Extra Credit 21
- Page ID
- 83529
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Calculate \(\mathrm{\Delta G^\circ}\) for the following reactions and determine if spontaneous:
- \(\mathrm{NO_3^-(aq) + Al(s) + 4H^+(aq) \rightarrow NO(g) + Al^{3+}(aq) + 2H_2O (l)}\)
- \(\mathrm{F_2(g) + 2Li(s) \rightarrow 2F^- (aq) + 2Li^+ (aq)}\)
S19.7A
a. \(\mathrm{NO_3^-(aq) + Al(s) + 4H^+(aq) \rightarrow NO(g) + Al^{3+}(aq) + 2H_2O (l)}\)
1. Write out the half cell reactions and look at the table to find the standard reduction potential for each half reaction.
\(\mathrm{Al(s) \rightarrow Al^{3+}(aq) + 3e^-}\)
Note: This is the anode because it Al(s) loses electrons and therefore is oxidized; \(\mathrm{(E^\circ_{Al^{3+}/Al}) = -1.676}\)
\(\mathrm{3e^- + \mathrm{NO_3^-}(aq) + 4H^+ \rightarrow NO(g) + 2H_2O (l)}\)
Note: This is the cathode because \(\mathrm{NO_3^-}(aq)\) gains electrons and therefore is reduced; \(\mathrm{ (E^\circ_{NO_3^-/NO^-}) = 0.956}\)
2. Calculate the standard cell potential.
\(\mathrm{E^\circ_{cell}}\) = cathode - anode
\(\mathrm{E^\circ_{cell} = (E^\circ_{NO_3^-/NO^-}) - (E^\circ_{Al^{3+}/Al})}\)
\(\mathrm{E^\circ_{cell} = (0.956) - (-1.676)}\)
\(\mathrm{E^\circ_{cell} = 2.63\,V}\)
3. Calculate \(\mathrm{\Delta G^\circ}\).
\(\mathrm{\Delta G^\circ= -nFE^\circ_{cell}}\)
\(\mathrm{n=}\) number of electrons transferred \(\mathrm{= 3}\)
\(\mathrm{F=}\) Faraday's constant \(\mathrm{= 96,485\,C/mol\,e^-}\)
\(\mathrm{\Delta G^\circ=-[(3)(96,485\,C/mol\,e^-)(2.63\,V)]}\)
\(\mathrm{\Delta G^\circ= -761266\, J\, = -761kJ}\)
Because \(\mathrm{\Delta G^\circ}\) is negative, the reaction is spontaneous.
b. \(\mathrm{F_2(g) + 2Li(s) \rightarrow 2F^- (aq) + 2Li^+ (aq)}\)
1. \(\mathrm{2Li(s) \rightarrow 2Li^+ (aq) + 2e^-}\); \(\mathrm{(E^\circ_{Li^+/Li})} = -3.040V\)
\(\mathrm{F_2(g) + 2e^- \rightarrow 2F^-(aq)}\); \(\mathrm{(E^\circ_{F_2/F^-})} = 2.866V\)
2. \(\mathrm{E^\circ_{cell}}\) = cathode - anode
\(\mathrm{E^\circ_{cell} = 2.866 - (-3.040) = 5.906\,V}\)
3. \(\mathrm{\Delta G^\circ= -[(2\,mol\,e^-)(96,485\,C/mol\,e^-)(5.906\,V)]}\)
\(\mathrm{\Delta G^\circ=-1,140\,kJ}\)
Spontaneous
Q19.59C
Imagine an iron nail that is corroding in a solution. Predict the appearance of the nail under the given conditions using your knowledge of corrosion in voltaic cells. Assume standard temperature conditions.
- The head and tip of the nail are covered in magnesium.
- The nail is galvanized, but there is a break in plating.
- The nail is plated in copper, but there is a break in plating.
S19.59C
1. Magnesium is a sacrificial anode, meaning it is extremely willing to be oxidized. The \(\mathrm{E^\circ_{cell}}\) for magnesium is smaller than the \(\mathrm{E^\circ_{cell}}\) for iron, and it will be the metal that is targeted for oxidation. Thus, the head and tail of the nail would not corrode as long as magnesium is present.
