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

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)`

Step-by-step 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=(SO3^(2) )/(SO2^(2).O2 )`

For the reverse reaction,

`2SO3(g) <=> 2SO2(g)+O2(g)`

`Kc=(SO2^(2)O2 )/(SO3^(2). )`

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

Hence, the new Kc value is:

`Kc=(1)/(4.0*10^(24) ) \\ =2.5*10^(-25)`

Answer is :

Kc=2.5*10^-25.