Extra Credit 31

Q12.1.4

A study of the rate of dimerization of C₄H₆ gave the data shown in the table:

\[\ce{2C4H6⟶C8H12}\]

1. Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.
2. Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C₄H₆]. What are the units of this rate?
3. Determine the average rate of formation of C₈H₁₂ at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).

Solution 12.1.4

(1) \[rate=\dfrac{\Delta Concentration}{\Delta t} \]

STEP 1: The average rate looks at the difference in concentration final and concentration initial divided by the difference of the final time and the initial time. The equation is shown above.

\[rate = -\frac{1}{2}\frac{\Delta C4H6}{\Delta T} = 1\frac{\Delta [C8H12]}{\Delta T }\]

\[rate = -\frac{1}{2}\frac{\Delta [C4H6]}{\Delta T} = -\frac{1}{2}\frac{[-0.00496M]}{[1600sec]}\] \[= 1.55 \times 10^{-6}\frac{M}{sec}\]

STEP 2: Plug in values to their respective places in the equation.

\[rate = -\frac{1}{2}\frac{\Delta C4H6}{\Delta T} = 1\frac{\Delta [C8H12]}{\Delta T }\]

\[rate = -\frac{1}{2}\frac{\Delta [C4H6]}{\Delta T} = -\frac{1}{2}\frac{[5.04 × 10^{-5} - 3.37 × 10^{-3}]}{[3200-1600sec]}\] \[= 1.04 \times 10^{-6}\frac{M}{sec}\]

(2)

STEP 1: You need to understand exactly what the question is asking. It is asking for the instantaneous rate of a certain concentration at a particular moment. Ergo, we need to find the slope of the tangent line at the point.

STEP 2: You must plot a graph for this. I suggest you use Excel for this.

\[slope\ (m) = \frac{Y2-Y1}{X2-X1}\]

STEP 3: Familiarize with the slope formula; you probably learned this in middle school. This is what we use to find the slope.

\[= \frac{(4.5\times 10^{-3} - 2.5\times 10^{-3}M)}{(1600 - 4800\sec)}\] \[= (-6.25\times 10^{-6}) \times -\frac{1}{2}\] \[= 3.125\times 10^{-7}\frac{M}{Sec}\]

STEP 4: Plug in values to their respective places in the equation.

(3) \[x = average.rate.of.formation.of.C_{8}H_{12}\]

\[= 1.55 \times 10^{-6}\frac{M}{sec}= 1x\]

STEP 1: The rate is what we found in part 2. Now with the rate equation of the product, find X, average rate of formation

\[x = 1.55\times 10^{-6} \frac{M}{Sec}\]

STEP 2: Compute and you get the answer.

Q12.5.2

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

Solution 12.5.2

The most significant factors that determine the rate of the reaction are: (1) Order of reaction (2) concentration, and (3) the K constant

rate = k[A]^M[B]^N

Step 1: Here is a generic rate law for an elementary reaction. As you can see, if we change the exponents (the orders of the RXN), the rate will change. If we change the value of K and the concentration of any of the reactants, the rate will change. However, if the one of the reactant's order is zero, then changing the concentration will not affect the rate. This is because any number to the zeroth power will equal one.

K = temperature, ionic strength, surface area of an adsorbent, or light irradiation

Step 2: What determines K? Well, K can calculated by understanding the temperature, ionic strength, surface area of an adsorbent, or light irradiation of the reaction.

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.

Solution 21.3.6

Step 1: The question is asking you to create a roadmap of Technetium-99 in multiple steps.

Part 1: \[_{42}^{98}\textrm{Mo} +_{0}^{1}\textrm{n} \rightarrow _{42}^{99}\textrm{Mo}\]

Step 2: Here, a neutron increases the mass by one since mass number encompasses the number of protons and the number of neutrons.

Part 2: \[_{42}^{99}\textrm{Mo} \rightarrow _{-1}^{0}\textrm{B} + _{43}^{99}\textrm{Tc}\]

Step 3: Here, the isotope Mo-99 emits a beta particle. Beta particles increase the proton number as it emits an electron in the reaction. Remember, the # of protons and the mass number should be equal on both sides.

Part 3: \[_{43}^{99}\textrm{Tc} \rightarrow _{0}^{0}\textrm{Gamma} + _{43}^{99}\textrm{Tc}\]

Step 4: Here, a gamma ray is emitted. Gamma rays do not contain any particles (electrons, neutrons, protons) so it neither changes the mass nor the proton number.

Part 4: \[_{43}^{99}\textrm{Tc} \rightarrow _{-1}^{0}\textrm{B} + _{44}^{99}\textrm{Ru}\]

Step 5: Here, the isotope TC-99 emits a beta particle. Beta particles increase the proton number as it emits an electron in the reaction. Remember, the # of protons and the mass number should be equal on both sides.

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?

Solution 20.2.2

(1) The redox reaction would occur because a composition of a poor oxidant and poor reductant is spontaneous (i.e. ΔG is negative). For example the standard reduction potential of a poor oxidant such as such as \(Li^{+}\) is -3.04V. The standard reduction potential of a poor reductant such as \(I_{2}/I^{-}\) is +0.54. Therefore, the overall standard reduction potential of this reaction would be +3.58. Since the \(E^{o}_{cell}\) value is positive, the reaction is spontaneous and will occur.

(2) A redox reaction would not occur because a composition of a strong oxidant and weak reductant is non-spontaneous (i.e. ΔG is positive). Look at a redox table to find the \(E^{o}_{cell}\) of \(F_{2}/F^{-}\), a strong oxidant, and \(Cu/Cu^{+}\), a weak reductant. If we subtract the cathode and anode, the \(E^{o}_{cell}\) value is negative so it a non-spontaneous reaction and will not occur.

