Extra Credit 40

Q17.5.8

Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.

Q17.5.8 solution:

If you look at the Nernst equation$$E_{cell}=E^o-\frac{RT}{nF} lnQ$$ it can be seen that the E _cell is proportional to the temperature (T). If you were to have a low temperature then that would lower your overall E _cell thus resulting in a lower voltage, which of course would cause poor performance.

Q12.3.3

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?

Q12.3.3 solutions:

1. We can start with a general rate law. Next, multiply your concentration(x) by 3 because you are tripling the concentration. Next, multiply your rate by 9 because the rate is being increased by a factor of 9 when the concentration is tripled. Let's say that your initial concentration and rate were both 1 to make the calculation simpler. At this point, you can ignore the K because it is just a constant.Take the log of both sides to bring down n. Divide by log[3] to solve for n. $$\begin{align} R=k[x]^n\\R=k[3x]^n\\9R=[3x]^n\\log[9R]=n(log[3x])\\\frac{log[9]}{log[3]}=n\\n=2\end{align}$$

Since n is two then your reaction is a second order reaction with respect to that reactant.

2. Just like in the previous problem we can start off with a general rate law. Next, multiply your concentration(x) by 4 because you are quadrupling the concentration. Next, multiply your rate by 4 as well because the rate is being increased by a factor of 4 when the concentration is quadrupled. Let's say that your initial concentration and rate were both 1 to make the calculation simpler. At this point, you can ignore the K because it is just a constant. It can be seen right away that the concentration and the rate are equal. $$\begin{align} R=k[x]^n\\R=k[4x]^n\\4R=[4x]^n\\n=1\end{align}$$

Since n is one, then your reaction is a first order reaction with respect to that reactant.

Q12.5.12

In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?

1. the change in free energy per second
2. the change in temperature per second
3. the number of collisions per second
4. the number of product molecules

Q12.5.12

In order to be able to answer this question, we must first look at the equation for collision theory. $$Rate=[Rate\;of\:molecules\; with\;correct\;orientation]+[Fraction\;of\;molecules\; with\;required\; energy]+[collision\; frequency]$$

If we look at the formula we can see that the correct answer is C because according to the equation the number of collisions per second directly affect the rate of the reaction.

Q21.4.7

Which of the following nuclei is most likely to decay by positron emission? Explain your choice.

1. chromium-53
2. manganese-51
3. iron-59

In order to answer this question, you must take a look at the belt of stability. Another way to look at this question is to think about the neutron to proton ratio for each of these elements. An element is the most stable when the ratio of neutrons to protons is the closest to 1. $$\begin{align}
\ \mathbf C\mathbf r\mathbf- \mathbf5 \mathbf3\\
\ Chromium-53\ has\ 24\ protons\ and\ 29\ neutrons\ (53-24=29)\\
\frac{29}{24}=1.2\\
\ \mathbf M\mathbf n\mathbf -\mathbf 5\mathbf 1\\
\ Manganese-51\ has\ 25\ protons\ and\ 26\ neutrons\ (51-25=26)\\
\frac{26}{25}=1.04\\
\ \mathbf F\mathbf e\mathbf -\mathbf 5\mathbf 9\\
\ Iron-59\ has\ 26\ protons\ and\ 33\ neutrons\ (59-26=33)\\
\frac{33}{26}=1.27\\
\end{align}$$

Since manganese-51 has the smallest Neutron to proton ration (1.04) it will be the most stable.

Q20.2.11

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

1. Platinum wire is dipped in hydrochloric acid.
2. Manganese metal is added to a solution of iron(II) chloride.
3. Tin is heated with steam.
4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.

Q20.2.11 Solution:

Let's first take a look at the activity series and what is tells us. The activity series lists the metals from highest to lowest reactivity. This allows us to determine if a displacement reaction will occur between the two metals. If one is higher in the activity series than it will displace the one that is lower.

1. Platinum wire is dipped in hydrochloric acid- no reaction will occur.

2. Manganese metal is added to a solution of iron(II) chloride- Because Manganese is higher in the activity series it will displace the iron. \begin{align}
\mathbf N \mathbf e\mathbf t\ \mathbf i\mathbf o\mathbf n\mathbf i\mathbf c\ \mathbf e\mathbf q\mathbf u\mathbf a\mathbf t\mathbf i\mathbf o\mathbf n\mathbf :\\
\ Mn^{+2}+2Cl^-\rightarrow MnCl_2\\
\mathbf I\mathbf o\mathbf n\mathbf i\mathbf c\ \mathbf e\mathbf q\mathbf u\mathbf a\mathbf t\mathbf i\mathbf o\mathbf n\mathbf :\\
\ Mn^{+2}+Fe^{+2}+2Cl^{-}\rightarrow MnCl_2+ Fe^{+2}
\end{align}

3. Tin is heated with steam. According to the activity series tin will react and displace H_2. \begin{align}
\mathbf N \mathbf e\mathbf t\ \mathbf i\mathbf o\mathbf n\mathbf i\mathbf c\ \mathbf e\mathbf q\mathbf u\mathbf a\mathbf t\mathbf i\mathbf o\mathbf n\mathbf :\\
\ Sn^{+2}+ O^ {2-}\rightarrow SnO\\
\mathbf I\mathbf o\mathbf n\mathbf i\mathbf c\ \mathbf e\mathbf q\mathbf u\mathbf a\mathbf t\mathbf i\mathbf o\mathbf n\mathbf :\\
\ Sn^{+2}+H_2O(g)\rightarrow SnO+ 2H^{+}
\end{align}

4. Hydrogen gas is bubbled through a solution of lead(II) nitrate. No reaction

Q20.5.7

Occasionally, you will find high-quality electronic equipment that has its electronic components plated in gold. What is the advantage of this?

Q20.5.7 Solution:

An advantage to having gold in electronic equipment is because gold does not react with oxygen making resistant to corrosion. Corrosion will occur to a metal that has been refined to make it more stable.

Q24.6.4

For an octahedral complex of a metal ion with a d6 configuration, what factors favor a high-spin configuration versus a low-spin configuration?

Q24.6.4 Solutions:

The only thing in this case that will favor high-spin vs low spin is if it the metal ion is a strong field or weak field. If it is a strong field then it will be low spin because the pairing energy will be less than Delta O. If it is a weak field then it will be high spin because the pairing energy will be greater than Delta O.