Given the following processes at 37degreeC: A Left right arrow5 B Keq(AB) = 0.01

Question

Given the following processes at 37degreeC: A Left right arrow5 B Keq(AB) = 0.01 B Left right arrow C Keq(BC) = 1000 a. Determine Keq (AC), the equilibrium constant for the overall process ALeft right arrowC, from Keq(AB) and Keq (BC).(Clue: write the Keq constant expressions for all three processes. Can you express Keq (AC) in terms of the other two expressions?) b. Determine standard-state energy changes (triangleGdegree) for all three processes, and use triangleGdegree(AC) to determine Keq(AC). Make sure that this value agrees with that determined on part a, above.

Explanation / Answer

For A <==> B, Keq(AB) = 0.01;
For B <==> C, Keq(BC) = 1000, both at 37 o C

a)
Keq(AB) = [B]/[A];
Keq(BC) = [C]/[B]
Keq (AC) = [C]/[A] = [B]/[A] x [C]/[B] = Keq(AB) x Keq(BC) = 0.01 x 1000 = 10

b)
Go (AB) = -RT ln Keq(AB) = -(8.3314)(310) ln (0.01) = 11.89 kJ/mol
Go (BC) = -RT ln Keq(BC) = -(8.3314)(310) ln (1000) = -17.8 kJ/mol
Go (AC) = -RT ln Keq(AC) = -(8.3314)(310) ln (10) = -5.94 kJ/mol

Go (AC) = -RT ln Keq(AC)

Substuting the values We get Keq = 9.97 which is very close to 10 which we determined in part 1

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