Extra Credit 31

Q17.4.4

Q12.1.4

A study of the rate of dimerization of \(C_{4}H_{6}\) gave the data shown in the table:

2\(C_{4}H_{6}\rightarrow C_{8}H_{12}\)

S12.1.4

1. Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.

Solution:

a)

GIVEN:

initial time: \(0 s\)

final time: \(1600 s\)

initial [\(C_{4}H_{6}\)]: \(1.00\times 10^{-2} M\)

final [\(C_{4}H_{6}\)]: \(5.04\times 10^{-3} M\)

UNKNOWN:

rate of dimerization of \(C_{4}H_{6}\) : ?

Equations to use:

\(-1/2\times Δ[\(C_{4}H_{6}\)] /Δt\)

And plug in your values:

= (-1/2) × ((5.04 × 10−3 M - 1.00 × 10⁻² M) / (1600 s - 0 s)

= \(1.55 \times 10^{-6} M/s\)

b)

GIVEN:

initial time: \(1600 s\)

final time: \(3200 s\)

initial [\(C_{4}H_{6}\)]: \(5.04\times 10^{-3} M\)

final [\(C_{4}H_{6}\)]: \(3.37\times 10^{-3} M\)

UNKNOWN:

rate of dimerization of \(C_{4}H_{6}\) : ?

Equations to use:

-1/2 × Δ[\(C_{4}H_{6}\)] / Δt

And plug in your values:

= \(-1/2\times 3.37 \times 10^{-3} M - 5.04\times 10^{-3} M) / (3200 s - 1600 s) \)

= 4.19 × 10^-7 M/s (Original Solution)

= \(5.22 \times 10^{-7} M/s\) (Phase II edit)

2. Estimate the instantaneous rate of dimerization at \(3200 s)\ from a graph of time versus [\(C_{4}H_{6}\)]. What are the units of this rate?

GIVEN:

time: \(3200 s\)

[\(C_{4}H_{6}\)] at time: \(3.37\times 10^{-3} M\) and 2.53\times \(10^{-3}\)

UNKNOWN:

instantaneous rate of \(C_{4}H_{6}\) : ?

units: ?

Equations to use:

\(- 1/2 × ( d[\(C_{4}H_{6}\)] / dt) \)

Solve:

= \(- 1/2 \times ( (0.00253 - 0.00337) / (3200 s)) \)

= \(1.37\times 10^{-7} M/s \)

3. Determine the average rate of formation of \(C_{8}H_{12}\) at \(1600 s\) and the instantaneous rate of formation at \(3200 s\) from the rates found in parts (a) and (b).

rate of formation = products

at \(1600 s [\(C_{8}H_{12}\)] / Δt\)

= \(5.04\times 10^{-3} M) / 1600 s\)

= \(3.15\times 10^{-6} M/s\)

instantaneous rate of formation = products

at \(3200 s [\(C_{8}H_{12}\)] = 3.37\times 10^{-3} M\)

= \(1.053\times 10^{-6} M\)

Q12.5.2

When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?

S12.5.2

Solution: The rate of a reaction can be determined through ¹⁾ the concentration of the substances. ²⁾ Temperature also has an effect on the rate of reaction.

⇒ The reaction rate and concentration are proportional to one another, there fore if you increase the concentration we will observe an increase in the rate of reaction and vice versa.

⇒ An increase in temperature may result in an increased number of collisions as the velocity of the particles is now increased. When the velocity of the particles increases, there is an increase in the fraction of particles with a kinetic energy that exceeds the activation energy needed for the reaction to occur. This trend is shown in the Arrhenius equation, \(ln\frac{k_2}{k_1}= \frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})\), which demonstrates that the rate constants of chemical reactions vary with temperature. (Phase II edit)

Q21.3.6

Technetium-99 is prepared from ⁹⁸Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as ⁹⁹Tc^*. This excited nucleus relaxes to the ground state, represented as ⁹⁹Tc, by emitting a γ ray. The ground state of ⁹⁹Tc then emits a β particle.

Write the equations for each of these nuclear reactions.

S21.3.6

Solution:

Step 1: Technetium-99* = Molybdenum-98

⇒ ⁹⁹Tc^* = ⁹⁸Mo

Step 2: Molybdenum-98 + neutron = Molybdenum-99 + β

⇒ ⁹⁹Tc^* = ⁹⁸Mo + \(\ce{^{1}_{0}n}\) = ⁹⁹Mo + \(\ce{^{0}_{1}β}\)

Step 3: Molybdenum-99 + β + γ = Technetium-99

⇒ ⁹⁹Mo + \(\ce{^{0}_{1}β}\) + \(\ce{^{1}_{1}γ}\) = ⁹⁹Tc

​​​​Phase II Edit:

In a nuclear reaction equation, the mass number and electrical charge of the reactants and products must be balanced to conform to the conservation laws.

Step 1: Molybdenum-98 combines with a neutron to give molybdenum-99.

