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

`[A]_(eq)=0.11M`

`[B]_(eq)=0.21M`

`[C]_(eq)=0.19M`

Step-by-step 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(-(\Delta _RG )/(RT) )=exp[-(-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=(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.