Medal and fan If the reaction SO2(g) + 1/2 O2 SO3(g) has the equilibrium constant Kc = 56, then what is the value of Kc for the following reaction? 2SO3(g) 2SO^2(g) + O2(g) Thanks(:

Question

Medal and fan If the reaction SO2(g) + 1/2 O2 <-> SO3(g) has the equilibrium constant Kc = 56, then what is the value of Kc for the following reaction? 2SO3(g) <-> 2SO^2(g) + O2(g) Thanks(:

anonymous · Answer

Could you help me with this?(: @Abhisar

abhisar · Answer

5.69 X 10^-6

abhisar · Answer

Is there any option ?

anonymous · Answer

That's not one of the options.

abmon98 · Answer

is the answer112

anonymous · Answer

Can you show me how you got that?

anonymous · Answer

Actually, 112 isn't an option. -112 is.

anonymous · Answer

options are: -112, 56, 3.2*10^-4, and 8.9*10^-3.

abhisar · Answer

3.2 * 10^-4 will be according to me

abhisar · Answer

yes sorry for the previous answer

anonymous · Answer

Could you show me how you come to that answer?

abhisar · Answer

The new Kc will be square of the inverse of original Kc

anonymous · Answer

That would make it 3.2*10^4, not to the -4.

abhisar · Answer

|dw:1402822957247:dw|

abhisar · Answer

This is for initial reaction

abhisar · Answer

|dw:1402823088687:dw|

abhisar · Answer

and this for final reaction

abhisar · Answer

Is it clear now ?

anonymous · Answer

I think so...

abhisar · Answer

New Kc = (1/old Kc)^2

anonymous · Answer

Alright, thankyou, that makes sense(:

abhisar · Answer

$\Huge\text{Anytime !}$
$\huge\ddot\smile$