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Question on rates and there equations

username1393226

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(edited 8 years ago)

Reply 1

TheLegalDealer

Original post by T-GiuR

For this question are these answers correct in respective order;
http://i.imgur.com/MbZU7tF.jpg

- [A]^0 as the concentrations from experiment 1 - 2 the concentration was double yet the rate remained the same

- ^2 the rate increased by a factor of 4 when the concentration of B doubled.

- Rate = K^2

- K= 3x10^-4/ ²= 7.5x10^-5

Thanks

Your calculation value of 7.5x10^-5 for K is correct but watch out for what you right as you calculation as it should be 3x10-4 / 2 *2

Reply 2

username1393226

Original post by TheLegalDealer

Your calculation value of 7.5x10^-5 for K is correct but watch out for what you right as you calculation as it should be 3x10-4 / 2 *2

Would it not be 2^2 ?

Also would the units of K be Mol^-1 L^-1 s^-1 ?

Reply 3

TheLegalDealer

Original post by T-GiuR

Would it not be 2^2 ?

Also would the units of K be Mol^-1 L^-1 s^-1 ?

Yes , it would be it not be 2^2 . I just dont know how top do the squared option yet lool .

The units would be Mol^-1 L^3 s^-1

(edited 9 years ago)

Reply 4

username1393226

Original post by TheLegalDealer

Yes , it would be it not be 2^2 . I just dont know how top do the squared option yet lool .

The units would be Mol^-1 L^3 s^-1

Sorry I'm confused. Is it B^2 or B*B ?
And when using B, do we use the lower concentration instead of the higher one when calculating K ? i.e. We used 2 instead of 4 if that makes sense

Thanks

Reply 5

TheLegalDealer

Original post by T-GiuR

Sorry I'm confused. Is it B^2 or B*B ?
And when using B, do we use the lower concentration instead of the higher one when calculating K ? i.e. We used 2 instead of 4 if that makes sense

Thanks

Its B^2 ( as in B squared ) , and it doesn't make a difference as long as you use the concentration of B and initial rate from the same experiment

E.g. if we used experiment 3 , it would be 12x10-4 / 4 ^2 Giving the same answer of 7.5x10^-5Mol^-1 L^3 s^-1

Hope that helps ! :wink: