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    It's the March 2017 sheet, I'm completely stuck on Question 9.

    Normally there are written solutions somewhere but I can't find any.

    I'll rewrite the question here:

    The following equilibrium is established at temperature T when 1.00 moil of HI is contained. It is found that 22% of the HI has broken down. Calculate Kc.

    2HI(g) _> (reversible reaction) H2(g) + I2(g)

    Thank you!
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    Well as the HI is a 1 mol, and breaks down by 22% - that means there's a decrease of 0.22 moles of HI. It's then a 2:1 reaction - for every 2 moles of HI, there's 1 mole of H2 and I2. So is HI decreases by 0.22, H2 and I2 will increase by 0.11 (0.22/2). Calculate the final moles of each compound and then substitute into the Kc equation. Hope this helps (I think this is the correct way)
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    (Original post by Rajan17)
    Well as the HI is a 1 mol, and breaks down by 22% - that means there's a decrease of 0.22 moles of HI. It's then a 2:1 reaction - for every 2 moles of HI, there's 1 mole of H2 and I2. So is HI decreases by 0.22, H2 and I2 will increase by 0.11 (0.22/2). Calculate the final moles of each compound and then substitute into the Kc equation. Hope this helps (I think this is the correct way)
    I thought I would need to substitute the concentrations of each compound rather than their moles?
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    (Original post by Adamburgess38)
    I thought I would need to substitute the concentrations of each compound rather than their moles?
    I think in this case the volumes would cancel out, so its fine to use moles. You should get the same answer if you use concentration though. Sorry about that!
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    (Original post by Rajan17)
    I think in this case the volumes would cancel out, so its fine to use moles. You should get the same answer if you use concentration though. Sorry about that!
    Ah thank you
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    (Original post by Adamburgess38)
    Ah thank you
    No probs
 
 
 

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