Extra Credit 47

A12.3.10

The instantaneous rate of is the rate at which the concentration is changing at a specific point. Therefore that's how we got the equation below.

The equation is a derivative of the equation:

\[\frac{1}{[A]}t=kt+\frac{1}{[A^\circ ]}\]

Since this is a second order reaction, we know the rate law will be Rate=k[A]²

We are given k (the rate constant) which is 4.7x10^-8 L/mol/s and are asked to solve the equation with the concentration 5.55x10^-4 M. We simply plug in the values into Rate=k[A]²

\[Rate=k[A]^2=4.7x10^{-8}\frac{L}{mol\ \times\ s}[5.55x10^{-4} M]^2=1.45x10^{-14} M/s \]

Q12.6.2

In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction A+B⟶C ? Can we predict the effect if the reaction is known to be an elementary reaction?

A12.6.2

In general, we can't predict the effect of doubling the concentration of A on the rate of the overall reaction. It is impossible to determine the rate law and reaction order with the balanced equation. This has to be determined experimentally. However, if it is known to be an elementary reaction, we know that the rate law will be rate=k[A][B]. This is because it is known at in an elementary reaction, the reaction order of the reactants depend of the coefficent of the reactant in the reaction equation. The rate law came from the reaction \[A+B\rightarrow C\], which each reactant has a coefficient of 1.
The doubling of the concentration of A will double the rate of the reaction. This is of A being directly proportional to rate.

Q21.4.14

A 1.00 × 10⁻⁶ g sample of nobelium, No-254, has a half-life of 55 seconds after it is formed. What is the percentage of No-254 remaining at the following times?

1. 5.0 min after it forms
2. 1.0 h after it forms

A21.4.14

First, we have to determine the rate constant. Since this is a first order reaction, we can use the equation t_1/2 = 0.693/k.

Rearrange the first-order half- life equation to solve for k:

\[t_{1/2}=\frac{ln2}{\lambda }\]

\[k=\frac{0.693}{t_{1/2}} \]

\[k=\frac{0.693}{55 s}=0.0126 s^{-1} \]

1. Convert 5.0 min to seconds

\[(5.0min) (60 \frac{s}{min})=300 s \]

Using \[ln \frac {[A_t]}{[A_0]}=-kt \], where [A_t] is the amount at time t and [A₀] is the initial amount:

\[ln \frac {[A_t]}{[A_0]}=-(0.0126s^{-1})(300s) \]

\[\frac {[A_t]}{[A_0]}=e^{-(0.0126s^{-1})(300s)} \]

\[\frac {[A_t]}{[A_0]}=0.0228 \]

Multiply by 100% to make this into a percentage

0.00228X100=2.28%

Therefore, the percentage left at 5 minutes is 2.28%

2. Convert 1.0 h to seconds

\[60 min) 60 \frac{s}{min}=3600 s \]

This equation is the integrated rate law for reaction rate of a first order reaction. Using \[ln \frac {[A_t]}{[A_0]}=-kt \], where \[[A_t]\] is the amount at time t and \[[A_0]\] is the initial amount:

\[ln \frac {[A_t]}{[A_0]}=-(0.0126s^{-1})(3600s) \]

\[\frac {[A_t]}{[A_0]}=e^{-(0.0126s^{-1})(3600s)} \]

\[\frac {[A_t]}{[A_0]}=2.00x10^{-20} \]

The percentage remaining after 1 hour is 2.00x10^-18 %

Q20.3.2

If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?

A20.3.2

If two half reactions are physically separated, it possible for a redox reaction to occur if there is a complete circuit and an external electrical connection between the two circuits. The electrons can flow through this electrical connection. The name of the apparatus in which two half-reactions are carried out simultaneously is called a Galvanic cell., in which two half reactions are carried our simultaneously. There is a anode, which is where the oxidize half reaction occurs, and the cathode, where the reduced half reaction occurs. The electrons would flow from the anode to the cathode. There is also a salt bride so that spectator ions can migrate in between the two electrode. Though they are not a part of the reaction, they are used to maintain electrical neutrality.

This is what a Galvanic cell would generally look like.

Q20.5.13

For the cell represented as Al(s)∣Al³⁺(aq)∥Sn²⁺(aq), Sn⁴⁺(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?

