2}=k'[NO]_{3}[NO]}\]

NO₃ is now reactant of second reaction

\[\mathrm{CO_{(g)}+NO_{3(g)}\rightarrow CO_{2(g)}+NO_{2(g)}\;\;\;\;\;\;\;(2)} \]

This step has rate law of

\[\mathrm{Rate\;Law=k[NO_{3}][CO]}\]

Combining chemical reaction 1 and 2, now we find overall reaction.

\[\mathrm{CO_{(g)}+NO_{2(g)}\rightarrow CO_{2(g)}+NO_{(g)}\;\;\;\;\;\;\;(overall\;reaction)}\]

Now we just have to find out which is slowest. To do so we have to use the fact that at low temperature mechanism's rate is Δ[CO₂]/Δt = k′[NO₂]². Let's assume that first step is slowest step. Since first step is in equilibrium we can solve for [NO₃]

\[\mathrm{[NO_{3}]=\frac{K[NO_{2}]^{2}}{K[CO]}}\]

Using this we can see that overall rate law is

\[\mathrm{k'\times\frac{K[NO_{2}]^{2}}{K[CO]}\times[CO]=K'[NO_{2}]^{2}}\]

Which confirms that first step is indeed slowest step

Question 17.5.2

List some things that are typically considered when selecting a battery for a new application

Solution 17.5.2

Some factors that need to be considered include cost and toxicity of the materials used to make a battery, should it be primary or secondary battery, energy requirements, and total mass of the battery.

Primary batteries are not rechargeable. However, primary batteries have high specific energy, long storage time and instant readiness. Also, sometimes recharging is impractical or impossible. Thus, if new application needs long storage time and high specific energy, and used in context where recharging is not practical, it needs primary energy. On the other hand, secondary battery is much more cost-efficient and recharging battery has been made easily available with the development of technologies. But regular maintenance is required and you need to constantly recharge it in order to function. Primary batteries are only limited for specific applications. Secondary batteries can be used in numerous applications and have continuing research for much more applications.

Toxicity and cost of the materials used in battery are another important factors. Alkaline is one of most common material to make a battery. It is a lot more affordable since it's not made to be long-lasting. It is cost effective and environmentally friendly. Lead Acid battery is not only expensive but highly toxic thus cannot be disposed in landfills. Nickel-cadmium batteries are also expensive and contain cadmium which is harmful. Lastly, lithium-ion batteries are expensive and need a protection circuit but high cycle count and low maintenance reduce the cost per cycle.

Total mass of the battery is important as well. Depending on whether new application needs to be portable or not, total mass of the battery needs can be light or heavy. If new application is something like a phone, total mass of a phone would depend on mass of a battery so a battery needs to be light. Primary batteries are usually smaller and lighter and are more used for transportation. Secondary batteries are more bulky and large. Only in recent and more advanced Lithium batteries are more lightweight like primary batteries.

Question 20.2.5

Of the following elements, which would you expect to have the greatest tendency to be oxidized: Zn, Li, or S? Explain your reasoning.

Solution 20.2.5

To answer this question, we need to take a look at the electrochemical series and compare E° values of Zn, Li and S. Each E° value shows whether the position of the equilibrium lies to the left or right of the H₂ equilibrium(the standard). The difference in the positions of equilibrium causes the number of electrons which build up on the metal electrode and the platinum of the hydrogen electrode to be different. That produces a potential difference which is measured as a voltage.

We find that Lithium has E° of -3.05V, Zinc of E°=-1.66V, and Sulfur of E°=0.14. Smaller the E° of element is, the better it acts as a reducing agent and likely to be oxidized and bigger the E° of element is, the better it acts as a oxidizing agent and likely to be reduced. We see that Lithium has smallest E° and is therefore more easily oxidized.

Question 20.4.24

Your lab partner wants to recover solid silver fro silver chloride by using a 1.0 M solution of HCl and 1 atm H₂ under standard conditions. Will this plan work?

Solution 20.4.24

First, we need to write out the described plan into chemical equation.

\[\mathrm{AgCl\;+H}_{2}\rightarrow \mathrm{HCl\;+Ag}\]

But we see that chemical equation is not balanced, so we need to balance. In order to balance it, we see that there are two Hs on the reactant side and one H on the product side. So we need to add a 2 to the HCl. Because now there are two Cls on the product side, we need to add a 2 to the AgCl. Lastly, there are 2 Ags on the reactant side, so we need to add a 2 on the Ag.

\[\mathrm{2AgCl\;+H}_{2}\rightarrow \mathrm{2HCl\;+2Ag}\]

Now, we see that H₂ needs to replace Ag from AgCl to form HCl. But we do not know if such reaction is spontaneous process or not. Therefore, we need to take a look at standard reduction potentials. We must do one more step before doing so. Identifying reduction and oxidation reaction. Remember OIL RIG. Oxidation Is Losing electrons and Reduction Is Gaining electron. The Ag from AgCl is reduced from a +1 charge(because Cl is -1) to a 0 charge. So it is becoming reduced. H₂ is being oxidized because it’s a 0 charge(being in its natural form) to a +1 charge.

