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Problem 120

Not all processes in which the system increases i…

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Problem 123

Borax (sodium tetraboratedecahydrate), a mineral found in dry lakebeds in California, is used as a preservative and in the manufacturing of soap and glass. By detervining the $K_{\text { sp }}$ of borax at different temperatures, we can determing the $K_{\text { sp }}$ of $\Delta G^{\circ}$ for the dissolution of borax:
$$\begin{array}{c}{\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}^{2-}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)} \\ {(\mathrm{Borax}) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text (Tetraborate)}\end{array}$$
The relationship:
$$\ln \left(K_{\mathrm{sp}}\right)=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R}$$
has the form of a linear equation $y=m x+b,$ where $y$ is the $\ln K_{\text { sp }}$ and $x$ is 1$/ T(T$ in Kelvin). The slope is equal to $-\Delta H / R,$ and the $y$ intercept is $\Delta S^{\circ} / R,$ where $R$ is the gas constant, 8.314 $\mathrm{J} /$ $\mathrm{K}$ mol. Determining $K_{\mathrm{sp}}$ at several different temperatures allows us to plot a graph of lnK versus 1$/ T$ as shown in Figure a
Knowing the values of $\Delta H^{\circ}$ and $\Delta S^{\circ}$ at a specific tempera-
ture allows the calculation of the change in Gibbs free energy
for the reaction $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$
$$\begin{array}{|l|l|}\hline Temperature \left(^{\circ} \mathrm{C}\right) & {K_{\mathrm{sp}}} \\ \hline 40.0 & {0.041} \\ \hline 45.0 & {0.083} \\ \hline 50.0 & {0.264} \\ \hline 55.0 & {0.486} \\ \hline 60.0 & {0.552} \\ \hline\end{array}$$
Use the information provided in the figure and table to do the following:
\begin{equation}\begin{array}{l}{\text { a. Plot a graph of } \ln K_{\text { sp versus } 1 / T(T \text { in Kelvin })} \text { . }} \\ {\text { b. Determine } \Delta H^{\circ} . \text { Is this process endothermic or exothermic? }} \\ {\text { c. Determine } \Delta S^{\circ} .} \\ {\text { d. Determine } \Delta G^{\circ} \text { at } 298 \mathrm{K} \text { . }} \\ {\text { e. Sketch a graph of lnK versus 1/T for an exothermic process. }}\end{array}\end{equation}

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Problem 119

If you roll 1 million dice, what will be the average of all the dice? If there is a room with 1 million dice and they all have a 1 on the top face, and there is an earthquake strong enough to roll dice around, what is the likelihood that after the earthquake all the top faces will sum to 1 million? To 6 million? How does this thought experiment illustrate the second law of thermodynamics?

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Video Transcript

All right, So in order to answer this question, let's start with the last part of it. How does this thought process this thought experiment illustrate the second law of thermodynamics. Well, the second law simply states that the total entropy of an isolated system naturally moves toward a state of greater disorder. This basically means that the greater the variability, the greater the entropy, which we're gonna label within us. In the case of rolling one million dice, there is no event such as rolling a one or two or three that is more likely than another. But isis six sides, you are just as likely to roll it one as you are to roll a four as you are to roll it to. The other part is this is that, um, these events are also mutually exclusive, which means that you are once you roll one dice, it's not going to affect the role of another dice. In this case, you have one million dice. One role isn't going to affect any of the 999,999 rules. To answer the first part of this question, you can't really calculate an average because there is so much variability in the situation and because all of the outcomes are equally likely to happen, there's no real way of calculating without actually being given an outcome. One possible outcome chance for the second part, and therefore the third part. In order to some to one million, all one million dice must roll up onto a one after being shaken. Similarly, to add up to six million all the dice Muslims on a six, regardless of what number that they start on in basic terms, The second law tells you that order isn't favored since it implies less variability or entropy. Therefore, in a system like this, or in any system that tends toward greater disorder, these two events are highly unlikely because there isn't much variability in these specific outcomes.

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