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.

1. Se(s) + Br₂(l) → H₂SeO₃(aq) + Br⁻(aq)
2. NO₃⁻(aq) + S(s) → HNO₂(aq) + H₂SO₃(aq)
3. Fe³⁺(aq) + Cr³⁺(aq) → Fe²⁺(aq) + Cr₂O₇²⁻(aq)

Solution 20.4.21

(A)

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

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

Step 1: Identify the two half reactions in the overall equation. One must be reducing and the other must be oxidizing.

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

Step 2: Balance both half reactions by adding coefficients, water, H⁺ ions, and electrons where needed.

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

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

Step 3: Make sure both half reactions have the same amount of electrons transferred. Multiply as needed.

\[ 4e^- + 2Br_{2} (l) \rightarrow 4Br^- (aq) \]

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

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

Step 4: Combine the half reactions.

\[Ecell^o = Ecathode^o - Eanode^o\]

\[Ecell^o = 1.066 V - (-0.4) V = 1.466 V\]

+Ecell^o = spontaneous

Step 5: Balancing the overall reaction is superfluous for determining its spontaneity, but we were still asked to do so. To determine the spontaneity of the reaction, simply subtract the cathode cell potential by the anode cell potential (find the values in a redox tower). If the value is negative, it is not spontaneous. If the value is positive, it is spontaneous. Remember, in this reaction Br₂ is being reduced whilst Se is being oxidized.

(B)

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

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

Step 1: Identify the two half reactions in the overall equation. One must be reducing and the other must be oxidizing.

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

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

Step 2: Balance both half reactions by adding coefficients, water, and H⁺ ions where needed.

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

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

Step 3: Make sure both half reactions have the same amount of electrons transferred. Multiply as needed.

\[4e^- + 6H^+(aq)+2NO_{3}^-(aq) \rightarrow 2HNO_{2}(aq) + 2H_{2}O(l)\]

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

\[2NO^-_3(aq)+H_2O(l)+2H^+(aq)+S(s)\rightarrow H_2SO_3(aq)+2HNO_2(aq)\]

Step 4: Combine the half reactions.

\[Ecell^o = Ecathode^o - Eanode^o\]

\[Ecell^o = 0.94 V - 0.15 V = 0.79 V\]

+Ecell^o = spontaneous

Step 5: Balancing the overall reaction is superfluous for determining its spontaneity, but we were still asked to do so. To determine the spontaneity of the reaction, simply subtract the cathode cell potential by the anode cell potential (find the values in a redox tower). If the value is negative, it is not spontaneous. If the value is positive, it is spontaneous. Remember, in this reaction NO₃^- is being reduced whilst S is being oxidized.

(C)

\(Cr^{3+}(aq) \rightarrow Cr_{2}O_{7}^{2-}(aq)\) (oxidation half reaction)

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

Step 1: Identify the two half reactions in the overall equation. One must be reducing and the other must be oxidizing.

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

\(7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)+6e^-\)

Step 2: Balance both half reactions by adding coefficients, water, and H⁺ ions where needed.

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

\(7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)+6e^-\)

Step 3: Make sure both half reactions have the same amount of electrons transferred. Multiply as needed.

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

\(7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)+6e^-\)

\(7H_2O(l)+2Cr^{3+}(aq)+6Fe^{3+}(aq)\rightarrow 6Fe^{2+}(aq)+Cr_2O^{2-}_7(aq)+14H^+(aq)\)

Step 4: Combine the half reactions.

\[Ecell^o = Ecathode^o - Eanode^o\]

\[Ecell^o = 1.33 V - 0.771 V = 0.559 V\]

+Ecell^o = non-spontaneous

Step 5: Balancing the overall reaction is superfluous for determining its spontaneity, but we were still asked to do so. To determine the spontaneity of the reaction, simply subtract the cathode cell potential by the anode cell potential (find the values in a redox tower). If the value is negative, it is not spontaneous. If the value is positive, it is spontaneous. Remember, in this reaction Fe³⁺ is being reduced whilst Cr³⁺ is being oxidized.

Remember that if a reaction is spontaneous, it means that the products are favored and the reaction will shift to the right.

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?

Solution 20.9.6

Electrolysis is the use of an electric current to stimulate a non-spontaneous reaction. Electrolysis can never occur with the stated conditions because the salts, Na⁺ and Cl^-, are not better oxidizing agents than H₂O. You can see this by viewing a table of standard reduction potentials. Sodium, Magnesium, and Aluminum have lower values than water. As a result, H₂O will be reduced instead of the metals. The only way electrolysis can take place is when the elements are less reactive (better oxidizing agents) than water.

Q14.5.2

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

1. electrostatic repulsions
2. bond formation in the activated complex
3. the nature of the activated complex

Before even answering the problem, it is vital to know the definition of every term.

• Activation energy - the minimum quantity of energy that the reacting species must possess in order to undergo a particular reaction.
• Electrostatic repulsions - the result of interaction between the electrical double layers surrounding particles or droplet.
• Activated complex - an intermediate state that is formed during the conversion of reactants into products
• Bond formation in the activated complex - a lasting attraction between atoms that enables the formation of chemical compounds in the activated complex.

With these definitions in mind, we can answer the question successfully.

(A) Activation energy signifies the minimum amount of energy reactants must have for a reaction to occur, but what does mean? Why does this "activation energy" have to exist? Well, it exists because a reaction can only occur if reactants have enough energy to overcome electrostatic repulsion, the result of interaction between the electrical double layers surrounding particles or droplet.

(B) An activated complex is an intermediate state that is established during the transformation of reactants into products. Activation energy is used to engender bond formation in an activated complex.

(C) The energy of the activated complex will be high if the beginning energy to the energy of the activated complex is also high.