2. If a metal is galvanized, that means it is coated with zinc, which is more active than the iron nail and will be oxidized. Even with a break in plating, the zinc is still more active and the iron won’t corrode.
3.If the metal is plated in copper, once the copper is broken, the iron is the more active metal and the nail will be oxidized and begin corrosion.
plating is a surface covering in which a metal is deposited on a conductive surface- Nika Khosravi Kia
Q21.4A
- How many unpaired electrons would you expect to find in the complex \(\ce{[Fe(CN)6]^3+}\)?
- How many unpaired electrons would you expect to find in the tetrahedral complex \(\ce{[MnBr4]^2-}\)? Would you expect more, fewer, or the same number of unpaired electrons as in the octahedral complex \(\ce{[Mn(NH3)6]^2+}\)?
S21.4A
a) The transition metal, iron, in this complex \(\ce{[Fe(CN)6]^3+}\) has 6 d electrons. However, it has a \(\mathrm{3^+}\) charge, so it loses a total of 9 electrons in order to get this value. The 6 d electrons are lost in order to cancel out the negative charge of the cyano ligand and another 3 electrons are lost (two 4s electrons and one 3p electron) to get the overall \(\mathrm{3^+}\) charge. It has one unpaired electron.
b) In the tetrahedral complex \(\ce{[MnBr4]^2-}\), Mn has a \(\mathrm{2^+}\) charge which means that it'll lose its two 4s electrons. Because the ligand Br is lower in the the spectrochemical series, it will be high spin and have 5 unpaired electrons.
In the octahedral complex \(\ce{[Mn(NH3)6]^2+}\), Mn has a \(\ce{2^+}\) charge which means that it'll lose its two 4s electrons. Because the ligand \(\ce{NH3}\) is high in the spectrochemical series, it wil be low spin have 1 unpaired electron.
Q24.11C
Use the table below to answer the following questions:
- Find the order of the reaction AA vs. BB
- Find the overall order
- Find the rate constant
| EXPT | [A][A],M | [B][B],M | Initial Rates, Ms-1 |
|---|---|---|---|
| 1 | 0.175 | 0.138 | 3.75 × 10-4 |
| 2 | 0.175 | 0.185 | 1.25 × 10-3 |
| 3 | 0.365 | 0.138 | 3.75 × 10-4 |
| 4 | 0.365 | 0.185 | 1.25 × 10-3 |
S24.11C
The reactions can be written as followed:
\(\mathrm{Reaction\: 1 =3.75 \times 10^{-4} = k \times [0.175]^m [0.138]^n}\)
\(\mathrm{Reaction\: 2 =1.25 \times 10^{-3} = k \times [0.175]^m [0.185]^n}\)
\(\mathrm{Reaction\: 3 =3.75 \times 10^{-4}= k \times [0.365]^m [0.138]^n}\)
\(\mathrm{Reaction\: 4 =1.25 \times 10^{-3}= k \times [0.365]^m [0.185]^n}\)
a) Order of reaction with respect to A
1. The first step in finding the order of reaction with respect to A is to look for reactions where the concentration of B remains constant while the concentration of A changes. This would mean that we use reactions 1 and 3 because the concentration of B remains constant at 0.138 but the concentration of A changes.
2. Next, divide the two reactions.
\(\mathrm{\dfrac{Reaction\: 3}{Reaction\: 1} = \dfrac{3.75 \times 10^{-4}}{ 3.75 \times 10^{-4}} = \dfrac{[0.365]^m[0.138]^n}{[0.175]^m[0.138]^n}}\)
After simplifying the equation, you should be left with
\(\mathrm{1= 2^m}\)
\(\mathrm{m=0}\)
The order of reaction with respect to A is zero.
Order of reaction with respect to B
1. To find the order of reaction with respect to B, we have to look for the reaction where the concentration of A remains constant while the concentration of B is changing. Looking at the reactions, we can see that reaction 1 and 2 satisfies this criteria.
