Rate law Watch

Barça
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#1
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I need some help with some questions regarding kinetics

1) (a) Write the rate law for the following reaction using x, y, and z for the order of the individual components:

5Br- (aq) + BrO3- (aq)+ 6H+ (aq)  3Br2 (l) + 3H2O (l)





(b) Determine the individual orders and overall order using the following
experimental data:


Br- (aq) (M) BrO3- (aq) (M) H+ (aq) (M) Rate (M/s)
0.33 0.33 0.33 4.3x10-4
0.33 0.66 0.33 4.3x10-4
0.33 0.66 0.66 1.7x10-3
1.32 0.66 0.66 6.8x10-3

Order of Br- (aq) = 1st

Order of BrO3- (aq) = 2nd

Order of H+ (aq) = 2nd

Overall order = 5th order

Is this correct?

(c) What are the units of k for this reaction?
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thelostchild
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for part A it would simply be


rate=k[Br^-]^x[BrO_{3}^-]^y[H^+]^z

and from the data youve given it looks like BrO_{3}^- is 0th order
H^+ is 2nd order
and
Br^- is 1st order

have another look and compare experiments 1+2

and the units of k will be
$k=\frac{rate}{[Br^-][H^+]^2}
so
$k=\frac{mol^1 dm^{-3} s^{-1}}{mol^3dm^{-9}}

$k=mol^{-2}dm^{6}s^{-1}
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Barça
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Cheers mate, appreciate that
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charco
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signature update needed..... Atletico Madrid 4 Barcelona 2
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(Original post by charco)
signature update needed..... Atletico Madrid 4 Barcelona 2
I highlighted that game in which the quote came from, nice try.
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Barça
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(Original post by thelostchild)
for part A it would simply be


rate=k[Br^-]^x[BrO_{3}^-]^y[H^+]^z

and from the data youve given it looks like BrO_{3}^- is 0th order
H^+ is 2nd order
and
Br^- is 1st order

have another look and compare experiments 1+2

and the units of k will be
$k=\frac{rate}{[Br^-][H^+]^2}
so
$k=\frac{mol^1 dm^{-3} s^{-1}}{mol^3dm^{-9}}

$k=mol^{-2}dm^{6}s^{-1}
By the way is the integrated rate law for a 1st order reaction this

ln([A]0) - ln([A]) = kt or this, rate = k[A]0 = k[A] ?
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thelostchild
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(Original post by Barca)
By the way is the integrated rate law for a 1st order reaction this

ln([A]0) - ln([A]) = kt or this, rate = k[A]0 = k[A] ?
it would be the first one, you have the differential equation
$

\frac{d[A]}{dt}=-k[A]
which when integrated gives the exponential decay
$[A]=[A]_0e^{-kt}

which like you said would rearrange to
$kt=\ln [A]-\ln [A]_0

theres a table on wikipedia which summerises it nicely
http://en.wikipedia.org/wiki/Rate_eq...C_1.2C_2_and_n
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Barça
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Thanks for that
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