usernamenew
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pls can someone explain why B is 2nd order? I thought it was 0?

i found that A is 1st order, so for B i looked at experiment 2 and 3, i looked at the effect A also had on the rate since both A and B were changing, i found the new rate with A was 1.8x10-5 when i did 0.90x2 because A is first order? then i divided this new rate by the rate they have given in experiment 3= 7.20x10-5 and i got 2.5x10-11, so i assumed this meant B is 0 order? is my working out wrong? pls can somoene show me how im supposed to do this
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BobbJo
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usernamenew
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(Original post by usernamenew)
pls can someone explain why B is 2nd order? I thought it was 0?

i found that A is 1st order, so for B i looked at experiment 2 and 3, i looked at the effect A also had on the rate since both A and B were changing, i found the new rate with A was 1.8x10-5 when i did 0.90x2 because A is first order? then i divided this new rate by the rate they have given in experiment 3= 7.20x10-5 and i got 2.5x10-11, so i assumed this meant B is 0 order? is my working out wrong? pls can somoene show me how im supposed to do this
heres the quesion
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usernamenew
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(Original post by BobbJo)
post question
ive posted it, sorry took me a while
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MathsGod100
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www.alevelrevisioncards.co.uk has great resources to use for exam preparation
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usernamenew
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(Original post by MathsGod100)
www.alevelrevisioncards.co.uk has great resources to use for exam preparation
thank u
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BobbJo
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(Original post by usernamenew)
ive posted it, sorry took me a while
from 1 and 2, order wrt A is 1

our rate equation is of the form
rate = k [A][B]n

use 1 and 3
from 1: 0.45 x 10^-4 = k x 0.15 x 0.24^n - (A)
from 3: 7.2 x 10^-4 = k x 0.6 x 0.48^n - (B)

eqn B/eqn A gives
7.2x10^-4/(0.45x10^-4)= (0.6 x 0.48^n)/(0.15 x 0.24^n)
we have eliminated k
now solve for n,
n = 2
order is 2nd order wrt B
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BobbJo
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(Original post by usernamenew)
thank u
mathematical method above

logic method
from 2 to 3
[A] doubled
this causes rate to double
but in fact rate in 3 is x8 the rate in 2
so change in concentration of B caused rate to x4 (quadruple)
doubling in concentration of B causes rate to x4
so order wrt B is 2nd order
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usernamenew
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(Original post by BobbJo)
mathematical method above

logic method
from 2 to 3
[A] doubled
this causes rate to double
but in fact rate in 3 is x8 the rate in 2
so change in concentration of B caused rate to x4 (quadruple)
doubling in concentration of B causes rate to x4
so order wrt B is 2nd order
thank u! i prefer this method, once i know that 0.90x10-5 x 2cubed = 7.20x10-2, how does this mean B is x4? is it because A is x2 so u do 8/2 = 4
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BobbJo
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(Original post by usernamenew)
thank u! i prefer this method, once i know that 0.90x10-5 x 2cubed = 7.20x10-2, how does this mean B is x4? is it because A is x2 so u do 8/2 = 4
yes
rate changes only due to change in concentration
both concentration of A and B changed

we know from 2 to 3 rate x8
from 2 to 3, [A] doubled. order wrt A is first order. so doubling [A] doubled rate
overall change in rate is x8
since we know change in [A] caused rate to double, other changes in rate must be due to [B]
so change in concentration of B caused rate to x4

change in [A] -> x2
change in [B] -> x4
overall -> x8

hence doubling [B] quadruples rate
so order wrt B is 2nd order
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usernamenew
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(Original post by BobbJo)
yes
rate changes only due to change in concentration
both concentration of A and B changed

we know from 2 to 3 rate x8
from 2 to 3, [A] doubled. order wrt A is first order. so doubling [A] doubled rate
overall change in rate is x8
since we know change in [A] caused rate to double, other changes in rate must be due to [B]
so change in concentration of B caused rate to x4

change in [A] -> x2
change in [B] -> x4
overall -> x8

hence doubling [B] quadruples rate
so order wrt B is 2nd order
thank u so much
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usernamenew
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(Original post by BobbJo)
yes
rate changes only due to change in concentration
both concentration of A and B changed

we know from 2 to 3 rate x8
from 2 to 3, [A] doubled. order wrt A is first order. so doubling [A] doubled rate
overall change in rate is x8
since we know change in [A] caused rate to double, other changes in rate must be due to [B]
so change in concentration of B caused rate to x4

change in [A] -> x2
change in [B] -> x4
overall -> x8

hence doubling [B] quadruples rate
so order wrt B is 2nd order
can i use this same method for finding order of F in another question? once i found E is 2nd order, from exp 1 to 2, u x9 for rate, and to work out F using exp 2 and 3, E is also changing there, u x2 from exp2 to 3 so the effect on the rate in exp 2 would be 3.78x10-3 x4? and to work effect of F its 3.78x10-3 x2= 7.56x10-3. how does this mean F is 1st order? do i do 9/4= 2.25
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BobbJo
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(Original post by usernamenew)
can i use this same method for finding order of F in another question? once i found E is 2nd order, from exp 1 to 2, u x9 for rate, and to work out F using exp 2 and 3, E is also changing there, u x2 from exp2 to 3 so the effect on the rate in exp 2 would be 3.78x10-3 x4? and to work effect of F its 3.78x10-3 x2= 7.56x10-3. how does this mean F is 1st order? do i do 9/4= 2.25
Yes method always works

from 2 to 3,
conc of E doubled, rate should x4
in fact, rate x 2
other changes in rate is due to change in conc of F
so change in conc of F caused rate to halve

[E] x 2 -> rate x 4 (2nd order)
[F] x 1/2 -> rate x 1/2
overall -> rate x 2

so halving [F] caused rate to 1/2
hence first order wrt F
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usernamenew
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(Original post by BobbJo)
Yes method always works

from 2 to 3,
conc of E doubled, rate should x4
in fact, rate x 2
other changes in rate is due to change in conc of F
so change in conc of F caused rate to halve

[E] x 2 -> rate x 4 (2nd order)
[F] x 1/2 -> rate x 1/2
overall -> rate x 2

so halving [F] caused rate to 1/2
hence first order wrt F
thank u!
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