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    (Original post by Adorable98)
    The natural log form of the Arrhenius equation (which you are given above) is of the same form as a straight line graph y = mx + c

    lnk = -Ea/RT + lnA
    y = mx + c

    If the gradient is 'm' what are y and x?
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    (Original post by Adorable98)
    What's this question from?

    I reckon it's choice D, as using logs with k will provide more suitable numbers for a scale, and 1/T is often used for rate (change in something per unit time)
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    (Original post by charco)
    The natural log form of the Arrhenius equation (which you are given above) is of the same form as a straight line graph y = mx + c

    lnk = -Ea/RT + lnA
    y = mx + c

    If the gradient is 'm' what are y and x?
    Yeah.... clear explanation.

    Nicely done
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    (Original post by Noj777)
    What's this question from?

    I reckon it's choice D, as using logs with k will provide more suitable numbers for a scale, and 1/T is often used for rate (change in something per unit time)
    Edexcel A2 chemistry Jan 2013 q3
    And thanks!!

    (Original post by charco)
    The natural log form of the Arrhenius equation (which you are given above) is of the same form as a straight line graph y = mx + c

    lnk = -Ea/RT + lnA
    y = mx + c

    If the gradient is 'm' what are y and x?
    I see, so y= ink K and x= 1/T

    Thank you!!
 
 
 
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