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    for task 7-Using the Arrhenius equation
    why does the answer to question 3 a) for the 20 degrees one contain the units mol-1 dm6 s-1 for the rate constant k

    how is this achieved?

    http://www.chemsheets.co.uk/Chemshee...20booklet).pdf

    page 16
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    3 a) Calculate the rate constant at 20C and 40C in a reaction where the Arrhenius constant (A) = 5.18 x 105 mol-1 dm3 s-1 and the activation energy (Ea) = 75 kJ mol-1.


    It says dm3 not dm6 (as you state).

    Also it is giving the unit for A, not k.

    Also also, where are the answers? You've given the link to the Q.
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    (Original post by Pigster)
    3 a) Calculate the rate constant at 20C and 40C in a reaction where the Arrhenius constant (A) = 5.18 x 105 mol-1 dm3 s-1 and the activation energy (Ea) = 75 kJ mol-1.


    It says dm3 not dm6 (as you state).

    Also it is giving the unit for A, not k.

    Also also, where are the answers? You've given the link to the Q.
    You have to work out k by rearranging the Arrhenius equation and the answer states the units are mol−1 dm6 s−1 at both 20 and 40 degrees
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    (Original post by esmeralda123)
    You have to work out k by rearranging the Arrhenius equation and the answer states the units are mol−1 dm6 s−1 at both 20 and 40 degrees
    If you know how to do it why are you asking?
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    (Original post by charco)
    If you know how to do it why are you asking?
    i don't know how to get the units
    i know how to get the actual value
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    (Original post by esmeralda123)
    i don't know how to get the units
    i know how to get the actual value
    Units are obtained by carrying out a dimensions analysis.

    Substitute the units for each concept into the equation and cancel down.
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    think of it like algebra.

     x^{2} \times x^{2} = x^{4}

    in this question

     k = Ae^{\frac{-Ea}{RT}} substitute the units in

     k = mol^{-1}dm^{3}s^{-1} \times e^{\frac{Jmol^{-1}}{JK^{-1}mol^{-1} \times k}}

    e has no units so we're just left with  mol^{-1}dm^{3}s^{-1}
 
 
 
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