A mass number refers to the number of protons and neutrons in a nucleus, so adding a neutron changes the mass number of molybdenum from 98 to 99. But since the number of protons remains the same, the element doesn't change.

\[{^{98}_{42}Mo} + {^{1}_{0}n}→{^{99}_{42}Mo}\]

Step 2: Molybdenum-99, an unstable isotope, emits a β particle to yield an excited form of technetium-99, represented as ⁹⁹Tc^*.

Since beta decay involves the conversion of a neutron to a proton and the emission of a high-energy electron, the mass number of the parent and daughter nuclei are the same, and the element changes from molybdenum to technetium as the atomic number increases by one.

\[{^{99}_{42}Mo}→{^{0}_{-1}\beta^-}+{^{99}_{43}Tc^*}\]

Step 3:

The excited nucleus, ⁹⁹Tc^*, relaxes to the ground state, represented as ⁹⁹Tc, by emitting a γ ray.

Gamma emissions often occur after an alpha or beta decay to return a nucleus in an excited state to its ground state. Since gamma rays are energy, their emission doesn't result in a change in the atomic or mass number between the parent and daughter nuclei.

\[{^{99}_{43}Tc^*}→{^{99}_{43}Tc}+ {^{0}_{0}\gamma}\]

Step 4: The ground state of ⁹⁹Tc emits a β particle.

Since beta decay involves the conversion of a neutron to a proton and the emission of a high-energy electron, the mass number of the parent and daughter nuclei are the same, and the element changes from technetium to ruthenium as the atomic number increases by one.

\[{^{99}_{43}Tc}→{^{99}_{44}Ru}+ {^{0}_{-1}\beta^-}\]

Q20.2.2

If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why?

S20.2.2

Solution:

I would not expect a redox reaction to occur because with both being poor reductants and oxidants there will be no exchange of electrons. A redox reaction involves the acceptance of electrons from the oxidizing agent and a loss of electrons from the reduction agent. With poor forms of both neither will occur in most common cases. If one is a strong oxidant and the other is a weak reductant I would expect an ionic bonding which involves a single exchange of electrons. This means the strong oxidant will accept the reductant's electrons and then become an anion while the reduction agent will become a cation after its loss of electrons.

Phase II Edit:

"I would not expect a redox reaction to occur because with both being poor reductants and oxidants there will be no exchange of electrons. A redox reaction involves the acceptance of electrons from the oxidizing agent and a loss of electrons from the reduction agent. With poor forms of both neither will occur in most common cases" (Phase I). If one element is a strong oxidant and the other is a weak reductant, I would expect a redox reaction to occur. The strong oxidant would be strong enough to cause the oxidation of the other species to occur, and any oxidation must always be accompanied by a reduction.

Q20.4.21

Each reaction takes place in acidic solution. Balance each reaction and then determine whether it occurs spontaneously as written under standard conditions.

S20.4.21

1. \( Se (s) + Br_{2} (l) \rightarrow H_{2}SeO_{3} (aq) + Br^{-} (aq) \)

Solution:

Step 1: Seperate Half Reactions

\( Se(s) \rightarrow H_{2} SeO_{3}(aq) \)

\( Br_{2}(l)\rightarrow Br^{-}(aq) \)

Step 2: Balance all elements

\( Se(s) \rightarrow H_{2} SeO_{3}(aq) \)

\( Br_{2}(l) \rightarrow 2Br^{-}(aq) \)

Step 3: Balance Oxygen and Hydrogen

\( Se(s) + 3H_{2}O \rightarrow H_{2}SeO_{3}(aq) + 4H^{+} \)

\( Br_{2}(l)\rightarrow 2Br^{-}(aq) \)

Step 4: Balance Charges

\( Se(s) + 3H_{2}O \rightarrow H_{2}SeO_{3}(aq)+ 4H^{+} + 4e^{-} \)

\( 2e^{-} + Br_{2}(l) \rightarrow 2Br^{-}(aq) \)

Step 5: Add reactions together

\( Se(s) + 3H_{2}O \rightarrow H_{2}SeO_{3}(aq)+ 4H^{+} + 4e^{-} \)

\( 2e^{-} + Br_{2}(l) \rightarrow 2Br^{-} (aq) \) X 2

= \( Se(s) + 3H_{2}O + 2Br_{2}(l) \rightarrow H_{2}SeO_{3}(aq) + 4H^{+} + 4Br^{-}(aq) \)

2. \(NO_{3}^{-}(aq) + S(s)\rightarrow HNO_{2}(aq) + H_{2}SO_{3}(aq) \)

Solution:

Step 1: Seperate Half Reactions

\(NO_{3}^{-}(aq) \rightarrow HNO_{2}(aq) \)

\( S(s) \rightarrow H_{2}SO_{3}(aq) \)

Step 2: Balance all elements

\(NO_{3}^{-}(aq) \rightarrow HNO_{2}(aq) \)

