A researcher is studying the effects of ibuprofen on a tissue sample. Ibuprofen
ID: 177017 • Letter: A
Question
A researcher is studying the effects of ibuprofen on a tissue sample. Ibuprofen in a non-steroidal drug used for relieving pain that helps with fever and reducing inflammation. It has two natural optical isomers, S and R, of which the S isomer is more biologically active. To establish a cause-effect relationship, the researcher needs to know how much of the S isomer she has in her sample. She is able to determine that under equilibrium her sample has 1 M of R. She also knows that under standard conditions the free energy change associated with the S-->R isomerization reaction is -1.7kcal/mol. What is the concentration of S in her sample? Assume T=300K.
--> a) 59mM
b) 17M
c) 2mM
d) 17 mM
The correct answer is A, but I really need a thorough expalanation as to how this is.
Explanation / Answer
We know that for the reaction, SR; Keq = [R]/[S]
or [S] = [R]/Keq
Also, we know that, Keq = e-G°/RT (please note that in this equation, R denotes the universal gas constant, 0.008314 kJ mol-1 K-1, and is not the same as R in the above equation)
Given that, G° = -1.7 Kcal/mol = -7.1128 KJ/mol; T = 300 K
Therefore, Keq = 17.3178
Also given that [R], i.e. concentration of R isomer = 1 M at equilibrium.
Therefore, [S] = 1/17.3178 = 0.0577 M = 57.7 mM
The concentration of S isomer is 57.7 mM.
(It is closer to 58 mM, NOT 59 mM! please check the question at source or with your instructor)