Carbon disulfide 1CS22 is a toxic, highly flammable substance. Th...

Question

Carbon disulfide 1CS22 is a toxic, highly flammable substance. The following thermodynamic data are available for CS21l2 and CS21g2 at 298 K: 
<TABLE> 
(e) Use the data in the table to calculate ΔS° at 298 K for the vaporization of CS21l2. Is the sign of ΔS° as you would expect for a vaporization?

Accepted Answer

Hello everyone. So in this video we're dealing with the vaporization of our phosphorus trike Lord were given a kind of a two part question. First one is to to determine the change of entropy. The second question is to see if the sign of our change of entropy is expected for the vaporization. So first things first, the vaporization of phosphorus trike Lord, let's go ahead and write out the equation for that. So of course um that's going to be the molecule in its liquid state, going to the molecule of its gaseous phase. So now dealing with our delta age of vaporization, that's equal to the date the delta H of our products minus the delta H of our reactant. And where we're going to find these values is going to be given in this problem this table right over here. Alright, so let's go ahead and get started with this calculations and putting its numerical values in for the delta H of vaporization. So we have one mole of our product and then the delta H corresponding to that would be negative 2 87. kila jewels per mole. And then for the delta age of my reactant again we have one mole of my reactant and the corresponding value for delta H is negative 19.7 killer jewels per mole. Now putting that into my calculator, I get that my delta age of vaporization is 32.7 killer jules. Alright, I'm just gonna go ahead and actually scroll down a little bit to give us a little bit more space. We're gonna go ahead and solve now for our delta G of vaporization. So again, that is equal to delta G. Of my products minus the delta G of my reactant. Again, we're gonna find those values. Give it to us in this sort of table right over here on top. So again, the delta G of vaporization is equal to Well, we have one mole of the products. The corresponding delta G value is negative 267.8 units being killed, jules per mole. That's being subtracted by it. Well, again, we have one mole of our reactant corresponding delta G value is negative 2 72.3 and it's being killed, joules per mole. We see here in this equation that the moles will cancel out, leaving us with the final answer being in units of just killing jewels. So putting the numerical values into my calculator, I get a value of 4.5 kg jewels. Now, again, it's just scroll down to give us more space here. So we see now we can finally go ahead and use the gibbs Free energy equation for vaporization. So delta G of vaporization is equal to delta H. Of vaporization minus temperature, multiplied by the delta s of vaporization. We're trying to solve for r delta s of vaporization here, let's go ahead and do some mathematical manipulation to isolate that once you do so we should get that the delta S of vaporization is equal to delta H of vaporization subtracted from the delta G. Of vaporization all over our temperature T. So now putting in my numerical values for this, we have 32.7 kg Joel's subtracted with 4.5 kg Joel's. Our temperature is 2 98 kelvin's. So now putting that into my calculator, I get that D delta us vaporization is 0.9463 killer jewels per kelvin. We want to convert the killer jewels unit into jewels. So do a direct conversion here. We have 1000 jewels for everyone. Kill a jewel. You can see here that the units of kilograms will counsel Leaving us with a final answer of 9 4.6 units being jewels per Kelvin. So, that's going to be my first answer for this problem. And the next, of course, is to see if the side of delta S. Is what we expected. So, let's go ahead and see again scrolling down. All right, So here we're going from a liquid into a gas. So, we know that of course gas occupies more volume, then liquid does because there's more emotional freedom. There's going to be a greater disorder of a system giving us a greater entropy than liquid. And if there's a greater entropy, of course our delta S of vaporization will be indeed positive. So, it is what we expected. But to put this in writing will say delta S of vaporization is going to be always positive. So, the answer is expected. Alright, and that's going to be my second answer for this problem. Thank you all so much for watching.

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