the following data were collected for the reaction of tbutylbromide (CH3)3CBr) w

Question

the following data were collected for the reaction of tbutylbromide (CH3)3CBr) with hydroxide ion at 55 degrees Celsius: equation: (CH3)3+CBr+OH-=> (CH3)3COH+Br-1 exp (CH3)3CBr initial [OH]initial Rate(M/S) 1 .10 .10 1.0*10^-3 2 . 20 .10 2.0*10^-3 3 .30 .10 3.0*10^-3 4 . 10 .20 1.0*10^-3 5 .10 .30 1.0*10^-3 Then Dtermine the rate law for the reaction. And then calculate the rate constant k for the reaction.

Explanation / Answer

for the reaction A + B ---> C + D rate = k x [A]^n x [B]^m where.. [ ] means concentration and n and m are reaction orders with respect to A and B the point of this problem is to use the data to determine n and m and then k ************ if you look at runs 1 to 2.. you'll notice that [H2] was constant at 0.1M and [NO] doubled.. so we can write this rate 1 = k x [NO#1]^n x [H2]^m rate 2 = k x [NO#2]^n x [H2]^m or.. rearranging.. rate 1 / [NO#1]^n = k x [H2]^m rate 2 / [NO#2]^n = k x [H2]^m since they both = k x [H2] they must = each other and rate 1 / [NO#1]^n = rate 2 / [NO#2]^n or... ([NO#2] / [NO#1])^n = rate 2 / rate 1 since NO#2 = 2 x NO#1... [2 x NO#1 / NO#1]^n = rate 2 / rate 1 2^n = 5.6 / 1.4 = 4 n must = 2 ********** now we know this... rate = k x [NO]^2 x [H2]^m ********** likewise.. from runs 1 to 3.. [NO] was constant and [H2] increased.. so that.. ([H2#3] / [H2#1])^m = 2.8 / 1.4 = 2 (2 x [H2#1] / [H2#1])^m = 2 (2)^m = 2 m = 1 and now we know... *** question part A *** rate = k x [NO]² x [H2] *** question part B *** plug in any of the three runs and calculate k = rate / ([NO]² x [H2]) k = 1.4x10^-4 mole/Lsec / [(0.1mole/L)² x (0.1 mole / L)] k = 0.14 L² / (mole² sec)

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