A reaction laq has a standard free-energy change of -5.10 kJ/mol at 25 C. What a

Question

A reaction laq has a standard free-energy change of -5.10 kJ/mol at 25 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? Number Number Number M M How would your answers above change if the reaction had a standard free-energy change of +5.10 kJ/mol? O There would be less A and B but more C O All concentrations would be higher o There would be no change to the answers O There would be more A and B but less C All concentrations would be lower.

Explanation / Answer

For the given reaction,

A + B --> C

dGo = -RTlnKeq

with,

dGo = -5.10 kJ/mol

R = gas constant

T = 25 + 273 = 298 K

we get,

5100 = 8.314 x 298 lnKeq

Keq = 7.8

Now,

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

                  A     +        B      <======> C

I               0.3             0.4                       0

C              -x               -x                       +x

E            0.3-x          0.4-x                     x

So,

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

7.8 = x/(0.12 - 0.7x + x^2)

7.8x^2 - 6.5x + 0.94 = 0

x = 0.19 M

thus equilibrium concentrations for,

[A] = 0.3 - 0.19 = 0.11 M

[B] = 0.4 - 0.19 = 0.21 M

[C] = 0.19 M

When, dGo = +5.10 kJ/mol

Keq = 0.13

therefore,

there would be more A and B but less C

--------------------------------------

Using the data given in the Table

dSvap = dHvap/Tb

           = 77.1 x 1000/1033

           = 74.64 J/K.mol

--------------------------------

for the buffer solution

pKa = -logKa

       = -log(3.4 x 10^-5) = 4.47

Using Hendersen-Hasselbalck equation,

let x amount of base is added

pH = pKa + log(base/acid)

4.13 = 4.47 + log(x/(0.2 M x 0.525 L - x))

0.05 - 0.46x = x

x = 0.05/1.46 = 0.034

Molar concentration of base present = 0.034 mol/0.525 L = 0.065

moles of acid present in buffer = 0.2 M x 0.525 L - 0.034 = 0.071 mol

when reaction is complete,

ratio (conjugate base/acid) = (0.034/0.071) = 0.48

moles of strong base initially added = x = 0.034 mol OH-

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