Extra Credit 14

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Q19.4B

Use thermal dynamic data from Appendix D to calculate a theoretical voltage of Electrolysis cell
\(\mathrm{Cu(s) + Zn^{2+}(aq) \rightarrow Cu^{2+}(aq) + Zn(s) \quad E^\circ_{cell} =-1.103\:V}\)

A19.4B

Equation: \(\mathrm{E^\circ_{cell} = E^\circ_{Cathode} - E^\circ_{Anode}}\)

\(E^\circ_{Cathode} \): \(\mathrm{Cu(s) \rightarrow Cu^2+(aq) +2e^-}\)

\(E^\circ_{Anode}\): \(\mathrm{Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)}\)

\(\begin{align}
\mathrm{E^\circ_{cell}} & = \mathrm{E^\circ_{cathode} - E^\circ_{anode}} \\
& = \mathrm{-0.762\,V- 0.341\,V} \\
& = \mathrm{-1.103\,V}
\end{align}\)

Q19.41B

When \(\ce{[Pb^2+]}\) is maintained at 1.1 M for reaction:

\(\mathrm{Pb(s) + Cu^{2+}(aq)\rightarrow Cu(s) + Pb^{2+}(aq)}\)

What is the minimum \(\ce{[Cu^2+]}\) for reaction to be spontaneous in forward direction?
Will the displacement of \(\ce{Cu^2+(aq)}\) by \(\ce{Pb(s)}\) go to completion? Explain reasoning

A19.41B

Oxidation: \(\mathrm{Pb(s) \rightarrow Pb^{2+}(aq) + 2e^- \hspace{4 pt} \quad E^\circ_{cell} = -0.125\:V}\)

Reduction: \(\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \quad E^\circ_{cell} = +0.340\:V}\)

a) Equation: \(\mathrm{E_{cell} = E^\circ_{cell} -\dfrac{0.0592}{n} \log \dfrac{[Ox]}{[Red]}}\)

\({=E_{cell} =0\: V}\) (minimum \(\ce{[Cu^2+]}\) for spontaneous )

\( E^\circ_{cell}\) = 0.340V- (-0.125V) = 0.465V

\(\mathrm{0\:V = 0.465\:V - \dfrac{0.0592}{2}\log\dfrac{[1.1 M]}{x}}\)

\(\mathrm{x = 2.15 \times 10^{-16}\,M}\)

Assume: \(\mathrm{[Cu^{2+}]_{initial} = 1.1\:M}\) and \(\mathrm{[Pb^{2+}]_{initial} = 0.0\:M}\),

then: \(\mathrm{[Cu^{2+}]_{final} = 2.15 \times 10^{-16}\: M}\) and \(\mathrm{[Pb^{2+}]_{final} = 1.1\:M}\).

Therefore, the reaction will go to completion.

Q20.57C

What chemical formulas would you expect for the metal carbonyls of

Chromium hexacarbonyl
Iron Pentacarbonyl
Nickel Tetracarbonyl

A20.57C

First we write the chemical symbol for the first element. The second element (carbonyl) has a prefix in front which determines the number of carbonyls, and is written as a subscript outside of the parentheses surrounding carbonyl.

Below are the prefixes for 1-10:

1: Mono-
2: Di-
3: Tri-
4: Tetra-
5: Penta-
6: Hexa-
7: Hepta-
8: Octa-
9: Nona-
10: Deca-

Cr(CO)₆
Fe(CO)₅
Ni(CO)₄
Q24.3C

In the reaction \(\mathrm{A \rightarrow B}\), \(\mathrm{[A]}\) is found to be 0.675M at \(\mathrm{t = 51.1\,s}\) and 0.605M at \(\mathrm{t = 61.5\,s}\). Find the average rate of the reaction during this time interval

A24.3C

\(\mathrm{Average Rate = -\left(\dfrac{\Delta[A]}{\Delta t}\right) = -\dfrac{0.605\,M-0.675\,M}{61.5\,s-51.1\,s} = 0.00673 = 6.73 \times 10^{-3}\, Ms^{-1}}\)

Q24.47A

For the reversible reaction \(\ce{A + B \leftrightarrow A + B}\) the enthalpy change of the forward reaction is +11 kj/mol. The activation energy of the forward reaction is 74 kj/mol.

