A reaction A_ (aq) + B(aq) C(aq) has a standard free-energy change of -300 kJ/mo

Question

A reaction A_ (aq) + B(aq) C(aq) has a standard free-energy change of -300 kJ/mol at 25 degree C. What are the concentration of A, B, and C at equilibrium if, at the beginning of the reaction, their concentration are 0.30 M, 0.40 M, and 0 M, respectively? How would your answers above change it the reaction had a standard free-energy change of +3.00 KJ/mol? All concentrations would be lower. There would be less A and B but more C. All concentration would be higher, There would be mare A and B but less C. There would be no change to the answers.

Explanation / Answer

a) G = -RTlnK

lnK = -G /RT

logK = -(-3000J/mol)/(8.314(J/mol K)×298.15K×2.303)

K = 3.36

3.36 = X/(0.3 - X)(0.4-X)

= X/(0.12-0.3X - 0.4X +X^2)

3.36X^2 + 0.4032 - 1.008X - 1.344X - X = 0

3.36X^2 - 3.352X + 0.4032 =0

X^2 - 0.9976X + 0.12=0

X = 0.1399

[A] = 0.30-0.1399= 0.16M

[B] = 0.40 - 0.1399 = 0.26M

[C] = 0.14M

b) When free energy is -3.00KJ/mol

logK = - (3000J/mol K)/8.3 14(J/mol K) × 298.15K × 2.303

= -0.526

K = 1×10^(-0.526)

=0.298

Equilibrium constant is decreased, So

There would be more A and B but less C

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