For all of the problems below, here is the reaction: A <=> B. Assume in each cas

Question

For all of the problems below, here is the reaction: A <=> B. Assume in each case that the temperature is 300 K.

1. If the Delta G0' for this reaction is - 10 kJ/mol, and there is twice as much B as A, will the reaction go forward as written?

2. If the Delta G0' for this reaction is + 10 kJ/mol, what is the ratio of B/A that will make the reaction at equilibrim? How does one tip the reaction in the direction of backwards?

3. At equilibrium, there is 5 times as much B as A. What is the G0' for this reaction?

4. If the Delta G0' for this reaction is + 5 kJ/mol and the concentration of B is 0.3M, what is the minimum concentration of A required to make the reaction favorable?

5. If the Delta G0' for this reaction is + 5 kJ/mol, what is the ratio of B to A at equilibrium?

6. If the Delta G for the reaction is 8 kJ/mol when there is 0.5M A and 0.2M B, what is the equilibrium concentration of A and B?

Explanation / Answer

1) Reaction quotient Kq = [B]/[A] = 2

Equilibrium constant, Keq = e^{-delta G0/(R*T)} = 55.11

Since , Kq < Keq, therefore reaction will proceed in the forward direction

2) delta G0 = -R*TlnKeq

or, 10000 = -8.314*303*ln{[B]/[A]}

or, [B]/[A] = 0.018

3) delta G0 = -R*TlnKeq = -8.314*300*ln(5) = -4014.3 J/mole = -4.0143 kJ/mole

4) delta G0 = -R*TlnKeq

or, 5000 = -8.314*300*ln(0.3/[A]}

or, [A] = 2.227 M

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