A level chemistry help

Hii84y4

I don't understand

(edited 2 years ago)

20220324_230603.jpg391.7KB

Reply 1

thomas.rhett

Original post by Hii84y4

I don't understand

The equilibrium is H2 + I2 <=> 2HI. K= [HI]^2 /

[I2].

So when you increase H2 under same temperature K doesn't change, so some but not all added H2 will react with I2 to form more HI until you have a new equilibrium where reaction quotient (Q) is again equal to K. Initially when you increase

the denominator of Q is greater so

X [I2] has to decrease and [HI] has to increase. When the mixture reaches equilibrium again the numerator is greater than it was before since there is more HI now. So the denominator must also be greater than before for the ratio to be equal to K. However since we have less [I2],

must be greater than it was before adding H2. So not all of the added H2 will react.

Reply 2

Hii84y4

Original post by thomas.rhett

The equilibrium is H2 + I2 <=> 2HI. K= [HI]^2 /

the denominator of Q is greater so

must be greater than it was before adding H2. So not all of the added H2 will react.

I'm sorry but i don't understand i feel bad especially as you gave such a thorough explanation :frown:

Reply 3

thomas.rhett

Original post by Hii84y4

I'm sorry but i don't understand i feel bad especially as you gave such a thorough explanation :frown:

Have you covered equilibria in school? Do you understand the H2 + I2 <=> 2HI equilibrium here?

Reply 4

Hii84y4

Original post by thomas.rhett

Have you covered equilibria in school? Do you understand the H2 + I2 <=> 2HI equilibrium here?

I'm a private candidate so I'm teaching myself

Reply 5

thomas.rhett

Original post by Hii84y4

I'm a private candidate so I'm teaching myself

Which part of the explanation don't you understand?

Reply 6

annaw328

Original post by Hii84y4

I'm a private candidate so I'm teaching myself

I think you should find some videos about equilibrium online first.

Reply 7

thomas.rhett

Original post by Hii84y4

I'm a private candidate so I'm teaching myself

H2 + I2 <=> 2HI

You can also look at it in terms of kinetics. When you increase

the _rate_ of forward reaction increases (unless the reaction in 0th order in H2). This will increase [HI] since now rate of forward reaction is greater than the backward reaction. Increase in [HI] increases rate of backward reaction and a new equilibrium is reached with more product or [HI] when rate of backward reaction equals forward reaction. Again there is less [I2] because some of it reacted to make more [HI] but since rate of forward reaction is greater than in the previous equilibrium

must be higher.

Reply 8

scimus63

I might be wrong but this looks like a simple question on Le Chatelier's principle. You are changing the conc of one of the substances in an equilibrium mixture.

try here for an explanation on Le Chateliers principle https://science-revision.co.uk/A-level_equilibrium_conditions.html

apologies if its wrong but your question is posted on its side!

Reply 9

scimus63

I might be wrong but this looks like a simple question on Le Chatelier's principle. You are changing the conc of one of the substances in an equilibrium mixture.

try here for an explanation on Le Chateliers principle https://science-revision.co.uk/A-level_equilibrium_conditions.html

or here if your doing gcse https://science-revision.co.uk/Le_Chatelier_principle.html

apologies if its wrong but your question is posted on its side!

Reply 10

Hii84y4

Original post by thomas.rhett

Which part of the explanation don't you understand?

Not sure if this is right but thinking about this question in terms of Kc, Kc will decrease but cannot change because Kc is only affected by temp so equilibrium shifts to the right to restore Kc. This means [2HI] increases and

and [I2] decreases but why is the

greater ?

(edited 2 years ago)

Reply 11

thomas.rhett

Original post by Hii84y4

and [I2] decreases but why is the

greater ?

at 2nd (new) equilibrium has to be greater than

at 1st equilibrium because [I2] at 2nd equilibrium is less than [I2] at 1st equilibrium. Kc is the same but [HI] is higher at 2nd equilibrium which means,