Extra Credit 41

Q17.6.1

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

1. Mg or Ca
2. Au or Hg
3. Fe or Zn
4. Ag or Pt

S17.6.1

1. Answer: Ca

Explanation: By looking at a Table of Standard Reduction Potentials (SRPs), we see that the reduction of Mg²⁺ to Mg (s) has an SRP of -2.36 V, and that the reduction of Ca²⁺ to Ca (s) has an SRP of -2.84 V. A negative reduction potential indicates that reduction reaction is nonspontaneous. This tells us that the reverse reaction, which would be the oxidation reaction, is spontaneous. A greater negative SRP corresponds to a greater positive oxidation potential. Thus, calcium metal is more likely to corrode (oxidize) than magnesium.

2. Answer: Hg

Explanation: By looking at a Table of SRPs, we see that Hg²⁺ has a reduction potential of 0.8535 V and that Au³⁺ has a reduction potential of 1.52 V. Since both reduction potentials are positive, the reduction reactions are spontaneous. Since Au has a a greater positive reduction potential, its corresponding oxidation potential is more negative, meaning that its oxidation is more thermodynamically unfavorable compared to Hg. Thus, Hg is more likely to corrode.

3. Answer: Zn

Explanation: By looking at a Table of SRPs, we see that Fe²⁺ has a reduction potential of -0.44 V and that Zn²⁺ has a reduction potential of -0.7618 V. Since Zn has a more negative reduction potential, it is a stronger reducing agent (i.e., more easily oxidized) than Fe. Since Zn is more likely to corrode than Fe, Zn is commonly used as a "sacrificial anode" for the cathodic protection of steel, an alloy of iron.

4. Answer: Ag

Explanation: By looking at a table of SRPs, we see that Ag⁺ has a reduction potential of 0.7996 V and that Pt²⁺ has a reduction potential of 1.2 V. Since Ag has a smaller reduction potential than Pt, it is a stronger reducing agent (or weaker oxidizing agent). Therefore, Ag is more easily oxidized and is more likely to corrode compared to Pt.

Q12.3.4

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

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

S12.3.4

1. Answer: Decreases reaction rate by a factor of four.

Explanation: The pressure of a gas (in atmospheres) is proportional to its molarity. The reaction rate indicates that the reaction is second order with respect to NO_2. Thus, decreasing the pressure by a factor of two (from 0.50 to 0.250 atm), decreases the reaction rate by a factor of 2². This can be mathematically demonstrated by

rate = k[NO₂]²

(1) rate = k[0.50]² = 0.25k

(2) rate = k[0.25]² = 0.0625k

Since k is a constant, we can see that 0.0625 is four times smaller than 0.25.

2. Answer: No affect on the rate of the reaction.

Explanation: Since CO does not appear in the rate law, the reaction rate is zero-order with respect to CO. This tells us that the reaction rate is independent of the concentration of CO. If we were to include CO in the rate law, we would get rate = k[NO₂]²[CO]⁰. By the fact that any nonzero integer raised to the zero power is 1, the reaction rate is only dictated by k[NO₂]² and CO can be excluded from the rate law.

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:

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

S12.5.13

Answer: 180 kJ/mol

Explanation: Two possible strategies for solving this problem:

1. Plot ln(k) vs. 1/Temperature (K) to determine the slope, which is equal to -E_a/R, where E_a is the activation energy and R is the ideal gas constant. Multiplying the slope by -R yields the activation energy.

2. Because the procedure described above yields a linear graph (i.e., constant slope), we can estimate the activation energy by using data from any two temperatures. This can be done by using the following form of the Arrhenius equation, using data from 575 K and 555 K.

ln(k₂/k₁) = -(E_a/R)(1/T₂ - 1/T₁)

ln(2.42 x 10^-6/ 6.23 x 10^-7) = - (E_a/0.0083145 kJmol^-1K^-1) (1/575K - 1/555K )

E_a = [-(0.0083145 kJmol^-1K^-1) ln(2.42 x 10^-6/ 6.23 x 10^-7)] / (1/575K - 1/555K)

E_a = 180 kJ/mol

Solving for E_a, we find that the reaction has an activation energy of 180 kJ/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.

a.

b.

c.

d.

e.

