Question

A reaction A(aq) + B(aq) ----> C(aq) has a standard free‑energy change of − 3.87 kJ / mol at 25 °C.

1. 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?

Answer 1

Answer: The concentration of A, B and C is 0.14 M, 0.24 M and 0.16 M respectively.

Step-by-step explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

`\Delta G^o=-RT\ln K_c`

where,

`\Delta G^o` = standard Gibbs free energy = -3.87 kJ/mol = -3870 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant =
`8.314J/K mol`

T = temperature =
`25^oC=[273+25]K=298K`

`K_c` = equilibrium constant in terms of concentration

Putting values in above equation, we get:

`-3870J/mol=-(8.314J/Kmol)* 298K* \ln K_c\\\\K_c=e^(1.562)=4.77`

The given chemical reaction follows:

`A(aq.)+B(aq.)\rightleftharpoons C(aq.)`

Initial: 0.30 0.40 0

At eqllm: 0.30-x 0.40-x x

The expression of
`K_c` for above equation follows:

`K_c=([C])/([A]* [B])`

We are given:

`K_c=4.77`

`[A]=0.30-x`

`[B]=0.40-x`

`[C]=x`

Putting values in above expression, we get:

`4.77=(x)/((0.30-x)* (0.40-x))\\\\x=0.16,0.75`

Neglecting the value of x = 0.75 because the equilibrium concentration cannot be greater than initial concentration.

So, concentration of A = (0.30 - x) = (0.30 - 0.16) = 0.14 M

Concentration of B = (0.40 - x) = (0.40 - 0.16) = 0.24 M

Concentration of C = x = 0.16 M

Hence, the concentration of A, B and C is 0.14 M, 0.24 M and 0.16 M respectively.