A reaction has a standard energy change of -4.66 kJ/mol at 25 degrees C. eaction

Question

A reaction has a standard energy change of -4.66 kJ/mol at 25 degrees C.

eaction has a standard free-energy change of-4.66 kJ/mol at 25°C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? Number Number Number How would your answers above change if the reaction had a standard free-energy change of +4.66 kJ/mol? O o O All concentrations would be higher There would be more A and B but less C. There would be no change to the answers. All concentrations would be lower There would be less A and B but more C. O

Explanation / Answer

First we get the equilibrium constant from the free energy:

-4.66 kJ/mol = - 0.008314472 kJ/K mol * 298K * ln Keq

1.88 = ln Keq

Keq = e^1.88 = 6.5535

Then we make our ICE table

So the equilibrium expresion will be:

6.5535 = x/ [0.3-x][0.4-x]

x = 0.1778

So:

[A]eq = 0.3 - 0.1778 = 0.1222 M

[B]eq = 0.4 - 0.1778 = 0.2222 M

[C]eq = 0.1778 M

If the value of the free energy turns positive, there would be more A and B, but less C.

A(aq) + B(aq) <-> C(aq) I 0.3M 0.4M 0 C -x -x +x E 0.3-x 0.4-x x

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