A reaction Alag)+ Blag) cla has a standard free-energy change of -3.77 kJ/mol at

Question

A reaction Alag)+ Blag) cla has a standard free-energy change of -3.77 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 How would your answers above change if the reaction had a standard free-energy change of +3.77 kJ/mol O All concentrations would be lower. 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. O There would be less A and B but more C

Explanation / Answer

1)

A)

T= 25.0 oC

= (25.0+273) K

= 298 K

?Go = -3.77 KJ/mol

?Go = -3770 J/mol

use:

?Go = -R*T*ln Kc

-3770 = - 8.314*298.0* ln(Kc)

ln Kc = 1.5217

Kc = 4.58

ICE Table:

[A] [B] [C]

initial 0.3 0.4 0

change -1x -1x +1x

equilibrium 0.3-1x 0.4-1x +1x

Equilibrium constant expression is

Kc = [C]/[A]*[B]

4.58 = (1*x)/((0.3-1*x)(0.4-1*x))

4.58 = (1*x)/(0.12-0.7*x + 1*x^2)

0.5496-3.206*x + 4.58*x^2 = 1*x

0.5496-4.206*x + 4.58*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 4.58

b = -4.206

c = 0.5496

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 7.622

roots are :

x = 0.7606 and x = 0.1578

x can't be 0.7606 as this will make the concentration negative.so,

x = 0.1578

At equilibrium:

[A] = 0.3-1x = 0.3-1* 0.1578 = 0.142 M

[B] = 0.4-1x = 0.4-1* 0.1578 = 0.242 M

[C] = +1x = +1* 0.1578 = 0.158 M

Answer:

[A] = 0.142 M

[B] = 0.242 M

[C] = 0.158 M

B)

if delta Go is positive, the reaction will not be spontaneous

And hence less amount of C will be there at equilibrium

There would be more amount of reactant

Answer: option 4

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