2. Divide.
\(\mathrm{\dfrac{Reaction\: 2}{Reaction\: 1} = \dfrac{1.25 \times 10^{-3}}{ 3.75 \times 10^{-4}} = \dfrac{[0.175]^m [0.185]^n}{[0.175]^m[0.138]^n}}\)
Simplify to get:
\(\mathrm{4= 2^n}\)
\(\mathrm{n=2}\)
The order of reaction with respect to B is 2.
b) To find the overall order of the reaction, simply add the reaction order of A and B.
Overall order \(\mathrm{= 0+2 = 2}\)
c) Since we already found the order of reactions for A and B, we can write an expression for the rate.
\(\mathrm{rate = k \times[A]^0[B]^2}\)
Simplify.
\(\mathrm{rate = k \times 1\times [B]^2}\)
To find k, plug in the values for B and the rate from any of the equations. Make sure the values you use for B and the rate are from the same equation.
\(\mathrm{3.75 \times 10^{-4} = k \times[0.138]^2}\)
\(\mathrm{k =0.0197\, M^{-1}s^{-1}}\)
Q24.51C
The rate constant for the reaction \(\ce{H2(g) + F2(g) \rightarrow 2HF}\) has been determined at the following temperatures 650 degrees K, \(\mathrm{k = 4.8 \times 10 ^{-4}\, M^{-1}s^{-1}}\); 700 degrees K, \(\mathrm{k = 3.0 \times 10^{-2}\,M^{-1}s^{-1}}\). Calculate the activation energy for the reaction.
S24.51C
Use the Arrhenius equation to solve for the activation energy.
(R= 8.314 J K-1 mol-1)- Nika Khosravi Kia
\(\mathrm{\ln\dfrac{k_1}{k_2} = \dfrac{E_a}{R} \left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)}\)
\(\mathrm{\ln\dfrac{4.80 \times 10^{-4}\, M^{-1}s^{-1}}{3.00 \times 10^{-2}\, M^{-1}s^{-1}} = \dfrac{E_a}{R} \left(\dfrac{1}{700 K} -\dfrac{1}{650 K}\right)}\)
\(\mathrm{(E_a)(-1.099 \times 10^{-4}) = -4.135R}\) \(\mathrm{R=8.314\: j/(mol\: K)}\)
\(\mathrm{E_a =\dfrac{(-4.135)(8.314\: J/(mol\: K))}{-1.099 \times 10^{-4}} = 313\: kJ/mol}\)
Q25.27C
A piece of wood is claimed to found in Qin Mausoleum and is offered for sale to an art museum. Radiocarbon dating of the object reveals a disintegration rate of 5.0 dis min-1 g-1. Do you think the object is real? Explain your answer. Given that the \(\mathrm{\lambda = \dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}}\) and the initial rate of decay is about 15 dis/min.
S25.27C
Radioactive decay is a first order reaction which means that we are able to use the law of first order kinetics in order to solve a problem. However, instead of solving for the rate constant like in kinetics, our goal is to solve for the radioactive decay constant symbolized by \(\mathrm{\lambda}\).
Use the formula from first order kinetics.
\(\mathrm{t_{1/2} = \dfrac{0.693}{\lambda}}\)
Rearrange the formula so that you can easily solve for the radioactive decay constant.
\(\mathrm{\lambda} = \dfrac{0.693}{t_{1/2}}\)
Plug in the values from the question and solve.
\(\mathrm{\lambda}=\dfrac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}\)
Now that we have the radioactive decay constant, we can use another equation derived from first order kinetics to solve for the age of the object.
\(\mathrm{\lambda}=-\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t}\) where \(\mathrm{[A]_t}\) = current disintegration rate and \(\mathrm{[A]_o}\) = initial disintegration rate.
Plug in the values and solve for t.
\(\mathrm{\ln\dfrac{5\, dis/min}{15\, dis/min} = -1.0986\,s=-\lambda t}\)
\(\mathrm{t = \dfrac{1.0976}{1.21\times10^{-4}\,y^{-1}}=9071\,years}\)
The object is around 9000 years old, and it is probably not from the Qin Mausoleum, which is around 3000B.C.