Reduction: \[\mathrm{2Ag^{+}+2e^{-}\rightarrow 2Ag\;\;\;\;\;E^{\circ}=0.22}\]

Oxidation:\[\mathrm{H_{2}\rightarrow 2H^{+}+2e^{-}\;\;\;\;\;E^{\circ}=0}\]

Now we use Nernst Equation under standard condition!

\[\mathrm{\Delta G^{\circ}=-nFE^{\circ}}\]

n is the number of electrons transferred in the reaction

F is the Faraday constant (96,500 C/mol)

E° is potential difference

\[\mathrm{E^{\circ}=E_{reduction}^{\circ}-E^{\circ}_{oxidation}}\]

\[\mathrm{E^{\circ}=0.22-0.00=0.22}\]

\[\mathrm{\Delta G^{\circ}=-2\times96500\times0.22=-42,460}\]

We see that Gibbs free energy is negative number so this reaction is spontaneous. Therefore, your lab partner's plan will work

Question 20.9.9

What mass of PbO₂ is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery?

Solution 20.9.9

From the description of the question, we find that lead is being reduced which means that lead is gaining electrons. Now we look at the standard reduction potential table and find the corresponding reaction of lead

\[\mathrm{Pb^{+2}_{(aq)}+2e^{-}\rightarrow Pb}_{(s)}\]

This means that for every 1 mole of lead reduced, 2 moles of electrons are transferred.

To convert convert to moles of electron we use formula \[\\\mathrm{n_{e}=\frac{It}{F}}\]

I = current (amps)

t=time (sec)

F= Fraday's constant (9.65x10^4 C/mole)

n_e= number of moles of electrons transferred

We cannot use 2 hours directly from the question because t in the formula is in seconds thus we have to convert 2 hours into seconds

\[\frac{2hr}{1}\times\frac{60min}{1hr}\times\frac{60sec}{1min} =7200 seconds\]

Because \[A=\frac{1c}{1sec}\]

We have \[\mathrm{n_e=\frac{7200Sec\times10\frac{C}{Sec}}{96,500\frac{C}{mole}}}=0.373 e^-\]

The seconds on the top(numerator) cancel out while the C cancels out from the top dividing by the bottom, leaving us with the number of moles of electrons.

But since every 1 mole of Pb means 2 moles of electrons transferred, we need to use this ratio in order to find how many moles of Pb was reduced.

\[0.373 e^-\times\frac{1molPb}{2mol\;e^-}=0.187\;mol\;Pb\]

Finally, we convert moles of Pb to grams of Pb by multiplying by molar mass, leaving us with the mass of Pb that was taken from the current/

\[0.187\;mol\;Pb\times239.2\frac{g}{mol} =44.73\;g\;PbO_{2}\;reduced\]

Question 21.4.1

What are the types of radiation emitted by the nuclei of radioactive elements?

Solution 21.4.1

α (helium nuclei), β (electrons), β⁺ (positrons), and η (neutrons) are the types particles of radiation emitted by the nuclei of radioactive elements. γ rays may be emitted.

α(Alpha particle) is composed of two protons and two neutrons. It is also known as helium. When an alpha particle is emitted, the atomic mass of an element goes down by 4 amu due to the loss of 4 nucleons while the atomic number of the atom goes down by 2 as the element loses 2 protons becoming a new element.

\[_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_{2}^{4}\alpha\]

\[_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+^{4}\textrm{He}^{2+}\]

β (Beta particle) are just electrons from the nucleus. The atomic number is designated by a -1 charge, displaying the opposite charge of a proton. The emission of a beta particle does not change the mass of the daughter nucleus but changes atomic number

\[_{55}^{137}\textrm{Cs}\rightarrow _{56}^{137}\textrm{Ba}^{+}+\beta ^{-}\]

β⁺ (positrons) are the antimatter counterparts of electrons. It has the mass of a electron but with a +1 charge instead. The emission of a positron does not change the mass of the daughter nucleus but changes atomic number

\[_{11}^{22}\textrm{Na}\rightarrow _{10}^{22}\textrm{Ne}+\beta^{+}\]

Gamma ray emission usually occurs with α and β emission. They are emitted by a nucleus in an excited state trying to go back to a more stable, ground state. Gamma rays have no charge or mass as they are just a photon(high energy electromagnetic particles). Gamma ray emission occurs because the nucleus is often unstable after α and β decay

\[_{90}^{234}\textrm{Th*}\rightarrow _{90}^{234}\textrm{Th}+\;\gamma\]

η particles are neutrons. It does not change atomic number but increases atomic number by one

\[_{235}^{92}\textrm{U}+_{0}^{1}\textrm{n}\rightarrow _{92}^{236}\textrm{U}\]