A reaction a standard free-energy change of-5.01 kj/mol at 25 degreeC. What are

Question

A reaction a standard free-energy change of-5.01 kj/mol at 25 degreeC. 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? How would your answers above change if the reaction had a standard free-energy change of +5.01 kJ/mol? There would be less A and B but more C. There would be no change to the answers. There would be more A and B but less C. All concentrations would be higher. All concentrations would be lower.

Explanation / Answer

dGo = -RT ln K

-5.01 x 1000 J/mol = -8.314 x 298 ln K

K = 7.5545 = [C]/[A][B]

initial [A] = 0.3 , [B] = 0.4 M , [C] = 0

[A] = 0.3-X , [B] = 0.4-X , [C] = X

K = 7.5545 = ( X) / ( 0.3-X)(0.4-X)

7.5545 ( X^2 -0.7X + 0.12) = X

X^2 -0.83237 +0.12= 0

X = 0.1855

[A] = 0.3-0.1855 = 0.1145 , [B] = 0.2145 , [C] = 0.1855

if dG +5.01 KJ   reaction will be favoured back side

hence more A and B will be present and less C

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