Q18.8
The following table contains data obtained by measuring the voltage between two metals in an experiment like the one you did with a lemon:
|
Voltaic Cell |
Anode (-) |
Cathode (+) |
Cell Voltage (v) |
|---|---|---|---|
|
\(\ce{Pb/Ni}\) |
\(\ce{Ni}\) |
\(\ce{Pb}\) |
0.10 volts |
|
\(\ce{Pb/Au}\) |
\(\ce{Pb}\) |
\(\ce{Au}\) |
0.80 volts |
|
\(\ce{Pb/Fe}\) |
\(\ce{Fe}\) |
\(\ce{Pb}\) |
0.25 volts |
|
\(\ce{Ni/Au}\) |
\(\ce{Ni}\) |
\(\ce{Au}\) |
0.90 volts |
|
\(\ce{Ni/Fe}\) |
\(\ce{Fe}\) |
\(\ce{Ni}\) |
0.15 volts |
|
\(\ce{Fe/Au}\) |
\(\ce{Fe}\) |
\(\ce{Au}\) |
1.05 volts |
- From the data in the table above, which metal is the strongest reducing agent?
- From the data in the table above, which metal is the weakest reducing agent?
- Using the reduction of lead (Pb) as a reference, construct a half-cell voltage table from the experimental data above.
S18.8
1. The strongest reducing agent is Iron (Fe) because it is always oxidized.
2. The weakest reducing agent is Gold (Au) because it is never oxidized.
3.
| Half - Reaction | \(\mathrm{E^\circ_{reduced}}\) (volts) |
| \(\mathrm{ Au^{3^+}+3e^- \rightarrow Au}\) | 0.80 v |
| \(\mathrm{ Pb^{2^+}+2e^- \rightarrow Pb}\) | 0.00 v |
| \(\mathrm{ Ni^{2^+}+2e^- \rightarrow Ni}\) | -0.10 v |
| \(\mathrm{ Fe^{2^+}+2e^- \rightarrow Fe}\) | -0.25 v |
Q21.4.5
Azomethane (CH3N2CH3) decomposes at 600 K to C2H6 and N2. The decomposition is first order in azomethane. Calculate t½ from the data in the following table:
|
Time (s) |
\(P_{\large{\mathrm{CH_3N_2CH_3}}}\) (atm) |
| 0 | 8.2 × 10−2 |
| 2000 | 3.99 × 10−2 |
| 4000 | 1.94 × 10−2 |
How long will it take for the decomposition to be 99.9% complete?
S21.4.5
1. Use this equation in order to calculate the value of k.
\(\mathrm{[A]_t}{[A]_o} =e^{-k\,t}\)
\(\mathrm{[A]_t}\) = final pressure
\(\mathrm{[A]_o}\) = initial pressure
\(\mathrm{t}\) = time elapsed
After rearranging the equation and plugging in the values from the table, we are able to solve for k.
\(\mathrm{k=}\,-\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t}\)
\(\mathrm{k=} -\dfrac{\ln \dfrac{1.94\times10^{-2}\, atm}{8.2\times10^{-2}\, atm}}{4000\, \text{s}} \)
\(\mathrm{k}=3.604\times10^{-4} \, \text{s}^{-1}\)
Because this is a first order reaction, we can use the following formula to calculate the half-life.
\(\mathrm{t_{1/2}}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{03.604\times10^{-4} \, \text{s}^{-1}} \approx 1923\, \text{s}\)
2. We can use the value of k we calculated above to solve for the time it takes for 99.9% decomposition. After 99.9% of (CH3N2CH3) decomposes, there is only 0.1% of it left.
\(\mathrm{ln \dfrac{[A]_t}{[A]_o}} = -kt\)
\(\mathrm{ln( \dfrac{0.1}{100}}) = -(3.604\times10^{-4} \, \text{s}^{-1})\,t\)
\(\mathrm{t = 19135 \,s}\)
Everything else seems to be correctly responed. -Nika Khosravi Kia