Extra Credit 43

Step-by-Step Explanation:

\[\frac{-d[NO]}{dt}=\frac{-d[O_{3}]}{dt}=\frac{d[NO_{2}]}{dt}=\frac{d[O_{2}]}{dt}\]

The rate of disappearance of each reactant is negatively proportional to the appearance of each product. The overall rate of the reaction is equivalent to the rate of disappearance of NO times the rate of disappearance of \(O_{3}\) times the rate constant, which are all given.

Plug the values in:

\[Rate =(2.2×10^{7}\frac{1}{Ms})[NO][O_{3}] *\]

\[Rate =(2.2×10^{7}\frac{1}{Ms})[3.3 10^{−6}M][5.9×10^{−7}M]\]

And solve for the overall rate of the equation:

\[Rate = 4.3×10^{-5}\frac{M}{s}\]

Since the overall rate of the reaction is \(4.3 × 10^{-5}\) M/s (and equivalent to the rate of formation), the rate of disappearance is the negative equivalent of the formation rate according to the balanced equation.

*Molarity (M) = moles/Liter

Step-by-Step Explanation:

1. Calculations \[k = Ae^{-Ea/RT}\] (Note: Temperature in Kelvin)

   1. \(E_{a}\)

      1. \(E_{a}\) is equivalent for both sets of temperature and k-values. Use the log of each and set them equal, solve for \(E_{a}\)
         1. \(ln(2.1 × 10^{−11}) = lnA - \frac{Ea}{(8.314)(300)}\)
         2. \(ln(8.5 × 10^{−11}) = lnA - \frac{Ea}{(8.314)(310)}\)
         3. Solve for A:
            1. lnA =\(ln(2.1 × 10^{−11}) - \frac{Ea}{(8.314)(300)}\)
            2. lnA =\(ln(8.5 × 10^{−11}) - \frac{Ea}{(8.314)(310)}\)
         4. Set them Equal:
            1. \(ln(2.1 × 10^{−11}) - \frac{Ea}{(8.314)(300)}\) = \(ln(8.5 × 10^{−11}) - \frac{Ea}{(8.314)(310)}\)
         5. Solve for \(E_{a}\)
            1. \(\frac{-Ea}{(8.314)(300)} + \frac{Ea}{(8.314)(310)} = ln(8.5 × 10^{−11}) - ln(2.1 × 10^{−11})\)
            2. \(E_{a}(\frac{-1}{(8.314)(300)} + \frac{1}{(8.314)(310)}) = ln(8.5 × 10^{−11}) - ln(2.1 × 10^{−11})\)
            3. \(E_{a}\) = \(\frac{ln(8.5 × 10^{−11}) - ln(2.1 × 10^{−11})}{-1/(8.314)(300) + 1/(8.314)(310)}\)
            4. \(E_{a}\) = 108103.5999 J = 108.103 KJ

   2. Frequency Factor

      1. \(2.1 × 10^{−11} = Ae^{-108103/(8.314)(300)}\)
         1. A = \(\frac{2.1 × 10^{−11}}{e^{-108103/(8.314)(300)}}\)
         2. A = \(1.4 x 10^{8}\)
      2. \(8.5 × 10^{−11} = Ae^{-108103/(8.314)(310)}\)
         1. A = \(\frac{8.5 × 10^{−11}}{e^{-108117.2/(8.314)(310)}}\)
         2. A = \(1.4 x 10^{8}\)

   3. Rate Constant (k)

      1. \(k = Ae^{\frac{-Ea}{RT}}\)
      2. k = \(1.4 x 10^{8}e^{-108103/(8.314)(47+273)}\)
      3. k = \(1.4 x 10^{8}e^{-40.633}\)
      4. k = \(3.16 x 10^{-10}\)

2. Time to reach equilibrium at 27 °C without a catalyst

   1. This reaction is first order, so we know that \(ln(\frac{[A]}{[A]_{o}})\) = -kt and we know the initial and final concentration, and k at 27 °C
      1. \(ln(\frac{[1.65 x 10^{-7}}{[.150]}) = -2.1 × 10^{−11}t \)
      2. t = \(\frac{ln([1.65 x 10^{-7}]/[.150])}{-2.1 × 10^{−11}}\)
      3. t = \(6.53 x 10^{11}\) s = \(1.09 x 10^{10}\) min = \(1.81 x 10^{8}\) hr = \(7.56 x 10^{6}\) days = 20717.4 years

Step-by-Step Explanation:

1. n/p = 6/4 = 1.5>1; The approximate slope of the middle the belt of stability is about 1.5 in the upper right hand of the graph and goes closer to 1 near the origin (assuming the belt interests with the origin and has a y intersect of zero, you can approximate using y= mx+ b). While it might be hard to see, the on the graph, this assumption can be used to evaluate that this isotope lays ABOVE the belt of stability (1.5> 1), which according to the chart indicates that it has \(ß^{-}\) decay; \(^{6}He \rightarrow ^{6}Li + _{-1}e\).
2. n/p = 30/30 = 1<1.2; Using a similar understanding, you can approximate the slope in that area of the curve and also compute the n/p ratio. This time, however, the ratio number is BELOW the approximated slope, which if you can't see from simply reading the graph, should be \(ß^{+}\) decay; \(^{60}Zn \rightarrow ^{60}Cu + _{1}e\).
3. n/p = 144/91 = 1.55> 1.5; The first thing you should note is that Z = 91, which is greater than 83, which should indicate that it has alpha decay. However, due the large number of neutrons in relation to protons, is lands just to the left of the curve, making it \(ß^{-}\) decay. In the middle chart, it lands in the small "orange island" in the upper right hand corner.
4. n/p = 148/93= 1.6>1.5; The first thing you should note is that Z = 93, which is greater than 83, which should indicate that it has alpha decay. However, due to the same reasoning as the previous element, it falls to the left of the belt and is thusly \(ß^{-}\) decay.
5. n/p = 9/9 = 1<= 1ish; again, its hard to see near the origin, but at around p = 8 or 9, the slope of the belt increases, making it likely that this isotope fall beneath it, making it \(ß^{+}\) decay.
6. n/p = 73/56 = 1.3<1.33; Since this falls really close to the belt, it is again hard to see. But approximations put the ratio just below the slope of the belt, making it \(ß^{+}\) decay.
7. The first thing you should note is that Z = 94, which is greater than 83, which should indicate that it has alpha decay. (However, the internet indicates that Electron Capture is much more likely, but according to the material Professor Larson gave us, alpha decay is correct)

* The notation for this problem is rather confusing. The isotope was written as Np, with an atomic number of 94 and and mass number of 241. However, the atomic number of Np is actually 93. Plutonium has an mass number of 94, in which case, the isotope would be Plutonium-241 and not Neptunium 241. Regardless, both should have a similar type of decay.

Step-by-Step Explanation:

1. Which is oxidized? Which is reduced?

   1. Remember the acronym : OIL RIG (Oxidation Is Losing Reduction Is Gaining)
   2. Cu metal has no charge, but after the reaction it has a positive charge. This means it has lost electrons, and therefore is Oxidized, because oxidation is the losing of electrons.
   3. Look at the oxidation states of each elemental atom in other compounds.
      1. \(HNO_{3}\)
         1. H: +1, N: +5 , O: -2
      2. NO
         1. N: +2, O: -2
   4. The oxidation number of Nitrogen decreases, which means electrons are gained which lowers the positive charge. This means \(HNO_{3}\) is reduced, because reduction is the gaining of electrons.

2. Balance the Chemical Equation: \(Cu_{(s)} + HNO_{3(aq)} \rightarrow Cu^{2+}_{(aq)}+ NO_{(g)​​​​​​}\)

   1. Make sure metals and non water related atoms are balanced (Basically anything besides H and O)
      1. There is one mole of Copper on each side and one mole of Nitrogen on each side
   2. Break the Reaction up in Half-Reactions:
      1. Oxidation: \(Cu_{(s)} \rightarrow Cu^{2+}_{(aq)}\)
      2. Reduction: \(HNO_{3(aq)} \rightarrow NO_{(g)​​​​​​}\)
   3. Balance Oxygen by adding \(H_{2}O\) molecules
      1. Oxidation: \(Cu_{(s)} \rightarrow Cu^{2+}_{(aq)}\)
      2. Reduction: \(HNO_{3(aq)} \rightarrow NO_{(g)​​​​​​} + 2H_{2}O_{(l)}\)
   4. Balance H by adding H+
      1. Oxidation: \(Cu_{(s)} \rightarrow Cu^{2+}_{(aq)}\)
      2. Reduction: \(3H^{+}_{(aq)} + HNO_{3(aq)} \rightarrow NO_{(g)​​​​​​} + 2H_{2}O_{(l)}\)
   5. Balance the charge by adding electrons to the opposing side
      1. Oxidation: \(Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-}\)
      2. Reduction: \(3e^{-} + 3H^{+}_{(aq)} + HNO_{3(aq)} \rightarrow NO_{(g)​​​​​​} + 2H_{2}O_{(l)}\)
   6. Multiply each half-reaction by a factor so that the electrons are equal
      1. Oxidation: \(3Cu_{(s)} \rightarrow 3Cu^{2+}_{(aq)} + 6e^{-}\)
      2. Reduction: \(6e^{-} + 6H^{+}_{(aq)} + 2HNO_{3(aq)} \rightarrow 2NO_{(g)​​​​​​} + 4H_{2}O_{(l)}\)
   7. Add the half-reactions and cancel terms that occur on both sides
      1. \(3Cu_{(s)} + 6H^{+}_{(aq)} + 2HNO_{3(aq)} + 6e^{-}\rightarrow 3Cu^{2+}_{(aq)} + 2NO_{(g)​​​​​​} + 4H_{2}O_{(l)} + 6e^{-}\)
   8. Answer: \(3Cu_{(s)} + 6H^{+}_{(aq)} + 2HNO_{3(aq)} \rightarrow 3Cu^{2+}_{(aq)} + 2NO_{(g)​​​​​​} + 4H_{2}O_{(l)}\)