S21.4.8

a. Answer: beta decay

Explanation: By undergoing beta decay, P-34 becomes S-34. Since sulfur has sixteen protons, this isotope will have an even numbers of protons and neutrons, which is generally more stable.

b. Answer: alpha decay

Explanation: Elements with an atomic number greater than or equal to 84 typically undergo alpha decay. Thus, U-239 is expected to decay to Th-235.

c. Answer: positron emission

Explanation: By looking at the belt of stability, elements with atomic numbers less than or equal to 20 generally have 1:1 rations of neutrons to protons. Since Ca-38 has a ratio less than 1,it lies below the belt of stability and is expected to undergo positron emission. By emitting a positron, Ca-38 decays to K-38, which has a 1:1 ratio of neutrons to protons.

d. Answer: beta decay

Explanation: H-3 lies above the belt of stability and is expected to undergo beta decay to form He-3. Since He has 2 protons, which is a magic number, it is more stable than H-3.

e. Answer: beta decay

Explanation: Pu-245 has a neutron:proton ratio of 1.6, while the ideal ratio is between 1 to 1.5. Therefore, it is expected to undergo beta decay to Am-245, by converting a neutron into a proton and emitting an electron. This lowers its n/p ratio.

S20.2.12

1. Answer: Iron displaces nickel ions from solution.

net ionic equation: Ni²⁺(aq) + Fe(s) ⟶ Fe²⁺(aq) + Ni(s)

complete ionic equation: Ni²⁺(aq) + 2Br^-(aq) + Fe(s) ⟶ Fe²⁺ (aq) + 2Br^-(aq) + Ni(s)

Explanation: Since iron has a more negative SRP, it is able to displace nickel ions from solution.

2. Answer: Zinc displaces H+ from HCl.

net ionic equation: Zn(s) + 2H⁺(aq) ⟶ Zn²⁺(aq)+ H₂(g)

complete ionic equation: Zn(s) + 2H⁺(aq) + 2Cl^-(aq) ⟶ Zn²⁺(aq)+ 2Cl^-(aq) + H₂(g)

Explanation: Since zinc has a more negative SRP, it is able to displace hydrogen ions from acidic solutions.

3. Answer: No reaction

Explanation: Since copper has a more positive SRP, copper is not reactive enough to displace zinc ions from solution.

4. Answer: Aluminum displaces silver ions from solution.

net ionic equation: 3Ag⁺(aq) + Al(s) ⟶ 3Ag(s) + Al³⁺(aq)

complete ionic equation: 3Ag⁺(aq) + 3NO₃^-(aq) + Al(s) ⟶ 3Ag(s) + 3NO₃^-(aq) + Al³⁺(aq)

Explanation: Since aluminum lies above silver in the activity series, it is reactive enough to displace silver ions from solution.

Phase 2 Edits: Depending on what table you use, the SRPs could be in increasing order or decreasing order, therefore a reaction could lie higher or lower than another reaction depending on the table used (libretext and sapling use these two differing tables). Thus, we can't generalize and say that solids will displace ions out of solution if the reaction for the solid lies above the reaction for the solution metal ions. This is only true for activity series that list SRPs from most negative to most positive. The same is true for the reverse situation. It is easier to distinguish whether a metal will displace metal ions from a solution by seeing if it has the most negative SRP and is the most reactive. From there, depending on the activity series table used, you can distinguish if the reaction for the solid metal in question lies above or below the reaction for the metal solution ions.

Q12.1

The following data were obtained for the reaction of methane with oxygen:

CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l)

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

S12.1

1. Answer: 1 mole CO₂ per mole CH₄ consumed.

Explanation: The stoichiometry of the balanced reaction shows that there is a 1:1 ratio between carbon dioxide and methane.

