A reaction A aq) Blaq has a standard free-energy change of -4.12 kJ/mol at 25 oC

Question

A reaction A aq) Blaq has a standard free-energy change of -4.12 kJ/mol at 25 oC 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 M M M How would your answers above change if the reaction had a standard free-energy change of +4.12 kJ/mol? There would be less A and B but more C. All concentrations would be higher. O There would be no change to the answers. All concentrations would be lower. O There would be more A and B but less C.

Explanation / Answer

A    +     B   <---------> C

0.3         0.4                   0(Initial conc)

0.3 - x     0.4 - x              x ( Final conc)

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

delGo = - RTln(K)

- 4.12 x 10^3 J/mol = - 8.314 J/mol-K * 298K * ln[( x / (0.3-x)(0.4-x)]

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

0.633 + 5.275 x^2 - 3.692 x = x

5.275 x^2 - 4.692 x + 0.633 = 0

x = 0.166

[A] = 0.3 - 0.166 = 0.134 M

[B] = 0.4 - 0.166 = 0.234 M

[C] = 0.166 M

If delGo = + 4.16 kJ/mol

K = 0.19

There would be more A and B but less C

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