Question

What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.30 M, 0.40 M, 0.40 M, and 0 M, 0 M, respectively?

Answer 1

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 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?

Answer: The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

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 = -4.88 kJ/mol = -4880 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 of the reaction

Putting values in above equation, we get:

$-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17$

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

                               $A+B\rightleftharpoons C$

Initial:                 0.30  0.40      0

At eqllm:         0.30-x   0.40-x    x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[C]}{[A][B]}$

We are given:

$K_c=7.17$

Putting values in above equation, we get:

$7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657$

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = $(0.30-x)=(0.30-0.183)=0.117M$

Equilibrium concentration of B = $(0.40-x)=(0.40-0.183)=0.217M$

Equilibrium concentration of C = $x=0.183M$

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.