Homework Help: Disturbing Equilibrium

1. Jul 12, 2012

storm13

1. The problem statement, all variables and given/known data

What is the effect on the system below which is already at equilibrium if extra hydrogen was added? How is the amount of I2 different?

2. Relevant equations

2HI ⇔ H2 + I2

3. The attempt at a solution
I am finding effects the addition or removal of a substance at chemical equilibrium not making a lot of sense. To me it's contradicting, especially when you refer to graphs.

For example the solution to this would be: the system tries to oppose this change by removing this extra Hydrogen. It does this by favouring the reverse reaction, thus equilibrium shifts to the left, and there is now more Hydrogen Iodide than previous. The amountof I2 has decreased as it has been used to react with the extra H2 to form HI.

What i don't understand is how equilibrium constant doesn't change. How is this possible? If you have more HI and less I2, mathematical the concentration fraction will be less. This is observed on many equilibrium vs time graphs with adding or removing substances. It clearly does show a differing Kc value if you added all the concnetrations.

2. Jul 12, 2012

Staff: Mentor

You forgot you still have more H2 - even if part was consumed, its concentration is still higher than it was initially. That means [H2][I2] is not lower than it was before hydrogen addition. Quite the opposite - it is larger.

3. Jul 12, 2012

chemisttree

Well, that would be the definition of the word "constant" right? If it changed it wouldn't be "constant", would it?

At equilibrium, the concentration of the products divided by the concentration of the starting materials is constant.

Keq = [products]/[starting materials]

It's a definition that happens to be true. Just learn it and use it.

4. Jul 12, 2012

storm13

Yes I understand that's the rule and you have to learn it but i what i really dont understand is why the graphs don't reflect this. Am i missing something?

Graph 1 in the attachment : the ratio at Equilibrium before and after the disturbance are not equal, whereas in graph 2 they are??

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• Equilibiurm Graphs.pdf
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5. Jul 12, 2012

storm13

Your hint helped me. Thank you. Im pretty sure i understand it now. The left over bit of H2 would defintely make the difference. I had been playing around with the values on the graphs and realised i was making assumptions of all the values, which meant the ratio was different. But a key concept I was missing was that the H2 still did have an increase.