Trouton’s rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about 88 J/mol‐K. b. Look up the normal boiling point of Br₂ in a chemistry handbook or at the WebElements website (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

Video transcript

hey everyone, we're told that trodden rule states that the standard molar entropy of vaporization for many liquids at their normal boiling point is about 88 joules per mole times kelvin. What is the normal boiling point of iodine? Assuming that trout and rule applies that it's standard entropy of vaporization remains constant with temperature. So here we have the vaporization of iodine. Writing out this reaction. We have iodine in its liquid state and this will vaporize and turn into its gaseous state. As we've learned, the standard entropy of vaporization is going to be equal to our products minus our reactant. So in this case we have our standard NLP of our iodine in its gaseous state minus the standard entropy of iodine in its liquid state. Now plugging in these values we have 62.4 kg joules per mole minus 13.5 kg jewels Permal which was provided to us in our question stem this will get us to a total of 48.9 kg joules per mole. Now to calculate our boiling point, we've learned that the formula is going to be the standard entropy of vaporization divided by the standard entropy of vaporization plugging in these values we have 48.9 kg joules per mole and we want to convert this into jewels so we know that we have 1000 jewels per one kg jule. Next we're going to divide this all By our standard molar entropy of vaporization which was 88 joules per mole kelvin and when we calculate this out and cancel out all of our units, we end up with a total of 5.6 times 10 squared Calvin, which is going to be our final answer. Now, I hope that made sense and let us know if you have any questions.