Question

A reaction A ( aq ) + B ( aq ) − ⇀ ↽ − C ( aq ) has a standard free‑energy change of − 5.24 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 1

Answer:

$[A]_{eq}=0.11M$

$[B]_{eq}=0.21M$

$[C]_{eq}=0.19M$

Explanation:

Hello,

In this case, for the given reaction, based on the information about its Gibbs free energy, we obtain the equilibrium constant as shown below:

$Kc=exp(-\frac{\Delta _RG }{RT} )=exp[-\frac{-5240J/mol }{(8.314J/mol*K)(298.15K)} ]=8.28$

Now, by means of the law of mass action in terms of the undergoing change $x$ due to the chemical reaction, we obtain:

$Kc=\frac{x}{(0.30-x)(0.40-x)} =8.28$

For which the solution for $x$ by solver is:

$x=0.19M$

Thus, the equilibrium concentrations result:

$[A]_{eq}=0.3M-0.19M=0.11M$

$[B]_{eq}=0.4M-0.19M=0.21M$

$[C]_{eq}=0.19M$

Best regards.