Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: gramentz on April 29, 2009, 10:27:37 AM
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A process has a standard enthalpy change of 1.98657 kJ/mol and a standard entropy change of 439.29385 J/mol K. Estimate the natural logarithm (ln(K)) of the equilibrium constant at 287.58K. Use R= 8.3144 J/mol K.
I think I have the proper equation to use? :delta: G= -RT(lnK). This means that I have to find :delta: G first, right? Then set that number equal to the natural log of K? I obtained a very large value for :delta: G and am not able to find ln K. I've been searching for how to do this one for a long time. Appreciate any advice/ help.
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It shouldn't be that large, can you show your working?
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This is how I was trying it:
:delta: G= :delta: H - T :delta: S
:delta: G= 1.98657-287.58(439.29385)
:delta: G= -124.345
:delta: G= -RT(lnK)
-124.345= - (8.3144)(287.58)(lnK)
~.052006=lnK?
final answer being 1.0533828417631?
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This is how I was trying it:
:delta: G= :delta: H - T :delta: S
:delta: G= 1.98657-287.58(439.29385)
:delta: G= -124.345
:delta: G= -RT(lnK)
-124.345= - (8.3144)(287.58)(lnK)
~.052006=lnK?
final answer being 1.0533828417631?
Check your units! If I had a distance of 1.7 km to travel, and have already gone 43 m, how much further do I need to go?
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This is how I was trying it:
:delta: G= :delta: H - T :delta: S
:delta: G= 1.98657-287.58(439.29385)
:delta: G= -124.345
:delta: G= -RT(lnK)
-124.345= - (8.3144)(287.58)(lnK)
~.052006=lnK?
final answer being 1.0533828417631?
I see your error. The units on :delta: G is in kJ, but on R it is in J. Correct that and your answer should look a lot better.
Also a slight error on R. The value out to that many decimals is 8.3145 not 8.3144.
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The only reason why I'm using the value of R given is because it says to in the instructions. The error tolerance for the answer can be +/- 1 percent.
-124.345= -(.0083144)(287.58)(lnK)
-124.345= -2.391055152 (lnK)
52.00423= lnK
= 3.8472763413229E+22