LO 6.25….AGAIN!

March 02, 2017

Having looked at the “thermal energy” nonsense in LO 6.25 for around 5 years, and having written this blog post in Feb 2016, and this blog post in Feb 2015 (you should read them both), I still feel lost when it comes how TE might ultimately be assessed. With only one MCQ ever written about this LO across ALL of the seven new exams that have been released by the CB in part or in whole, AND with that question NOT addressing anything regarding thermal energy (rather only asking if you know that negative ∆G°’s go with positive Ecell‘s and K’s greater than 1, and yes I already knew that on the old course), I decided to attempt to write a question that might (somehow) address the LO in terms of thermal energy, and what the CB/TDC might be getting at. Shooting in the dark here, feedback encouraged.


(a) At 298 K, a chemical reaction is found to have an equilibrium constant equal to 1.2 x 109.

(i) Calculate a value for ∆G° for the reaction at 298 K.

(ii) What do the value of the equilibrium constant, and the value of ∆G°, suggest about the reaction in terms of relative amounts of products and reactants present in the reaction mixture at equilibrium? Explain your answer.

(b) An entirely different chemical reaction to the one in (a) but also at 298 K, has ∆G° with a magnitude that is almost identical to the magnitude of thermal energy. What does such a magnitude for ∆G° suggest about the reaction in terms of,

(i) The numerical value of ∆G°.

(ii) The approximate value of K.

(iii) The approximate, relative number of reactants and products present at equilibrium.



(i) ∆G° = – R T ln K = – (8.314/1000)(298) ln K

∆G° = -52 kJmolrxn-1

(ii) A very large preponderance of products in the equilibrium mixture, since K >> 1, and ∆G° = a negative value, suggesting a thermodynamically favored forward reaction.


(i) ∆G° would range between +2.4 and -2.4 kJmolrxn-1 (or close to zero).

(ii) K would be approximately 1.

(iii) They are approximately the same.


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