2. Answer: 0.020 M

Explanation: The concentration of CH₄ used up after 10 minutes is the difference in concentration between the initial concentration and the concentration at 10 minutes. These values are obtained from the given table.

Δ[CH₄] represents the change in the concentration of CH₄, which is equivalent to quantity consumed.

Δ[CH₄] = [CH₄]_t=10 - [CH₄]_t=0

= 0.030 M - 0.050 M

= - 0.020 M

The concentration decreases by 0.020 M because 0.020 M of methane is consumed.

3. Answer: 0.030 M

Explanation: Since the ratio of carbon dioxide and methane is 1:1, the amount of methane consumed is equal to the amount of carbon dioxide produced. Since at 20 minutes the concentration of methane has decreased from 0.050 M at t=0 to 0.020 M at t=20 minutes, we see that 0.030 M of methane is consumed in 20 minutes.

Δ[CH₄] represents the change in the concentration of CH₄, which is equivalent to quantity consumed.

Δ[CH₄] =[CH₄]_t=20 - [CH₄]_t=0

= 0.020 M - 0.050 M

= -0.030 M

The absolute value is also equivalent to the quantity of CO₂ produced by reaction stoichiometry. Assuming reaction is contained in a 1-liter tank,

(0.030 mol CH₄ /1-L)(1 mol CO₂/1 mol CH₄) = 0.030 M CO₂

4. Answer: rate = Δ[CO₂]/Δt

Explanation: Since there is no coefficient in front of the CO₂ term, the rate of the reaction over a time interval is proportional to the change in the concentration of CO_2, Δ[CO₂], divided by the change in time, Δt. This rate tells us how the concentration of carbon dioxide changes with respect to time. Specifically, since CO₂ is a product, the rate tells us how fast CO₂ (g) is produced as methane reacts with oxygen.

5. Answer: 0.0010 M/min

Explanation:

In part 4 we determined that,

rate = Δ[CO₂]/Δt

We divide the difference in the concentrations over the interval of time between 10 and 20 minutes by the change in time. The concentration of carbon dioxide at t=20 min was determined in part 3.

rate = Δ[CO₂]/Δt

= ([CO₂]_t=20 - [CO₂]_t=10)/(20 min - 10 min)

= (0.030 M - 0.020 M)/10 min

= 0.0010 M/min

6. Answer: 0.0012 M/min

Explanation:

The average rate of the reaction is proportional to the change in the concentration of methane over the interval 0 to 30 min. We have to include a negative sign because the rate of change of methane is negative since it is being consumed, and reaction rates should be expressed as positive values.

rate = -Δ[CH₄]/Δt

= -([CH₄]_t=30 - [CH₄]_t=0)/(30 min - 0 min)

= -(0.015 M -0.050 M)/(30 min)

= 0.0012 M/ min

7. Answer: reaction rate = -Δ[O₂]/2Δt =Δ[CO₂]/Δt

Explanation: By looking at the reaction stoichiometry, we can see that oxygen is being consumed twice as fast as carbon dioxide is being produced. Therefore, in order to express the rate as equal to the rate of carbon dioxide, we must divide the rate of change in oxygen by 2. In addition, since the concentration of oxygen is decreasing as the concentration of carbon dioxide is increasing, we must include a negative in front of the rate of change of oxygen in order to have the rates be positive and equal.

8. Answer: Oxygen is being consumed at a rate of 0.0020 M/min

Explanation:

In step 7 we obtained that,

reaction rate = -Δ[O₂]/2Δt =Δ[CO₂]/Δt

Between ten and twenty minutes:

⇔ Δ[O₂]/Δt = -2([CO₂]_t=20 - [CO₂]_t=10)/(20 min - 10 min)

⇔ Δ[O₂]/Δt = -2(0.030 M - 0.020 M)/(20 min - 10 min)

⇔ Δ[O₂]/2Δt = -0.0020 M/min

Therefore, Oxygen is being consumed at a rate of 0.0020 M/min.