I was working on my lab for chemistry and was only able to get this far. Im not
ID: 1083962 • Letter: I
Question
I was working on my lab for chemistry and was only able to get this far. Im not sure how to do part 2b and 2c. Included my observed data. Thank you
Explanation / Answer
2b.
To calculate this , you have to follow this
let intial concentration is C1 , Initial volume Is = V1
Concentration after dilution = C2 , volume after dilution = V2 = (V1+ volume added)
now
C1V1 = C2V2
then , C1 = C2V2/V1
2d.
To calculate the rate
from the equation , K[I-]x[BrO3-]y[H+]z (1)
calculate the rate of reaction by changing one reactant and keeping the other reactants constant.
You can follow the given procedure
let Rate = 2 *10-3 , when concentration of I- = P , [BrO3-] = Q , and [H+] = R
if concentration of [I-] changes to 2p, let rate is 4*10-3 ( concentration of others are constant)
and if concentration of [BrO3-] changes to 2Q then the rate is 8*10-3 ( concentration of others are constant)
and if concentration of [H+] changes to 2R then the rate is 2*10-3
now you can put the data in the equation 1
2*10-3 = K. Px. Qy. Rz (i)
4*10-3 = K .(2P)xQy. Rz (ii)
8*10-3 = K. Px. (2Q)y. Rz (iii)
2*10-3 = K. Px. Qy. (2R)z (iv)
now divide (i) and (ii)
2/4 = Px/(2P)x
or, 1/2 = (1/2)x ; or, x = 1
again divide (i) by (iii)
2/8 = Qy/ (2Q)y
or, 1/4 = (1/2)y ; or, y = 2
again divide (i) by (iv)
1/1 = Rz/ (2R)z
or, 1 = (1/2)z
or, Z = 0
hence, order of the reaction is = 1+2+0 = 3
,By following the above process you can calculate the order of the reaction.
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