Extra Credit 9

Question 19.1.7

Which of the following elements is most likely to form an oxide with the formula \(MO_3\): Zr, Nb, or Mo?

Source: http://www.ptable.com/Images/periodic%20table.png

Problem Solving:

An oxide of the form \(MO_3\) has will have a metal with oxidation number of

M + 3(-2) = 0

M - 6 = 0

M = 6

Looking at the periodic table, we must decide upon the element that is most likely to have an oxidation number of 6. Zr is from group IVB and period 5. Nb is from group VB and period 5. Mo is from group VIB and period 5. All of the suggested elements are in period 5 so we shall look at the groups. Zr has 4 valence electrons, Nb has 5 valence electrons, and Mo has 6 valence electrons. The only metal that can easily form a 6+ ion is Mo because the other two transition metals do not have enough valence electrons to easily do so.

Question 19.2.9

Predict whether the carbonate ligand \(CO_3^{2-}\) will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.

Problem Solving:

Most ligands are monodentate. The ligands that are polydentate are going to be big. Bidentates like the oxalate ligand or the ethlynediamine ligand are over 6 atoms. Judging purely by size, the carbonate ligand is likely to be monodentate.

Question 12.3.22

The following data has been determined for the reaction:

$$I^{-} + OCl^{-} \to IO^{-} + Cl^{-}$$

Determine the rate equation and the rate constant for this reaction.

Problem Solving:

Step 1: Determine the rate equation

In order to determine the rate equation, we must first find the order of each reactant.

For \(I^{-}\):

When \(I^{-}\) is doubled and \(OCl^{-}\) is kept constant as seen when comparing trial 1 with trial 2, we may observe that the rate is approximately doubled. In equation form,

$$r = k[I^{-}]^x[OCl^{-}]^y \; and \; 2r = k(2[I^{-}])^x[OCl^{-}]^y$$

Substituting r into the second equation yields us the following:

$$2(k[I^{-}])^x[OCl^{-}]^y) = k(2[I^{-}])^x[OCl^{-}]^y$$

$$2[I^{-}]^x = (2[I^{-}])^x$$

$$2[I^{-}]^x = 2^x[I^{-}]^x$$

$$2 = 2^x$$

$$x = 1$$

The reaction is first order in respect to \(I^{-}\).

For \(OCl^{-}\):

Knowing that the reaction is first order in respect to \(I^{-}\), the equation may be simplified as follows:

$$r = k[I^{-}][OCl^{-}]^y$$

Comparing either reactions 1 and 3 or reactions 2 and 3 will allow us to find the order of the reaction with respect to \(OCl^{-}\). Because reactions 1 and 3 have a simpler multiplier, we will use that one.

$$r = k[I^{-}](5[OCl^{-}])^y \; and \; .6r = k(3[I^{-}])[OCl^{-}]^y$$

Again, substituting in r, we may solve for the order of \(OCl^{-}\) as follows:

$$.6(k[I^{-}](5[OCl^{-}])^y) = k(3[I^{-}])[OCl^{-}]^y$$

$$.6(5[OCl^{-}])^y = 3[OCl^{-}]^y$$

$$.2(5^y)[OCl^{-}]^y = [OCl^{-}]^y$$

$$.2(5^y) = 5^{-1}(5^y) = 5^{y-1} = 1$$

$$y = 1$$

The reaction has an order of 1 in respect to \(OCl^{-}\).

Putting this together, we my find the general rate equation for this reaction to be:

$$r = k[I^{-}][OCl^{-}]$$

With the equation solved, it is now possible to plug in the values for the concentrations of the reactants and the rate in order to solve for the rate constant. We may use any of the three reactions given to do this. Using reaction 1, we get

$$3.05\times 10^{-4} = k(0.10)(0.050)$$

$$k = 6.1\times10^{-2}$$

Finding the units for k, we note that the rate equation has an overall order of 2, which means the units for k will be in \(M^{-1}s^{-1}\).