A20.5.13

We can figure out how many electrons are transferred by determining the balanced equation. First, we have to separate the reaction into two half reactions.

\[Al(s) \rightarrow Al(aq)^{3+} \]

\[Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) \]

Next, we balance the charges by adding the appropriate number of electrons.

\[Al(s) \rightarrow Al(aq)^{3+} +3e^- \]

\[Sn^{2+}(aq)+2e^- \rightarrow Sn^{4+}(aq) \]

Multiply the top equation by 2 and the bottom equation by 3 to obtain the same amount of electrons in each half reaction.

\[2x(Al(s) \rightarrow Al(aq)^{3+} +3e^-) \] = \[2Al(s) \rightarrow2 Al(aq)^{3+} +6e^- \]

\[3x(Sn^{2+}(aq)+2e^- \rightarrow Sn^{4+}(aq)) \] =\[3Sn^{2+}(aq)+6e^- \rightarrow 3Sn^{4+}(aq) \]

Add the resulting equations together and cancel out the like terms.

\[3Sn^{2+}(aq)+2Al(s)+6e^- \rightarrow 3Sn^{4+}(aq)+Al(aq)^{3+}+6e^-\]

The electrons cancel out and we get

\[3Sn^{2+}(aq)+2Al(s) \rightarrow 3Sn^{4+}(aq)+Al(aq)^{3+}\]

Therefore, the number of electrons transferred in the redox reaction is 6.

The equation used to calculate standard cell potentials is

\[E°_{cell}=E°_{cathode}-E°_{anode}\]

We have to determine which of the reactions are the cathode and anode.

Looking back at our half reactions, we see that we have:

\[Al(s) \rightarrow Al(aq)^{3+} +3e^- \]

Since Al is losing electrons, it is being oxidized. This means it is the anode.

\[Sn^{2+}(aq)+2e^- \rightarrow Sn^{4+}(aq) \]

Sn is gaining electrons which means it has been reduced- therefore it is the cathode.

We now have to refer to a standard reduction potential table to obtain E°cathode and E°anode.

E°cathode=0.154 E°anode=-1.676

\[E°_{cell}=E°_{cathode}-E°_{anode}=0.154-(-1.676)= 1.83V\]

Since E°cell is positive, this is a spontaneous reaction. If the Standard Cell potential was negative, it would be non-spontaneous.

To calculate ΔG°, we use the equation \[ΔG°=−nFE°cell\]

n=moles of electrons transferred=6

F=Faraday's constant=96,468 J/(V⋅mol e^-)

E°cell=1.83 V

\[ΔG°=−nFE°cell=-(6 mol\, e^-)(96,468 \frac {J}{(V)(mol)})=-1,059,416\, J\]

Q24.6.8

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

1. [Cu(NH₃)₄]²⁺
2. [Ni(CN)₄]²⁻

A24.6.8

1. [Cu(NH₃)₄]²⁺

To determine the structure, we have to look at the coordination number of the complex. By looking at the number of ligands, we see that we have a coordination number of 4, so the structure will either be tetrahedral or square planar. To decide, we need to find the electron configuration of Cu. Copper's configuration is [Ar]4s¹3d¹⁰. The charge on the complex is 2+, and since NH₃ has a neutral charge, the we know that it is Cu²⁺ in the complex. The electron configuration of Cu²⁺ is [Ar]3d⁹. We do this by removing 1 electron from the 4s subshell and 1 from the 3d subshell. Therefore, copper has 9 electrons. This is a square planar complex.

From this spectrum, we can see that NH₃ is a strong ligand, therefore making this low spin.

Next, we have to look at the energy diagram. Since this is low spin, we fill up the bottom row, which is the t2g level, before we move to the top, the eg level. A strong spin would mean that we place one electron in both eg and t2g levels before pairing them up. This gives:

There is one unpaired electron.

2.[Ni(CN)₄]²⁻

The coordination number of this complex is also 4, which means it will have a square planar or tetrahedral structure. CN has a -1 charge so we know Ni has a +2 charge. The electron configuration of Ni²⁺ is [Ar]3d⁸. Since the outermost orbital has a d⁸ configuration, this is square planar. From the spectrum, we can see that CN is a strong field ligand, so this is a low spin complex. This is because strong field ligands have large splitting energy because it takes less energy to pair electrons in lower energy orbitals than in high energy orbital.

The energy diagram of Ni²⁺ is

There are two unpaired electrons.

Q14.4.6

Iodide reduces Fe(III) according to the following reaction:

2Fe³⁺(soln) + 2I⁻(soln) → 2Fe²⁺(soln) + I₂(soln)

Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?