What is the activation energy for the reverse reaction?

A24.47A

Because the enthalpy change of the forward reaction is positive, it is endothermic reaction.

Therefore, Ea (reverse) = Ea (foward) - enthalpy change

Activation energy for the reverse reaction: 74 kj/mol - 11 kj/mol = 63 kj/mol

Q25.25A

Suppose that a sample containing \(\ce{^{28}_{12}Mg}\) has an activity 1000 times the detectable limit. How long would an experiment have to run with this sample before the radioactivity could no longer be detected?

A25.25A

Using the equation for first-order kinetics, the following equation can be derived:

\[\mathrm{\ln\left(\dfrac{N_t}{N_0}\right) = -\lambda t}\]

where "N" is the amount of radioisotope remaining after time "t" has elapsed. "N_o" is the initial amount of radioisotope at the beginning of the period, and "k" is the rate constant for the radioisotope being studied. The units of measure for N and No can be in grams, atoms, or moles. It does not matter as long as they are like measures. The units of measure for time are dependent upon the unit of measure for the rate constant. The ratio of "N/N_o" gives the percentage activity as compared to the activity at time zero. This equation has a variety of applications.

From what is given to us in the problem, we know that our half-life would be 21 hours (in order to find the half-life for each isotope, we can refer to a page like https://www.isotopes.gov/catalog/pro...oduct_index=29) and our first step would is to solve for lambda (λ) using:

Equation: \[\mathrm{\lambda = \dfrac{0.693}{t_{1/2}}}\]

\[\mathrm{\lambda = \dfrac{0.693}{21\,h} = 0.033\,h^{-1}}\]

Once lambda is solved for, we plug its value of 0.033 h^-1 to the first-order kinetics equation listed above. N₀is given to us in the question and is 1000. After substituting the variables for the proper values and solving for t (time), we get:

Equation: \[\mathrm{\ln\left(\dfrac{N_t}{N_0}\right) = -\lambda t}\]

\[\mathrm{ln\left(\dfrac{1}{1000}\right) = (0.033\,h^{-1})(t)}\]

\[\mathrm{t = -\dfrac{\ln\left(\dfrac{1}{1000}\right)}{(0.033\,h^{-1})} = 209.33\,h}\]

Q18.1

Calculate the oxidation number for nitrogen in the following substances:

\(\ce{NH3}\)
\(\ce{N2}\)
\(\ce{NO2}\)
\(\ce{NO3-}\)

A18.1

You assign oxidation numbers to the elements in a compound by using the Rules for Oxidation Numbers.

Rules of Oxidation Numbers:

The oxidation number of a free element is always 0.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.
The oxidation number of a Group 1 element in a compound is +1.
The oxidation number of a Group 2 element in a compound is +2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

a. -3 - Using Rule #3, we know that each H has a +1 charge and that the overall charge of the compound should be 0. In order for our overall charge to equal 0, our Nitrogen (N) must have a charge of -3 to balance the +3 charge of the three Hydrogens.

b. 0 - N₂ is considered a free element because it is a chemical element that is not combined with or chemically bonded to other elements. Going based off of Rule #1, the oxidation number of a free element is always 0.

c. +4 - Using Rule #4, we know that each O has a -2 charge and that the overall charge of the compound should be 0. In order for our overall charge to equal 0, our Nitrogen (N) must have a charge of +4 to balance the -4 charge of the two oxygens.

d. +5 - Using Rule #9, we know that the sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion being -1 in this case. The charge of each oxygen is -2, so our Nitrogen should have a charge of +5 so the overall charge remains at -1.

Q21.3.7

Write a balanced nuclear reaction for the formation of each isotope.

²⁷Al from two ¹²C nuclei
⁹Be from two ⁴He nuclei

A21.3.7

a. _{6}^{12}\textrm{C}+_{6}^{12}\textrm{C}+_{-1}^{0}\textrm{e}+_{2}^{4}\textrm{He}\rightarrow _{13}^{27}\textrm{Al}+_{1}^{0}\textrm{n}

b._{2}^{4}\textrm{He}+_{2}^{4}\textrm{He}+_{0}^{1}\textrm{n}\rightarrow _{4}^{9}\textrm{Be}

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