Answer:
[tex][A]_{eq}=0.11M[/tex]
[tex][B]_{eq}=0.21M[/tex]
[tex][C]_{eq}=0.19M[/tex]
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:
[tex]Kc=exp(-\frac{\Delta _RG }{RT} )=exp[-\frac{-5240J/mol }{(8.314J/mol*K)(298.15K)} ]=8.28[/tex]
Now, by means of the law of mass action in terms of the undergoing change [tex]x[/tex] due to the chemical reaction, we obtain:
[tex]Kc=\frac{x}{(0.30-x)(0.40-x)} =8.28[/tex]
For which the solution for [tex]x[/tex] by solver is:
[tex]x=0.19M[/tex]
Thus, the equilibrium concentrations result:
[tex][A]_{eq}=0.3M-0.19M=0.11M[/tex]
[tex][B]_{eq}=0.4M-0.19M=0.21M[/tex]
[tex][C]_{eq}=0.19M[/tex]
Best regards.
