Question

The value for the equilibrium constant for the reaction 2SO2(g)+O2(g) rightleftharpoons 2SO3(g) is 4.0*10^ 1 24 at 298K What would be the value for the equilibrium constant for the following reaction at the same temperature? 2SO3(g) rightleftharpoons2502(g)+O2(g)

Answer 1

Answer:

The value for the equilibrium constant for the reaction $2SO2(g)+O2(g) 2SO3(g) Kc=4.0*10^24$ at 298K .

What would be the value for the equilibrium constant for the following reaction at the same temperature?

$2SO3(g) 2SO2(g)+O2(g)$

Explanation:

The Kc value for the reverse reaction of the first reaction that is:

$2SO2(g)+O2(g) 2SO3(g) Kc=4.0*10^{24}$

$Kc=\frac{SO3^{2} }{SO2^{2}.O2 }$

For the reverse reaction,

$2SO3(g) 2SO2(g)+O2(g)$

$Kc=\frac{SO2^{2}O2 }{SO3^{2}. }$

So, it is the inverse of the first Kc value.

Hence, the new Kc value is:

$Kc=\frac{1}{4.0*10^{24} } \\ =2.5*10^{-25}$

Answer is :

Kc=2.5*10^-25.