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Equilibrium constants help..

Multiple choice question 4(b) of the Edexcel June 2013 paper...

The first part of the question gave the equilibrium reaction:
H2 (g) + I2 (g) <--> 2HI (g)

the second part of the question says:
The equation for the equilibrium reaction between hydrogen and iodine may also be written as
1/2 H2 (g) + 1/2 I2 (g) <--> HI (g)

This change to the equation, compared to that in part (a),

A: has no effect on the value of Kc
B: halves the value of Kc
C: doubles the value of Kc
D: square roots the value of Kc

the answer is D, but I don't understand why... I thought the only thing that changed Kc was temperature.
Well the only thing that changes Kc is temperature, as long as the expression for Kc remains the same.

Writing an equation differently will change the expression for Kc, from:

[HI]2/([H2][I2]) into [HI]/([H2](1/2)[I2](1/2))

(Sorry if the formatting doesn't look good).
Reply 2
Avatar for KanKan
KanKan
11
If you think about it mathematically:

Kc=[HI]2[H2][I2]Kc = \frac{[HI]^2}{[H_2][I_2]}

and so it follows:

Kc=[HI]2[H2][I2]\sqrt{Kc} = \frac{\sqrt{[HI]^2}}{\sqrt{[H_2]}\sqrt{[I_2]}}

Can you see it from here?
Reply 3
Avatar for Basmaa
Basmaa
OP
Original post by KanKan
If you think about it mathematically:

Kc=[HI]2[H2][I2]Kc = \frac{[HI]^2}{[H_2][I_2]}

and so it follows:

Kc=[HI]2[H2][I2]\sqrt{Kc} = \frac{\sqrt{[HI]^2}}{\sqrt{[H_2]}\sqrt{[I_2]}}

Can you see it from here?


I get that... But it results in a different Kc value for the same reaction at the same temperature. That's what I don't understand.


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Reply 4
Avatar for KanKan
KanKan
11
Original post by Basmaa
I get that... But it results in a different Kc value for the same reaction at the same temperature. That's what I don't understand.


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The relationship between Kc and all the reactants/products are still the same. In that sense, Kc has not changed - just the value has been square rooted, in line with the rest of the equation.
Reply 5
Avatar for Borek
Borek
4
Value of Kc depends on the way reaction is written, but once you decide on the reaction equation, Kc changes with temperature only.

You can think about it as different reaction equation was in fact describing a different reaction.
Reply 6
Avatar for Basmaa
Basmaa
OP
Original post by Borek
Value of Kc depends on the way reaction is written, but once you decide on the reaction equation, Kc changes with temperature only.

You can think about it as different reaction equation was in fact describing a different reaction.


Oh, I get it now. Thank you so much!


Posted from TSR Mobile

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