Consider the evaporation of methanol at 25.0 °C :
CH3OH(l ) ¡ CH3...

Question

Consider the evaporation of methanol at 25.0 °C :
CH3OH(l ) ¡ CH3OH( g)
b. Find ΔG 
r at 25.0 °C under the following nonstandard 
conditions: 
i. PCH3OH = 150.0 mmHg 
ii. PCH3OH = 100.0 mmHg 
iii. PCH3OH = 10.0 mmHg

Accepted Answer

Hey everyone today, we're being asked to find the Delta G of the vaporization of bromine at 25°C at two different pressures of bromine. So breaking the question down, we're being asked to find the delta G of the reaction at different pressures. Now, that alone should hint that we are not dealing with standard delta G. We're actually dealing with non standard gibbs free energy. So we have to use the formula that delta G of the reaction is equal to the standard delta G of the reaction plus R. T. Natural log of Q. Where are as a gas constant. T. Is the temperature in kelvin? Delta G. Standard is the standard gibbs free energy of the reaction and Q is the reaction quotient and Q is one of the most important parts here. So let's break that down. The reaction quotient is nothing but the either concentration or pressure of the products raised to the power of their coefficients divided by the pressure or concentration of the reactant raised to the power of their coefficients. However, one very important thing to note is that this does not apply to liquids or solids, it only applies two gasses only gasses. Oops, So with that in mind, if we go ahead and take a look at the balance equation given the vaporization of bromine, which is BR two liquid uh heating up and becoming br to gas, it is vapor rising. The only, the only gas here is the bruning gas which is a product and it has a coefficient of one, which means the reaction quotient. Now here is simply equal to the pressure of the bromine gas. Keep in mind that we need pressure in ATMs in ATM atmospheres. But with that in mind we now know that we can use this to go ahead and find Q once we get to the actual solving of the problem. However, there's still a couple more terms that we need to do. The first is very easy. It is the temperature we need temperature in kelvin, but it is given in uh degrees Celsius. And to convert this, it's also very easy. We simply take the 25 degrees Celsius And we added to 273.15, which gives us 298. Kelvin. And that is the value for T. So t. Is equal to that. The final value that we need to find now is the uh standard delta G of the reaction. Now we can do this also quite simply utilizing the standard gibbs free energy of formation. So, let's write this out. The standard delta G of the reaction is equal to the difference in the standard gibbs free energy of formation of the products and the reactant. Since products, products and the reactant reactant. So we only have one product and one reactant. The product is the br two gas. The propane gas. So the standard entropy of formation is very simple. It is a experimentally calculated value that we can find in tables or in in the back of a textbook, for example. But the value here will be 82.4 kg joules per mole for the burning liquid. However, for the product or sorry, the reactant, the actual delta G of formation, the standard delta G information will actually be zero. It'll be zero kg per mole because for elements that exist in their natural state, such as 02 or B R two in this case, this is their most stable state and it is what they would appear like in nature if they weren't reacting with other elements as such, their standard delta G information will simply be zero. They both have a moller coefficient of one. So we don't have to really worry about the Mueller conversions or anything. The uh delta G of the reaction. The standard delta G of the reaction would therefore just become reaction is simply 84 kg joules per mole minus zero or 80 82.4, sorry, 82.4 kg joules per mole minus zero, which is simply equal to 82.4 kg joules per mole. And we're done. So those are the key values that we need to find out that are that will apply to both conditions one and two for the different temperatures. Or pressures. My bad with these in hand, we can go ahead and start taking a look at the actual questions. So, let's write this out. We'll take a look at the first one where the partial pressure blooming blooming gas specifically is 100 or 1.5 mm HG millimeters of mercury. So to work with this, all we need to do is convert millimeters H. G. Into atmospheres, which is very simple. All we do is take the PBR so let's write this out. Q. Is equal to P. B. R. Two CASS, It's equal to 1.5 zero mm H. G. And if we recall one atmosphere is equivalent 2 760 HG. Which gives us a answer of 1. times 10 to the -3 atmospheres. So with all this in mind, we can go ahead and start solving for the delta G. Of the reaction because now we have the value of Q. So let's write this out. Delta G of the reaction is equal to 82. kg joules per mole 82.4 kg joules Permal. All right, just like this. Permal Plus R. is given as a value of 8. 314 jewels per mole kelvin jewels per more kelvin. However, we need our answer and kill a jewel. So we need to convert this to kill a jewels as well and recall that we have one. Kill a jewel for every 10 to the third jules 10 to the power of three jewels. So our jewels will cancel out. Then we multiply this by the temperature Which we determined to be 298.15 Kelvin to 98.15 Kelvin. And finally multiplied by the value of Q. Which in this case is or sorry, multiplied by the natural log of human. I'll write this below just so we have a little more space times the natural log of 1.9737 times 10 to the negative three atmospheres, which gives us a final value Of 67.0 kg jewels Permal. So that is the value of delta G. For the first scenario where we have a partial pressure of 1.5 mm hg for Boromir. Let's go ahead and take a look at the second scenario where it's 0.150 millimeters Hg. So we take a very similar approach. The only thing different here and let me make sure to write this in red this time. The only difference here is that the partial pressure is different. So Q. Is equal to the partial pressure bruning gas cast Which is equal to 0. mm h.g. Multiplied by one atmosphere for every 7 60 H. G. So this will cancel out and you will be left with a value of 1.9737 times 10 to the negative 4th atmospheres subbing all this back into our equation. Everything else will be the same except for the Q. Let's write this out is equal to and I'll just sort of speed through it. Since we just went through this earlier, we have the standard delta G. And we have our which we have to convert to kill a jules. Since it's given in jewels over mole kelvin into the conversion factor of one. Kill a jewel for every 10 to the third jewels, it'll cancel out. We have a temperature that we determined at 25°C, which is just 208.15 Kelvin multiplied by the natural log of Q. Which was determined to be 1. times 10 to the negative fourth A. T. M. Which gives a final value of 61.2 kg jewels Permal. So when the partial pressure of browning is 1.5 mm H. G. The delta G of the vaporization of grooming is 67 kg joules per mole. And when the partial pressure is 0.15 millimeters mercury, we get a delta G of 61.2 kg joules per mole. I hope this helps. And I look forward to seeing you all in the next one

Consider the evaporation of methanol at 25.0 °C : CH3OH(l ) ¡ CH3OH( g) b. Find ΔG r at 25.0 °C under the following nonstandard conditions: i. PCH3OH = 150.0 mmHg ii. PCH3OH = 100.0 mmHg iii. PCH3OH = 10.0 mmHg

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