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    Can someone pleas help me on this question and show me why you do the steps you do. I've tried every possible way now and I'm just not getting the answer.


    4.The reaction for the formation of hydrogen iodide does not go to completion but reaches an equilibrium: H2(g) + I2(g) == 2HI(g)
    A mixture of 1.9 mol of H2 and 1.9 mol of I2 was prepared and allowed to reach equilibrium in a closed vessel on 250 cm3 capacity. The resulting equilibrium mixture was found to contain 3.0 mol of HI. Calculate the value of Kc.

    The answer btw is 6.6 mol-2dm6
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    Okay, 1.9 + 1.9 moles were put in, making 3.8 moles altogether.

    So if 3.0 moles ended up as HI then the other 0.8 must be H2 and I2 (so 0.4 of each)

    Converting this to concentration in mol dm-3 by dividing by the volume (0.25 dm3) gives 12 for [HI], 1.6 for [H2] and 1.6 for [I2].

    Therefore Kc = ([HI]^2) / ([H2][I2]) = (12^2) / (1.6*1.6) = 144/2.56 = 56.25 arbitrary units

    I know that's not the answer you gave but I think that answer is wrong, the units definitely are because the Kc should be unitless in this case
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    1.9+1.9 = 3.8 moles
    There was 3 moles of HI, so 3.8-3= 0.8 moles left for H2+I2 (so half it.. 0.4 moles of H2 and 0.4 moles of I2 at equilibrium)

    Moles = conc*Vol so Conc = Moles/Volume

    250cm3 Volume - convert this to dm3 (250/1000) = 0.25dm3

    Conc of HI:
    3/0.25 = 12M
    Conc of H2:
    0.4/0.25 = 1.6M
    Conc of I2: 1.6M

    Kc = Products/Reactants
    Kc = [HI]^2/[H2][I2] (we square the HI conc as we have 2 moles)
    Kc = [144]/[1.6][1.6]
    Kc = [144]/[1.6x1.6]
    Kc = 144/2.56 = 56.25

    Same as above yup
    For units:
    We use concentration to find Kc... conc is moldm-3
    Kc = [HI]^2/[I2][H2]
    Kc = [moldm-3]^2/[moldm-3][moldm-3]
    Kc = mol2dm-6/[moldm-3][moldm-3]
    They cancel each other out, so there are no units for Kc
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    Thanks both of you
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    (Original post by Phalange)
    1.9+1.9 = 3.8 moles
    There was 3 moles of HI, so 3.8-3= 0.8 moles left for H2+I2 (so half it.. 0.4 moles of H2 and 0.4 moles of I2 at equilibrium)

    Moles = conc*Vol so Conc = Moles/Volume

    250cm3 Volume - convert this to dm3 (250/1000) = 0.25dm3

    Conc of HI:
    3/0.25 = 12M
    Conc of H2:
    0.4/0.25 = 1.6M
    Conc of I2: 1.6M

    Kc = Products/Reactants
    Kc = [HI]^2/[H2][I2] (we square the HI conc as we have 2 moles)
    Kc = [144]/[1.6][1.6]
    Kc = [144]/[1.6x1.6]
    Kc = 144/2.56 = 56.25

    Same as above yup
    For units:
    We use concentration to find Kc... conc is moldm-3
    Kc = [HI]^2/[I2][H2]
    Kc = [moldm-3]^2/[moldm-3][moldm-3]
    Kc = mol2dm-6/[moldm-3][moldm-3]
    They cancel each other out, so there are no units for Kc
    I'm sooo sorry but I wrote the wrong question, thats why neither of you got 6.6 would you please help me with this one.

    In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
    N2(g) + 3H2(g) == 2NH3(g)
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    (Original post by SuperhumanTouch)
    Okay, 1.9 + 1.9 moles were put in, making 3.8 moles altogether.

    So if 3.0 moles ended up as HI then the other 0.8 must be H2 and I2 (so 0.4 of each)

    Converting this to concentration in mol dm-3 by dividing by the volume (0.25 dm3) gives 12 for [HI], 1.6 for [H2] and 1.6 for [I2].

    Therefore Kc = ([HI]^2) / ([H2][I2]) = (12^2) / (1.6*1.6) = 144/2.56 = 56.25 arbitrary units

    I know that's not the answer you gave but I think that answer is wrong, the units definitely are because the Kc should be unitless in this case
    Sorry, its because I wrote
    the wrong question
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    (Original post by interact)
    I'm sooo sorry but I wrote the wrong question, thats why neither of you got 6.6 would you please help me with this one.

    In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction.
    N2(g) + 3H2(g) == 2NH3(g)
    It's the same principal!

    So there was 1/3 mole of N2 and H2 left at equilibrium
    1/3 of 9 = 3
    1/3 of 27 = 9

    So 6 moles N2 reacted to produce ammonia. Using mole ratio 1:2, 6x2 = 12 moles ammonia made

    Kc = [NH3]^2 / [N2][H2]^3
    [NH3] = 12/10 = 1.2
    [N2] = 3/10 = 0.3
    [H2] = 9/10 = 0.9

    Kc = [1.2]^2 / [0.3][0.9]^3 = 6.58 = 6.6 (1 dp rounded)
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    (Original post by Phalange)
    It's the same principal!

    So there was 1/3 mole of N2 and H2 left at equilibrium
    1/3 of 9 = 3
    1/3 of 27 = 9

    So 6 moles N2 reacted to produce ammonia. Using mole ratio 1:2, 6x2 = 12 moles ammonia made

    Kc = [NH3]^2 / [N2][H2]^3
    [NH3] = 12/10 = 1.2
    [N2] = 3/10 = 0.3
    [H2] = 9/10 = 0.9

    Kc = [1.2]^2 / [0.3][0.9]^3 = 6.58 = 6.6 (1 dp rounded)
    Thanks alot. I know this is a really basic/stupid question but how do you know to use the number of moles N2 to work out the no. of moles of ammonia. How do you know not to use the no. of moles of H2
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    Damn thought this was about the film
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    (Original post by interact)
    Thanks alot. I know this is a really basic/stupid question but how do you know to use the number of moles N2 to work out the no. of moles of ammonia. How do you know not to use the no. of moles of H2
    Ok so you have 18 moles H2 and 6 moles N2 reacting

    N2(g) + 3H2(g) == 2NH3(g)

    For N2... 1:2 ratio so we have 12 moles NH3
    For H2... 3:2 ratio so we have (18/3)*2 = 6*2 = 12 moles NH3
    You can use either way. What is important to remember is that they both react TOGETHER to form ammonia so you don't need to use the ratios separately then add them together
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    (Original post by Phalange)
    Ok so you have 18 moles H2 and 6 moles N2 reacting

    N2(g) + 3H2(g) == 2NH3(g)

    For N2... 1:2 ratio so we have 12 moles NH3
    For H2... 3:2 ratio so we have (18/3)*2 = 6*2 = 12 moles NH3
    You can use either way. What is important to remember is that they both react TOGETHER to form ammonia so you don't need to use the ratios separately then add them together
    Thanks alot!!
 
 
 
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