Answer:
The value for the equilibrium constant for the reaction [tex]2SO2(g)+O2(g) <=> 2SO3(g) Kc=4.0*10^24\\[/tex] at 298K .
What would be the value for the equilibrium constant for the following reaction at the same temperature?
[tex]2SO3(g) <=> 2SO2(g)+O2(g)[/tex]
Explanation:
The Kc value for the reverse reaction of the first reaction that is:
[tex]2SO2(g)+O2(g) <=> 2SO3(g) Kc=4.0*10^{24}\\[/tex]
[tex]Kc=\frac{SO3^{2} }{SO2^{2}.O2 }[/tex]
For the reverse reaction,
[tex]2SO3(g) <=> 2SO2(g)+O2(g)[/tex]
[tex]Kc=\frac{SO2^{2}O2 }{SO3^{2}. }[/tex]
So, it is the inverse of the first Kc value.
Hence, the new Kc value is:
[tex]Kc=\frac{1}{4.0*10^{24} } \\ =2.5*10^{-25}[/tex]
Answer is :
Kc=2.5*10^-25.