Thus, the rate equation will be:

$$r = (6.1\times10^{-2}\;M^{-1}s^{-1})[I^{-}][OCl^{-}]$$

Question 12.7.2

Compare the functions of homogenous and heterogenous catalysts.

Definitions:

A homogenous catalyst is in the same state as the reactant(s). This makes it very easy to evenly distribute the catalyst in the solution. This property allows homogenous catalysts to maximize collision rate and thusly maximize reaction catalyzation.

A heterogenous catalyst is in a different state from the reactant(s). This allows for easier retrieval of the catalyst in comparison to the homogenous catalyst. This ability allows for more control over the reaction and more flexibility in industrial use.

Question 21.4.25

A\({\;}_5^8B\) atom (mass = 8.0246 amu) decays into a\({\;}_4^8Be\) atom (mass = 8.0053 amu) by loss of a \(\beta^{+}\) particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?

Quick Review:

In order to do this type of problem, we will need to use an equation relating mass to energy. The equation that we will use for this will be Einstein's famous equation $$E = mc^2$$ As is, it will not be of much use to us, but we can modify the equation slightly to calculate for change, which will take the form of $$\Delta{E} = \Delta{m}c^2$$ Where \(\Delta{E}\) is the change in energy, \(\Delta{m}\) is the change in mass, and c is the speed of light \((3\times10^8 \;m/s)\).

Problem Solving:

The first thing that we will do is write out the equation to better visualize the problem.$$\ce{_5^8}B \to {\!}_4^8Be + {\!}_{1}^{0}{\beta}^+$$ The problem gives us the mass, so using the numbers given, we will calculate \(\Delta{m}\).

$$\Delta{m} = m_{products} - m_{reactants}$$

$$\Delta{m} = (8.0053 + 0.00055) - (8.0246)$$

$$\Delta{m} = 8.00585 - 8.0246$$

$$\Delta{m} = -0.01875\;amu$$

Now that we have obtained \(\Delta{m}\), we may now calculate for the energy using the equation $$\Delta{E} = \Delta{m}c^2$$ However, this equation uses kg, not amu. If a mole of this reaction were to occur, \(\Delta{m} = -0.01875 g\). We may use this fact to convert amu to kg. Using this fact, we obtain the following:

$$\Delta{E} = (-1.875\times10^{-5}\;kg/mol)(3\times10^8\;m/s)^2$$

$$\Delta{E} = (-1.875\times10^{-5}\;kg/mol)(9\times10^{16}\;m^2/s^2)$$

$$\Delta{E} = -1.6875\times10^{12} J/mol$$

Now that we have our answer in J/mol, we may now use Faraday's constant (\(96485 J\:mol^{-1}\:V^{-1}\)) to convert to electron volts.

$$\Delta{E} = \frac{-1.6875\times10^{12} J\:mol^{-1}}{96485 J\:mol^{-1}\:V^{-1}}$$

$$\Delta{E} = -1.7490\times10^7\;V = -17.490\; MeV$$

Problem 20.3.12

Write the spontaneous half reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.

a. \(Pb(s)|PbSO_4(s)|SO_4^{-2}(aq)||Cu^{2+}(aq)|Cu(s)\)

b. \(Hg(l)|Hg_2Cl_2(s)|Cl^{-}(aq)||Cd^{2+}(aq)|Cd(s)\)

Quick Review:

The left side of the cell diagram is usually the anode, which will be oxidized and release electrons. The right side of the cell diagram is usually the cathode, which will be reduced and absorb the electrons. In addition, the electron transfer from anode to cathode in a galvanic cell causes the solid electrode at the anode to lose mass since it is transferring electrons to the cathode, which is gaining mass at its electrode.

