Question

Given the data below for the reaction, 2 A + 2 B + 4 C => D + E + 3 F,

Experiment	Initial conc of A, mol/L	Initial conc of B, mol/L	Initial conc of C, mol/L	Initial rate, mol/L.s
1	0.1	0.2	0.4	2 x 10-3
2	0.2	0.2	0.4	4 x 10-3
3	0.3	0.4	0.4	6 x 10-3
4	0.4	0.6	0.2	2 x 10-3

Calculate the value of k to 3 significant figures.

Answer 1

Answer - Given, reaction – 2 A + 2 B + 4 C ----> D + E + 3 F,

Assume rate law is - Rate = k [A]^x [B]^y [C]^z

In this rate law there are x,y and z are the order with respect to A, B and C

Rate1 = k [A]₁^x [B]₁^y [C]₁^z

Rate2 = k [A]₂^x [B]₂^y [C]₂^z

Rate3 = k [A]₃^x [B]₃^y [C]₃^z

Rate4 = k [A]₄^x [B]₄^y [C]₄^z

Order with respect to A

Rate2/ Rate1 = k [A]₂^x [B]₂^y [C]₂^z / k [A]₁^x [B]₁^y [C]₁^z

4*10^-3/ 2.0*10^-3 = (0.2)^x /(0.1)^x * (0.2)^y /(0.2)^y *(0.4)^z /(0.4)^z

   2.0 = (2)^x

So, x = 1

Order with respect to B

Rate3/ Rat2 = k [A]₃^x [B]₃^y [C]₃^z / k [A]₂^x [B]₂^y [C]₂^z

6.0*10^-3 /4.0*10^-3 = (0.3) /(0.2) * (0.4)^y /(0.2)^y *(0.4)^z /(0.4)^z

1.5 = (1.5)_­ *(2)^y

So, y = 0

Order with respect to C

Rate4/ Rate3 = k [A]₄^x [B]₄^y [C]₄^z / k [A]₃^x [B]₃^y [C]₃^z

2.0*10^-3 /6.0*10^-3 = (0.4) /(0.3) * (0.6)⁰ /(0.4)⁰ *(0.2)^z /(0.4)^z

1.5 = (1.3)_­ *(0.5)^z

So, (0.5)^z = 1.15

z = 0

So the order with respect to A, B and C are 1, 0 and 0 respectively.

Overall order of reaction = 1+ 0 +0 = 1

So, rate law

Rate = k [A]

Now we need to put the values and calculate k

2.0*10^-3 M.s^-1 = k (0.1 M)

0.000335 Ms^-1 = k*0.00304

k = 2.0*10^-3 Ms^-1 /0.1 M

k = 2.00*10^-2 M s^-1