Extra Credit 36

Q17.5.4

Consider a battery with the overall reaction:

\[Cu(s)+2Ag^+(aq)⟶2Ag(s)+Cu^{2+}(aq)\]

1. What is the reaction at the anode and cathode?
2. A battery is “dead” when it has no cell potential. What is the value of \(Q\) when this battery is dead?
3. If a particular dead battery was found to have \([Cu^{2+}] = 0.11\, M\), what was the concentration of silver ion?

S17.5.4

1. anode: \[Cu(s)⟶Cu^{2+}(aq)+2e^-\] with \(E^∘(anode)=0.337\, V\)

   cathode: \[Ag^+(aq) + e^-⟶ Ag(s)\] with \(E^∘(cathode)=0.7996\, V\)

2. We can use Nernst equation to solve the problem:

   \[E=E^∘-\frac{0.05916}{n}\log Q\]

   Where \(E^∘\) is \(0.7996-0.337=0.4626\, V\), and \(n\) is \(2\), which is the number of electrons being transferred in the reaction.

   The student needs to add in all the steps:

   Plugging everything in, we get:

   \[E=E^∘- \frac{0.05916}{n}\log Q\]

   \[0=0.4626 - \frac{0.05916}{2}\log Q\]

   *E is equal to zero because a dead battery has no cell potential.

   \[0.4626 = \frac{0.05916}{2}\log Q\]

   \[15.63894523 = \log Q\]

   \(\ce{\ce{10^{15.63894523}}}\) = \(\ce{10^{log Q}}\)

   \[Q=4.35457 \times 10^{15}\]

3. For this reaction, \[Q=\frac{[Cu^{2+}]}{[Ag^+]^2}\]

   \[[Ag^+]=\sqrt{\frac{[Cu^{2+}]}{Q}}\]

   \[[Ag^+]=5.61 \times 10^{−9}\, M\] (this answer is incorrect)

   Correct Answer for part c for 17.5.4:

   c. For this reaction, \[Q=\frac{[Cu^{2+}]}{[Ag^+]^2}\]

   \[[Ag^+]=\sqrt{\frac{[Cu^{2+}]}{Q}}\]

   \[[Ag^+]=\sqrt{\frac{0.11}{4.35457 \times 10^{15}}}\]

   \[[Ag^+]=5.026 \times 10^{−9}\, M\]

Q12.3.3(student did not answer the correct question)

Tripling the concentration of a reactant increases the rate of a reaction nine times. With this knowledge, answer the following questions:

1. What is the order of the reaction with respect to that reactant?
2. Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times. What is the order of the reaction with respect to that reactant?

S12.3.3

1. Since tripling the concentration of a reactant increases the rate of a reaction nine times, it is a second order reaction.

2. Since increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times, it is a fourth order reaction.(this answer is incorrect)

   Correct Answer for part b for 12.3.3:

   b. If the concentration of the reactant quadruples (increase by a factor of four) and the rate of the reaction increases four times, then it means that it's a first order reaction.

   Correct Answer for part b for 12.3.3:

   Q12.2.3

   In the PhET Reactions & Rates interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature?

   Answer:

   If the temperature is increased, then the reactant particles move quickly, they have more energy, the particles collide more often and more collisions happen, so the rate of the reaction increases. If the concentration is increased, then the reactant particles become more crowded, there is a higher probability of the particles colliding, and the rate of the reaction increases.

Q12.5.8

In an experiment, a sample of NaClO₃ was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher?

S12.5.8(student did not answer question)

Correct Answer for 12.5.8:

b. We are given how much decomposed and the time, \(\ce{t}\), it took for that amount to decompose. We are given that 90% has decomposed which means that there is only 10% (100-90 = 10) left. From this information, we can solve for \(\ce{k}\) by using the integrated rate law and plugging in the known values:

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-kt}\)

For \(\ce{N_{0}}\) and \(\ce{N}\), we can put in whatever values we want as long as the ratio between the two is 0.10. So, I will use \(\ce{N_{0}}\) = 100 g and \(\ce{N}\) = 10 g. Plugging everything in, we get:

\(\ce\ln({\frac{10}{100})}\) =\(\ce{(-k)(48)}\)

\(\ce\ln({0.10}\) =\(\ce{-48k}\)

\(\ce{-2.302585093}\) = \(\ce{-48k}\)

\(\ce{k}\) = 0.04797

Next, we have to use the following equation:

\(\ce{k}\) = \(\ce{Ae^{\frac{-E_{a}}{RT}}}\)

One more equation to note is that:

\(\ce{A}\) = \(\ce{kN}\)

We can substitute this equation into the equation involving activation energy and solve for it:

\(\ce{k}\) = \(\ce{kNe^{\frac{-E_{a}}{RT}}}\)

We can simplify this equation even more by:

\(\ce{1}\) = \(\ce{Ne^{\frac{-E_{a}}{RT}}}\)

Here, we know R = 8.3145 \(\ce{J}\)\(\ce{mol \cdot K}\); for T, we can choose any temperature so I will let T = 298 K, and we will plug we will plug in the value for \(\ce{N_{0}}\) for \(\ce{N}\) and solve for activation energy:

\(\ce{1}\) = \(\ce{100e^{\frac{-E_{a}}{(8.3145)(298)}}}\)

\(\ce{0.01}\) = \(\ce{e^{\frac{-E_{a}}{(8.3145)(298)}}}\)

\(\ce{ln({0.01})}\) =\(\ce{\frac{-E_{a}}{(8.3145)(298)}}\)

\(\ce{E_{a}}\) = \(\ce{11410.33}\) \(\ce{J}\)

Now that we know the activation energy, \(\ce{E_{a}}\), we can use the following equation:

\(\ce\ln({\frac{k_{2}}{k_{1}})}\) = \(\ce{\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})}\)

For \(\ce{T_{2}}\), it's 298 + 20 = 318 K because in the problem it stated 20 degrees Celsius higher and here we don't need to convert the 20 degrees Celsius into Kelvin because it's a difference. We need to solve for \(\ce{k_{2}}\). Plugging everything in, we get:

\(\ce\ln({\frac{k_{2}}{0.04797})}\) = \(\ce{\frac{11410.33}{8.3145}(\frac{1}{298}-\frac{1}{318})}\)

\(\ce\ln({\frac{k_{2}}{0.04797})}\) = \(\ce{\frac{11410.33}{8.3145}(\frac{1}{298}-\frac{1}{318})}\)

\(\ce\ln({\frac{k_{2}}{0.04797})}\) = \(\ce{0.289633}\)

\(\ce{\frac{k_{2}}{0.04797}}\) = \(\ce{e^{0.289633}}\)

\(\ce{k_{2}}\) = 0.06399

With this \(\ce{k}\) value, we can use the integrated rate law to find out the time, \(\ce{t}\) it takes to decompose when the temperature is 20 degrees higher:

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-kt}\)

Plugging everything in, we get:

\(\ce\ln({\frac{10}{100})}\) =\(\ce{(-0.06399)(t)}\)

\(\ce\ln({0.10}\) =\(\ce{-0.06399t}\)

\(\ce{-2.302585093}\) = \(\ce{-0.06399t}\)

\(\ce{t}\) = 35.98

This decomposition would have taken 35.98 min if the sample had been heated 20 °C higher.