A reaction A(aq) + B(aq) rightwardsharpoonoverleftwardsharpoon C(aq) has a stand

Question

A reaction A(aq) + B(aq) rightwardsharpoonoverleftwardsharpoon C(aq) has a standard free-energy change of -4.43 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 +4.43 kJ/mol? All concentrations would be lower. 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.

Explanation / Answer

dG = -4.43 kJ/mol

find A,B,C

given

[A] = 0.3

[B] = 0.4

[C] = 0

in equilirium

[A] = 0.3 - x

[B] = 0.4 - x

[C] = 0 + x

Find K:

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(4430/(8.314*298)) = 5.9777

substitute

K = [C]/([A][B])

5.9777 = (x)/((0.3 - x)(0.4 - x))

folve for x

5.9777 (0.12 - 0.7x + x^2) = x

0.1672x =  (0.12 - 0.7x + x^2)

x^2 - (0.8672)x + 0.12 = 0

x = 0.1728

[A] = 0.3 - 0.1728 = 0.1272

[B] = 0.4 - 0.1728 = 0.2272

[C] = 0 + 0.1728 = 0.1728

Q2.

if dG = 4.43, then this will favour REACTANTS

then

more A,B and less C

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