Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a ...

Question

Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a 
normal boiling point of 46.3 °C. Find ΔHvap for carbon disulfide.

Accepted Answer

everyone in this example, we need to calculate the heat of vaporization for carbon tetrachloride. If its vapor pressure at 23 degrees Celsius is 100 Tour and it has a normal boiling point of 76.7 degrees Celsius. So we're going to recall the following formula where we would take the natural log of the quotient where we would have our second pressure or final pressure divided by our initial pressure. We're setting this equal to negative one times our heat of vaporization, which is then going to be divided by r gas constant R. And then we want to go ahead and multiply this by our Inverse of our final temperature. T 2 - the inverse of our initial temperature T one. So given from the info from the prompt for our initial pressure P one. We would say That that should be our standard value of tour. And then according to the prompt, our final pressure is 100 tour. We're also given our final temperature from the prompt as 23°C. And sorry, let's fix this. So 23°C And we want to go ahead and convert this to Kelvin. So we're going to add to 73.15. And that's going to give us a kelvin temperature of 96.15 Kelvin as our final temperature for our initial temperature. We're going to assume standard temperature Or rather we're going to take that boiling point. Sorry. So that's given us 76.7°C and we're going to add to 73.15 to convert this to Kelvin. So that we have its initial temperature being 349. Kelvin. So because we want to isolate for our heat of vaporization, we're going to say that our heat of vaporization is equal to the negative natural log of Our final pressure, P two divided by our initial pressure, multiplied by a gas constant R. And then we're going to divide this by our inverse final temperature subtracted from our inverse initial temperature. So plugging in everything that we know what we should have is the negative L. N of P two which we again above stated is tour. However, we're converting this to Or sorry, I was thinking of temperature. So we would plug this in as 100 tour And then in our denominator we have our initial pressure which we said is going to be 760 tours, which is our standard pressure. And then we're going to recall our gas constant. R. So recall that our gas constant R is 8.314, sorry, So 8.314 units of jewels divided by moles, times kelvin. And then now we have our denominator where we have our inverse final temperature which we converted to Kelvin as to 96.15 Kelvin Subtracted from our inverse initial temperature which we converted to 349.85 Kelvin. So our next step is to cancel out our unit. So let's get rid of tours. We're going to also cancel out our units of kelvin. Leaving us with leaving us with units of joules per mole. Now we still have these brackets and our denominators. So we're going to start by taking that difference in our denominator. So starting off at the bottom here, we're going to now have In our numerator we still have the negative Ln of now. 100 tour Over 7 60 tour. So we're going to keep that part in brackets. We're going to have the gas constant are still there and then in our denominator. And sorry, we don't actually need to carry those tour units because we've canceled them out In our denominator. Since we're going to do that part first, this difference between these two fractions should give us a value equal to 0.000519. And so what we want to do next is take the L. N. In our numerator. And so what we would have is so far two .02815. And this is going to be multiplied by r gas constant 8. joules per mole. And then we're dividing by our numerator, which was that difference which we said was equal to 0.000519. And so now this should be much easier to calculate. So we can get our heat of vaporization equal to a value of 3, 2489.5 jules Permal. And then we want to go ahead and convert from jules per mole, two kg, joules per mole. So we're going to multiply by the conversion factor, which tells us that we have 10 to the third power jewels for one kg jule, because our prefix kilo tells us we have 10 to 3. And so now we're able to cancel out our units of jewels, leaving us with kayla jewels per mole as our final unit for the heat of vaporization, which is the proper unit. And this will give us a value of sorry, 32 0.5 kg joules per mole. And so to complete this example as our final answer, we found our heat of vaporization of 32.5 kg joules per mole for our carbon tetrachloride. So what's highlighted in yellow is our final answer? I hope that everything I went over was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

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