Question

A reaction A(aq) + B(aq) = C(aq) has a standard free-energy change of -3.06 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? How would your answers change if the reaction had a standard free-energy change of +3.06 kJ/mol? O There would be less A and B but more C. All concentrations would be higher. There would be no change to the answers. All concentrations would be lower. There would be more A and B but less C.

Answer 1

Answer – We are given,

Reaction with standard free energy –

A(aq) + B(aq) + C (aq) AG° = -3.06– mol

The initial concentration

[A] = 0.30 M , [B] = 0.40 M, [C] = 0.0 M , T = 25+273.15 = 298.15 K

First, we need to calculate the equilibrium constant from the given standard free energy and we know the relationship between these two is as follows:

AG° = -RTINK

We need the standard free energy in J, hence the conversion kJ to J is as follows

1 kJ = 1000 J

-3.06 kJ = -3060 J

By plugging the values in the above formula

-3060 ) = -8.314 x 298.15 K x In K

In K= 7 -3060) -8.314mmo x 298.15 K = 1.23

Now antiln from both side

K = 3.44

Now we need to put the ICE chat

	A	B	C
I	0.30	0.40	0.0
C	-x	-x	+x
E	0.30-x	0.40-x	+x

Now equilibrium expression


K [A] [B]


3.44 = to 30 – x)(0.40 - x)


3.44 (00.30 - x)(0.40 – x)] = x


3.44x2 - 2.408x + 0.413 = x

3.44x2 – 3.408x + 0.413 = 0

By solving the quadratic equation, the value of x is 0.141

Hence the concentration of A, B, and C are as follows :

[A] = 0.30-x

      = 0.30 – 0.141

       = 0.159 M

[B] = 0.40-x

      = 0.40 – 0.141

       = 0.259 M

[C] = x = 0.141 M

When we take the standard free energy is positive then the value of the equilibrium constant is less than 1 and the reaction is favored toward the reactant side means the left side.

As the reaction favored toward the reactant side then the concentration of reactant A and B are more than product A at equilibrium.

The correct option is – There would be more A and B but less C.