Question

A reaction: A (aq) + B (aq) <-----> C (aq) has a standard free energy change of -3.05 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?

A = ? M

B= ? M

C= ? M

How would your answer above change if the reaction had a standard free energy change of +3.05 kJ/mol?

A.) There would be no change to the answers.

B.) All concentrations would be higher.

C.) All concentrations would be lower.

D.) There would be more A and B, but less C.

E.) There would be less A and B, but more C.

Answer 1

Answer – In this question there are given, standard free energy change ∆G^o = -3.05 kJ

Reaction - A(aq) + B (aq) <-----> C (aq)

[A] = 0.3 M , [B]= 0.40 M, T = 25+273 = 298 K

First we need to calculate equilibrium constant from the standard free energy change

We know formula

∆G^o = -RTlnK

-3.05*10³ J = - 8.314 J/mol.K * 298 K * ln K

-3.05*10³ J / - 8.314 J/mol.K * 298 K = ln K

So, ln K = 1.23

Taking antiln from both side

K = 3.43

Now we need to write the ICE table –

   A(aq) + B (aq) <-----> C (aq)

I 0.30       0.40               0.0

C -x           -x                   +x

E 0.30-x 0.40-x               +x

We know

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

3.43 = x / (0.30-x)(0.40-x)

3.43 [(0.30-x)(0.40-x)] = x

3.43x² -2.401x+0.412 = x

3.43x²-3.401x+0.412 = 0

So using the quadratic equation

x = 0.162 M

so, at equilibrium

[A] = 0.30-x = 0.30-0.162 = 0.138 M

[B]= 0.40-x = 0.40-0.162 = 0.238 M

[C] = x = 0.162 M

If the reaction had a standard free energy change of +3.05 kJ/mol then equilibrium constant gets decreased and hence there is product formed less and reactant remaining more

So answer is - D.) There would be more A and B, but less C.