A reaction A (aq) + B (aq) C (aq) has a standard free-energy change of -3.05 kJ/

Question

A reaction A (aq) + B (aq) C (aq) has a standard free-energy change of -3.05 kJ/mol at 25 degree 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? [A] = [B] = [C] = How would your answers above change if the reaction had a standard free-energy change of +3.05 kJ/mol? There would be no change to the answers. There would be more A and B but less C. There would be less A and B but more C. All concentrations would be higher. All concentrations would be lower.

Explanation / Answer

ln Keq = -(Delta g)/RT

Delta g = 3050 j/mol

R = 8.31j/k/mol

T = 298k

Solving keq = 3.42

Keq = [ C ] / [A][B]

3.42 = [C] /[A][B]

Use ice diagram to figured out for

[C] =x

[ A] =0.3-x

[B] =0.2-x

3.42= x/(0.3-x)(0.2-x)

3.42(0.06-0.5x+x^ 2)-x=0

3.42x^2 -2.71x+0.205 = 0

X= 0.0846

[C] = 0.0846

[A] =0.3-0.0846 =0.215

[B] =0.2 - 0.0846 =0.1154

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