Suppose for the reaction in the \"previous problem\" that [A] in the cell were 1
ID: 844057 • Letter: S
Question
Suppose for the reaction in the "previous problem" that [A] in the cell were 1*10-8 M and [B] were 5*10-4 M.
a) In which direction would the reaction proceed spontaneously?
b) What would be the final concentration of [A] and [B] at equilibrium?
I started with
[B]/[A] = (1*10-8) / (5*10-4) = 5*104
[A]+[B] = 5*104
but I don't know how to move on to the next step.
Please explain how you came up with the answers.
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***Previous problem
For the reaction A->B, delta G0'=-7.1kcal/mol. what is the equilibrium ratio of B/A at 37'C? (R=1.987cal/K-mol)
so I did
delta G0'= -RT*ln(B/A)
-7.1 kcal/mol =1.987 * 10-3 kcal/mol * 310K * ln(B/A)
ln(B/A) = 11.5
B/A = 101370
Explanation / Answer
(a) Initially A=1*10-8 M and B=5*10-4 M
Reaction quotient, Q = B/A = 50000
Now, as you have already calculated,
At equilibrium,
Equilibrium constant, K = B/A = 101370
Since, K>Q the reaction will proceed in the forward direction.
(b) As you have already calculated,
At equilibrium, B/A = 101370 ....... Equation (1)
And initially A=1*10-8 M and B=5*10-4 M
Hence, A+B=5.0001*10-4 M ........Equation (2)
A+B remains constant throughout the reaction.
Solving equations (1) and (2),
A=4.93*10-9 M
B=5*10-4 M