At a certain temperature the following data were collected for the reaction 2ICl

Question

At a certain temperature the following data were collected for the reaction
2ICl + H2 I2 + 2HCl.
Initial Concentrations (mol L-1)
Initial Rate of Formation of
[ICl] [H2] I2 (mol L-1 s-1)
0.10 0.10 0.0015
0.20 0.10 0.0030
0.10 0.0500 0.00075

Determine the rate law and the rate constant (with correct units) for the reaction.
(A)R = k[ICl][H2]
(B)R = k[ICl]2[H2]
(C)R = k[ICl]2[H2]2
(D)R = k[ICl][H2]0
(E)R = k[ICl][H2]2


What is the value of the rate constant? ________
The unit of the rate constant is L mol-1 s-1

Explanation / Answer

let order of reaction with respect to ICl be m and with respect to H2 be n then rate , r = k [ICl]^m [H2]^n now when we change conc. of ICl or H2 ...r changes but rate constant k remains the same ... from expt-1 [ICl] = 0.1 [H2] = 0.1 r = 0.0015 putting in rate expression... 0.0015 = k X [0.1]^m [0.1]^n .....(1) similarly from experiment -2 ... 0.003 = k X [0.2]^m [ 0.1]^n .....(2) dividing (1) from (2)... 0.003/0.0015 = k X [0.2]^m X [0.1]^n / k X [0.1]^m X [0.1]^n 2 = [0.2/0.1]^m 2 = 2^m m = 1 now from experiment -3 0.00075 = k X [0.1]^m [0.05]^n .....(3) dividing (3) from (1)... 0.0015/0.00075 = k X [0.1]^m X [0.1]^n / k X [0.1]^m [ 0.05]^n 2 = [0.1/0.05]^n 2 = 2^n n = 1 so order with respect to ICl is 1 and order with respect to H2 is 1 so r = k [ICl] [H2] now choose any of the three experiments ....say experiment -2 0.003 = k X 0.2 X 0.1 k = 0.003/0.02 = 0.15 L/mole/s now in order to verify that this is the correct answer choose another experiment as rate constant k remains the same even if the conc. changes... lets choose experiment -3... 0.00075 = k X 0.1 X 0.05 k = 0.00075 / 0.005 = 0.15 L/mole/s

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