Position of Equilibrium help

TheNoobishKnight

Hi guys, a bit stuck on this question. Can anyone please explain it to me?

The answer for the first part is 16.8(no units) which I got right. I just can't seem to understand the second part.

Reply 1

charco

Study Forum Helper

Original post by TheNoobishKnight

Hi guys, a bit stuck on this question. Can anyone please explain it to me?

The answer for the first part is 16.8(no units) which I got right. I just can't seem to understand the second part.

Use the equilibrium constant from part (a) and the equilibrium law equation to estimate the values at equilibrium of the components.

kc = [HI]² / [H₂][I₂]

and you know that x moles of H₂ and I₂ have reacted at equilibrium and formed 2x moles of HI

... but the simple rule is: if you add to the left you make more of the right ...

(edited 9 years ago)

Reply 2

TheNoobishKnight

Original post by charco

Yup that's what I thought. More H2 is added, then more Hi needs to be made to make up for it. But I couldn't understand what happened to I2?

Reply 3

Borek

Iodine is consumed in the reaction till reaction quotient equals the equilibrium constant. It won't ever got down to zero.

Reply 4

TheNoobishKnight

Original post by Borek

Iodine is consumed in the reaction till reaction quotient equals the equilibrium constant. It won't ever got down to zero.

Okay. I am still a bit confused about what happens to I2, especially in different scenarios but I will ask my teacher. Thanks for the help buddy :smile:

Like you add more to left, get more on right. But why is I2 being used up when more H2 is added, isn't meant to stay the same to further reduce more shifting of the Position of Equilibrium?

(edited 9 years ago)

Reply 5

charco

Study Forum Helper

Original post by TheNoobishKnight

Okay. I am still a bit confused about what happens to I2, especially in different scenarios but I will ask my teacher. Thanks for the help buddy :smile:

Like you add more to left, get more on right. But why is I2 being used up when more H2 is added, isn't meant to stay the same to further reduce more shifting of the Position of Equilibrium?

It has no choice!

If you add hydrogen the equilibrium is disturbed and the system must respond by moving to the right hand side.

It can only do this by reacting hydrogen and iodine together. The hydrogen needs the iodine to react with...

Hence, increase the hydrogen and the system reacts hydrogen and iodine together to make more hydrogen iodide and restore the equilibrium situation.

Reply 6

TheNoobishKnight

Original post by charco

OHHHHHHHHHHH NOW I get it.

A needs to make C but to do that it needs to B. So there is more A(added), less B and more C(to reduce the effect of the increase of the left side).

You rock :smile:

Reply 7

charco

Study Forum Helper

Original post by TheNoobishKnight

OHHHHHHHHHHH NOW I get it.

A needs to make C but to do that it needs to B. So there is more A(added), less B and more C(to reduce the effect of the increase of the left side).

You rock :smile:

Not so hard once you see the light :wink: