How Do You Determine the Equilibrium Constant for Modified Reactions?

[shaa]

Hello, I'm new to the forums and did in fact search for my thread before posting. I couldn't find anything that helped me with what I'm looking for. With that said, the problem I'm having is:

1) The value of K_eq for the equilibrium H₂+I₂ → 2 HI is 794 at 25°C. What is the value of K_eq for the equilibrium below?
1/2 H₂+1/2 I₂ → HI

2) The equilibrium constant for reaction 1 is K. The equilibrium constant for reaction 2 is _____.
1: SO₂+1/2 O₂ → SO₃
2: 2SO₃ → 2SO₂+O₂

I apologize for not denoting an equation at equilibrium, but I couldn't find the right arrow in the "Quick Symbols" box. Both equations are at equilibrium and involve only matter in the gaseous state.

 

[shaa]

I've gotten the first question, considering that the second reaction is half of the first. Resulting in: x^1/2, and x being the original K_c.

I know what the answer to the first question is, but I'm having difficulty understanding why it is what it is.

 

[Borek]

This is a simple math. Write reaction quotient for each reaction.

 

[shaa]

It states that the equilibrium constant for reaction 1 is = k. The equilibrium constant for reaction 2 is = 1/k².

The answer makes complete sense given a general form of an equilibrium reaction, but why wouldn't the first reaction be = 1/k or the second be = -1/k²?

 

[Borek]

I don't understand why you think it should be. If you write reaction quotients for each reaction and compare them, it is obvious what value the new equilibrium constant has.

You can as well ask why $\sqrt 2$ doesn't equal $\pi$.

 

[shaa]

Hahahah. Of course. It's the inverse reaction and it double. Thus 1/k².
Thanks? :P

 

[Redbelly98]

I apologize for not denoting an equation at equilibrium, but I couldn't find the right arrow in the "Quick Symbols" box. Both equations are at equilibrium and involve only matter in the gaseous state.

One of these might help; feel free to copy-and-paste for future use, or bookmark the links in my sig:
← or ↔

p.s. Welcome to Physics Forums.