4.6 Application of rates and equilibrium
Posted by: intechemistry on: September 23, 2010
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5 Responses to "4.6 Application of rates and equilibrium"
November 4, 2010 at 2:26 pm
deltaS(tot) = deltaS(sys) – (deltaH)/T
For an endo reaction, deltaH is +’ve, and as the T increases, then (deltaH)/T becomes smaller, deltaS(tot) = deltaS(sys) minus deltaH/T <– the now smaller number, so deltaS(tot) gets larger, i.e. more positive
Note deltaS(tot) may overall be positive or negative depending on deltaS(sys), but the point is, deltaS(tot) is more positive than it was before, hence the reaction becomes more feasible.
Dream up some numbers for deltaS(sys), deltaH and some T values and see for yourself what happens.
“if the surrounding are hot, the entropy increase is small” Is supposed to be a learning aid. If it doesn’t work for you, then abandon it, but it’s just a summary of the above. If T is large, deltaH/T will be small so the contribution to deltaS(tot) from deltaH/T is small. Note: I’m talking about the change /i> in entropy, i.e. relative entropy, not absolute entropy… As T goes up total entropy does indeed increase, but the change of entropy for the change in temp is relatively/comparatively small.
February 9, 2012 at 11:48 pm
salam sir..
in GF A2(pg76),it says “in a reversible reaction, an increase in temperature incerase the rate of the forward and back reactions,but does not do so equally. The endothermic reaction has higher activation energy and so its rate increased more than that of the exothermic reaction.This is one reason why the position of equilibrium shifts in the endothermic direction when the temperature of the system is raised”.
My questions regarding to these statements:
1) why endo has higher activation energy than exo(i mean what is the evidence)
2)how this reason(reason given in 1st question) cause equilibrium shifts to endo direction when temperature increase.(as i know, equlibrium shift to endo direction as temperature increase in order to absorb the heat release to surrounding.Am i right sir?)
February 9, 2012 at 11:56 pm
Salam 
Glad to see you’re really going through the book
Humm…. gotta think about this one. The occasional student asks this question and I can never remember what I said about it 
begins thinking…
February 10, 2012 at 12:03 am
I have a way of explaining it. Are you free tomorrow during the 9-50 break? I’ll be at block W then. Either way, SMS or call me to let me know.
(ask your class rep for my number if necessary)
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November 1, 2010 at 9:33 am
salam sir.
why, for an endothermic reaction with a higher entropy, it is only spontaneous at high temperature?
i dont really understand the concept – “if the surrounding are hot, the entropy increase is small”