Van't Hoff Plots

[@ItSamiraclelol]

This clears things up so much for me that my chem prof. couldn't explain!! Thank you

[@Sillypiee_Epsilonn]

Love from Bangladesh 🇧🇩

[@Nicole-cm6tc]

Thank you so much! SUPER HELPFUL video before my exam!

[@evaburghardt98]

This is such a useful video! Thanks so much for sharing and clearing this up!!

[@rking3808]

what would be the units of the delta H naught and delta S naught if we solve it?

[@siameexiong1046]

thank you so much! Very well explained!!

[@OhxMyxBarnacle]

How do we know what to put as our delta H naught and delta S naught in the linear equation? I'm confused as to how to find the values of slope and y intercept!

[@Chemistry123]

+Brit Mal From a research point of view, you would normally run your reaction and measure the equilibrium constant, K, at various different temperatures, T. You would then make a plot of ln(K) vs 1/T, and find the straight line that best fits your data (using software). From a student perspective, this "best fit line" for the data is often given to you in an exam-style question. So, all you need to do is to recognize how the slope of the given linear equation relates to the enthalpy and how the intercept relates to the entropy.

[@Roughen1991]

This is really good! Very logical and easy to follow!

[@mr_stardust0.032]

That was a really good explanation 🙌 :)

[@hoxis_dk]

thank you for this video, really good explanation!

[@catherineroloff3923]

What would be the final units?

[@Chemistry123]

The y values are unitless and the x values are in units of 1/K, so the units of the slope (y/x) are K. Thus, for ΔH° if we multiply the slope (in K) by R (in J/molK), the final units would be J/mol. The y-intercept has the same units as the y axis, so is unitless. For ΔS°, then, if we multiply the y-intercept (unitless) by R (in J/molK), the final units are J/(mol K).

[@Mohammad12765]

Very nice

[@guillermojarne2803]

Very well explained! Thank you :)

[@SirTrollingham]

Great video, much appreciated.

[@danielraygoza3933]

Thank you #finals

[@stephen6713]

This is gonna sound dumb I'm sure but I just don't get how when you divide both sides of the equation by RT (to cancel out the RT in with lnK on the left) do you have to divide the RHS by it two times? Wouldn't the right hand side just be (-dH+dS)/R?

[@Chemistry123]

You're absolutely correct that you just divide by -RT once. You divide the right side by -RT so that it's -(dH-TdS)/RT (you've just forgotten the T in your version). What I've done in the video is separate out the fraction so that it's -dH/RT + TdS/RT and then the T cancels on the top and bottom of the second fraction to give -dH/RT + dS/R.

[@reemmadani7932]

was so helpful thanks a lot :)

[@laurenfagan6072]

soooooooooo helpful THANK U!

[@petergriffin742]

Awesome explaining. heheheheheh

[@jacobushenricusvanthoff7635]

noice

[@manjaysharma6136]

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