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    3. Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
    CH3OH(g) + CO(g) --> CH3COOH(g)

    a) A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

    for this I got value of Kc=1.53

    then there was this part
    c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted

    and I got value of 2.91

    and then it says using ur values is reaction exo or endo
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    (Original post by otrivine)
    3. Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
    CH3OH(g) + CO(g) --> CH3COOH(g)

    a) A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

    for this I got value of Kc=1.53



    then there was this part
    c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted

    and I got value of 2.91

    and then it says using ur values is reaction exo or endo
    If 77.6% has reacted = 0.776 * 17.6 = 13.66 hence remaining = 3.94 mol

    The same moles of CO react hence CO remaining = 5.94

    Moles of ethanoic acid = 13.66

    Hence kc = 2.92

    If your kc value is higher at a lower temperature the reaction is more to the RHS (side of the products) at a lower temperature. The forward reaction is exothermic.
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    (Original post by charco)




    If your kc value is higher at a lower temperature the reaction is more to the RHS (side of the products) at a lower temperature. The forward reaction is exothermic.
    However, is the calculation right, because the volume is 5dm3 as well for the second reaction, as I got 2.91?
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    (Original post by otrivine)
    However, is the calculation right, because the volume is 5dm3 as well for the second reaction, as I got 2.91?
    edited above ...
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    (Original post by charco)
    edited above ...

    I got 2.91 correct to 3 sig figures
    i used the exact values when putting in Kc expression
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    (Original post by otrivine)
    I got 2.91 correct to 3 sig figures
    Near as makes no difference. I rounded to 2 dp.
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    (Original post by charco)
    Near as makes no difference. I rounded to 2 dp.

    Oh , I used the exact figures.

    Lastly,a) MnO4- + U3+ → Mn2+ + UO2+

    my overall balanced equation came to


    H2O + 5/2 U3+ + Mno4- ---> 5/2UO2 + + 2H+ + Mn2+


    is fine
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    (Original post by otrivine)
    Oh , I used the exact figures.

    Lastly,a) MnO4- + U3+ → Mn2+ + UO2+

    my overall balanced equation came to


    H2O + 5/2 U3+ + MnO4- ---> 5/2UO2 + + 2H+ + Mn2+


    is fine
    I should get rid of the fractions ...

    MnO4- + 8H+ + 5e --> Mn2+ + 4H2O
    U3+ + 2H2O --> UO2+ + 4H+ + 2e

    Multiply by 2 in equation 1 and by 5 in equation 2

    2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O
    5U3+ + 10H2O --> 5UO2+ + 20H+ + 10e
    ----------------------------------------------------------- add together
    2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+

    and cancel common terms

    2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+
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    (Original post by charco)
    I should get rid of the fractions ...

    MnO4- + 8H+ + 5e --> Mn2+ + 4H2O
    U3+ + 2H2O --> UO2+ + 4H+ + 2e

    Multiply by 2 in equation 1 and by 5 in equation 2

    2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O
    5U3+ + 10H2O --> 5UO2+ + 20H+ + 10e
    ----------------------------------------------------------- add together
    2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+

    and cancel common terms

    2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+
    with fractions , do you lose marks?
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    (Original post by otrivine)
    with fractions , do you lose marks?
    That would depend entirely on the markscheme. To be on the safe side I would only ever use fractions in thermodynamics where you are restricted to mole quantities by the standard conditions.
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    (Original post by charco)
    That would depend entirely on the markscheme. To be on the safe side I would only ever use fractions in thermodynamics where you are restricted to mole quantities by the standard conditions.
    Ok will make sure thank you
 
 
 
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