Problem Solving:

a. \(Pb(s)|PbSO_4(s)|SO_4^{-2}(aq)||Cu^{2+}(aq)|Cu(s)\)

On the left side of the cell diagram, we have:

Anode (Oxidation): \(Pb(s) + SO_4^{-2}(aq) \to 2e^- + PbSO_4(s)\)

On the right side of the cell diagram, we have:

Cathode (Reduction): \(Cu^{2+}(aq) + 2e^{-} \to Cu(s)\)

Overall, the reaction would be $$Pb(s) + SO_4^{-2}(aq) + Cu^{2+}(aq) + 2e^{-} \to 2e^- + PbSO_4(s) + Cu(s)$$

$$Pb(s) + SO_4^{-2}(aq) + Cu^{2+}(aq) \to PbSO_4(s) + Cu(s)$$

b. \(Hg(l)|Hg_2Cl_2(s)|Cl^{-}(aq)||Cd^{2+}(aq)|Cd(s)\)

On the left side of the cell diagram, we have:

Anode (Oxidation): \(2Hg(l) + 2Cl^{-}(aq) \to 2e^- + Hg_2Cl_2(s)\)

On the right side of the cell diagram, we have:

Cathode (Reduction): \(Cd^{2+}(aq) + 2e^{-} \to Cd(s)\)

Overall, the reaction would be $$2Hg(l) + 2Cl^{-}(aq) + Cd^{2+}(aq) + 2e^{-} \to 2e^- + Hg_2Cl_2(s) + Cd(s)$$

$$2Hg(l) + 2Cl^{-}(aq) + Cd^{2+}(aq) \to Hg_2Cl_2(s) + Cd(s)$$

Problem 20.5.24

The biological molecule abbreviated as \(NADH\) (reduced nicotinamide adenine dinucleotide) can be formed by reduction of \(NAD^{+}\) (nicotinamide adenine dinucleotide) via the half-reaction \(NAD^{+} + H^{+} + 2e^{-} \to NADH\); \(E° = -0.32V\).

$$acetate + CO_2 + 2H^{+} + 2e^{-} \to pyruvate + H_2O \;\;\; E° = -0.70V$$

$$pyruvate + 2H^{+} + 2e^{-} \to lactate \;\;\; E° = -0.185V$$

a. Would \(NADH\) be able to reduce acetate to pyruvate?

b. Would \(NADH\) be able to reduce pyruvate to lactate?

c. What potential is needed to convert acetate to lactate?

Problem Solving:

a. In order to see if \(NADH\) would be able to reduce acetate to pyruvate, we simply must calculate the overall standard reduction potential of the system. In order for \(NADH\) to reduce something, \(NADH\) would have to be oxidized, which would form the reaction $$NADH \to NAD^{+} + H^{+} + 2e^{-} \;\;\; E° = 0.32V$$ Please note that in order to calculate \(E°\) for the reverse reaction, we simply just change the sign.

With this reaction, we may now pair it with the desired reaction and calculate the overall \(E\) to determine if such a pairing of reactions would be spontaneous.

$$E = \Sigma{E°} = 0.32V + (-0.70V) = -0.38V \;<\; 0$$

We note that the resulting overall reduction potential is less than 0, which indicates that \(NADH\) would not be able to reduce acetate to pyruvate.

b. Repeating this process for b, we find that $$E = \Sigma{E°} = 0.32V + (-0.185V) = 0.135V \;>\;0$$

This value is positive, which indicates that \(NADH\) would be able to reduce pyruvate to lactate.

c. In order to convert acetate to lactate, we would have to combine the standard reduction potentials of both half reactions and the paired reaction would have to make the overall standard reduction potential for all 3 reaction positive.

$$E = xV + (-0.70V) + (-0.185V) > 0$$

$$xV - 0.885V > 0$$

$$xV = 0.885V$$

We would require a reaction with a potential greater than 0.885V in order to convert acetate to lactate.

The overall approach to this problem would be to calculate the overall standard potential of the system by Cathode_{reduction potential} - Anode_{reduction potential} where reduction occurs at the cathode and oxidation at the anode. The overall standard potential should be positive if it were to be favored and negative if it is not favored.