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Problem 71

Consider the reaction: $$\mathrm{CH}_{3} \mathrm…

Problem 70

Consider the evaporation of methanol at $25.0^{\circ} \mathrm{C} :$
$$\mathrm{CH}_{3} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$
\begin{equation}\begin{array}{l}{\text { a. Find } \Delta G_{\mathrm{r}}^{\circ} \text { at } 25.0^{\circ} \mathrm{C} \text { . }} \\ {\text { b. Find } \Delta G_{\mathrm{r}} \text { at } 25.0^{\circ} \mathrm{C} \text { under the following nonstandard }} \\ {\text { conditions: }} \\ \quad {\text { i. } P_{\mathrm{CH}_{3} \mathrm{OH}}=150.0 \mathrm{mm} \mathrm{Hg}} \\ \quad{\text { ii. } P_{\mathrm{CH}_{3} \mathrm{OH}}=100.0 \mathrm{mmHg}} \\ \quad{\text { iii. } P_{\mathrm{CH}_{1} \mathrm{OH}}=10.0 \mathrm{mm} \mathrm{Hg}} \\ {\text { c. Explain why methanol spontaneously evaporates in open air }} \\ {\text { at } 25.0^{\circ} \mathrm{C} .}\end{array}\end{equation}


PART $\mathrm{A} : \Delta G=4.3 \mathrm{kJ}$
PART B-I: $\Delta G^{\circ}=0.3 \mathrm{kJ}$
PART B-II: $\Delta G=-0.7 \mathrm{kJ}$
PART B-III: $\Delta G=-6.7 \mathrm{kJ}$
PART C: See explanation.



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Video Transcript

So we have this reaction. Methanol is a liquid going to methanol as a gap, the rage. And we need to figure out what Don't touch Jeannot iss at 25 degrees C. So we need to calculate Don't age first, which we confined by doing. Don't the agent the age information of products minus H permission reactors. So for the gas, it's negative to one. And then subtract your reactions, which is negative to 38.6, which is the liquid form the methanol. And I get to you 37.6 Coastal's and you can find the these two values in your textbook in the appendix. Same thing for Dr S not, um 2 39.9 minus 1 26.8 which equals Go in there. Teen jewels for Calvin. We want to cover that coach ALS, since that's a jizz in killed Jules. So that's really dear 0.11 three killer jewels for coming. So if that when concha adults do not using the reaction, doesn't h night minus Tito the s not where t is to 98 Kelvin. So you plug bent in using the values we just got Ah, 37.6, honest to 98 terms. 0.113 and I get to you 3.926 Killer Jules. Next. They want you to figure out what Delta G is a nonstandard state conditions such as, Ah, 1 50 millimeters of mercury. And so I'm just gonna cover this atmosphere's since that's what we're used to working in. So it would be 0.197 atmospheres. Find out the G using this equation. You could find your textbook. Don't Jeannot plus Artie. Ln of keys. Where Q. Is the partial pressure of methanol in this case 0.197 So you plug it in 3.96 plus R, which is a constant is 8.31 for Jules Per Kelvin Moore Cullen lessons for Inco, Jules and this one. We'd put this in Killer Jules, Do you point? A 314 times room temperature said to 98 comes the natural log of 0.17 This comes out to be negative. 0.0 None. None. Killer. Jules, look. So you do the same thing for the other pressures. So 200 millimeters of Mercury, which is equal to 0.132 atmospheres. You plug into that same equation in Delta G equals negative 1.0, non one killer Jules for 10 movies. Mercury, which is Europe. Wait 01312 atmospheres. Delta G equals minus 6.7 and six. Killer Jules. And so they want to say why methanol spontaneously evaporates in air a room temperature, even though Delta Gina is positive. And so that's because the proper pressure of methanol is not one atmosphere. It's more likely less than that, UM could be is low as high the highest it could be. We point win on seven atmospheres because that's about when this is close to zero. So this is about closest. Negative is it'll get. So that's how you know the reaction is done. Spontaneous. So has to do with the partial pressure of methanol that determines whether it evaporates in the air, not at room temperature

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