\( S(s) \rightarrow H_{2}SO_{3}(aq) \)

Step 3: Balance Oxygen and Hydrogen

\( 3H^{+} + NO_{3}^{-}(aq)\rightarrowHNO_{2}(aq) + H_{2}O \)

\(3H_{2}O + S(s) \rightarrow H_{2}SO_{3}(aq) + 4H^{+}\)

Step 4: Balance Charges

\(2e^{-} + 3H^{+} + NO_{3}^{-}(aq) \rightarrow HNO_{2} (aq) + H_{2}O \)

\(4e^{-} + 3H_{2}O + S(s) \rightarrow H_{2}SO_{3}(aq) + 4H^{+}\)

Step 5: Add reactions together

\(2e^{-} + 3H^{+} + NO_{3}^{-}(aq) \rightarrow HNO_{2} (aq) + H_{2}O \) × 2

\(4e^{-} + 3H_{2}O + S(s) \rightarrow H_{2}SO_{3}(aq) + 4H^{+}\) × 1

= \(6H^{+} + 2NO_{3}^{-}(aq) + 3H_{2}O + S(s) \rightarrow 2HNO_{2} {aq) + 2H_{2}O + H_{2}SO_{3}(aq) + 4H^{+}\)

3. \(Fe³⁺(aq) + Cr³⁺(aq) → Fe²⁺(aq) + Cr₂O₇²⁻(aq) \)

Solution:

Step 1: Seperate Half Reactions

\(Fe^{3+}(aq)\rightarrow Fe^{+2}(aq) \)

\(Cr^{3+}(aq)\rightarrow Cr_{2}O_{7}²⁻(aq) \)

Step 2: Balance all elements

\(Fe^{3+}(aq)\rightarrow Fe^{+2}(aq) \)

\(2Cr^{3+}(aq)\rightarrow Cr_{2}O_{7}²⁻(aq) \)

Step 3: Balance Oxygen and Hydrogen

\(Fe^{3+}(aq)\rightarrow Fe^{+2}(aq) \)

\(7H_{2}O + 2Cr^{3+}(aq)\rightarrow Cr_{2}O_{7}²⁻(aq) +14H^{+}\)

Step 4: Balance Charges

\(e^{-} + Fe^{3+}(aq)\rightarrow Fe^{+2}(aq) \)

\(7H_{2}O + 2Cr^{3+}(aq)\rightarrow Cr_{2}O_{7}²⁻(aq) +14H^{+} +6e^{-} \)

Step 5: Add reactions together

\(e^{-} + Fe^{3+}(aq)\rightarrow Fe^{+2}(aq) \) X 6

\(7H_{2}O + 2Cr^{3+}(aq)\rightarrow Cr_{2}O_{7}²⁻(aq) +14H^{+} +6e^{-} \) X 1

Step 5: Add reactions together

= \(6Fe^{3+}(aq) + 7H_{2}O + 2Cr^{3+}(aq)\rightarrow 6Fe^{2+}(aq) + Cr_{2}O_{7}²⁻(aq) +14H^{+} \)

Phase II Edit:

To determine whether the reaction occurs spontaneously, calculate the E°cell by using the equation \(E°_{cell}= E°_{cathode}- E°_{anode}\).

Q20.9.6

Electrolysis is the most direct way of recovering a metal from its ores. However, the Na⁺(aq)/Na(s), Mg²⁺(aq)/Mg(s), and Al³⁺(aq)/Al(s) couples all have standard electrode potentials (E°) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead?

S20.9.6

Solution:

The standard reduction potential is the likelihood that a species will be reduced, that is, it will gain electrons. The more positive the standard reduction potential then the higher chance it has of being reduced. At the top of this activity series list that lists these standard potentials starts positive where it is at a position of strongest reduction; where as towards the bottom, where it is less likely, it is a negative value. There fore if the value of the standard electrode potential of the metal ores is more negative than that of water it will be oxidized by such rather than be reduced and gain electrons, it will be stripped of them and thus become cations.

Q14.5.2

For any given reaction, what is the relationship between the activation energy and each of the following?

S14.5.2

1. electrostatic repulsions

Solution:

Electrostatic repulsion is the result of interaction between two particles of the same charge who repel each other away. The activation energy is the minimum energy for a reaction to proceed, comparable to a threshold. Electrostatic repulsion may cause a reaction to halt as it prevents a charge from acrewing and particles from colliding as they repel.

2. bond formation in the activated complex

Solution:

Bond formation bond is a lasting attraction between atoms that enables the formation of chemical compounds. This formation occurs after the activation energy has been reached as it allows the reaction to proceed.

3. the nature of the activated complex

Solution: The activated complex is the structure that forms at the maximum energy point along the reaction mechanism/pathway. The complex will only form it the activation energy has been reached. The nature can vary depending on how much past the activation energy the reaction has surpassed.