Question

For the reaction        A(g) + B(g) + C(g)  " D(g)     the following data were obtained at 25°C.

Exp’t	Initial [A] (M)	Initial [B] (M)	Initial [C] (M)	Initial Rate (mole∙L–1s–1)
1	0.0500	0.0500	0.0100	6.25 x 10–3
2	0.1000	0.0500	0.0100	1.25 x 10–2
3	0.1000	0.1000	0.0100	5.00 x 10–2
4	0.0500	0.0500	0.0200	6.25 x 10–3
5	0.2000	0.3000	0.3000	?????

Which choice gives correct kinetics information for this reaction?   Which choice is closest to the rate constant for this reaction?  Which choice is closest to the missing initial rate for Exp’t 5?

1st-order wrt A; 2nd-overall     k = 50.0 Lmol–1s–1     0.00400 mol∙L-1s-1

1st-order wrt A; 1st-order wrt B       k = 2.5 Lmol–1s–1     0.200 mol2L–2s–1

2nd-order wrt B; 3rd-order overall   k = 50.0 L2mol–2s–1     0.900 mol∙L-1s-1

2nd-order wrt B; 1st-order wrt C     k = 25 L2mol–2s–1     0.200 mol∙L-1s-1

Answer 1

answer :

2nd-order wrt B; 3rd-order overall  

k = 50.0 L2mol–2s–1    

0.900 mol∙L-1s-1     

rate= k [A]^x [B]^y [C]^z

from data

6.25x10^-3 =k [0.05]^x [0.05]^y [0.01]^z ------------->(1)

1.25 x10^-2 =k [0.1]^x [0.05]^y [0.01]^z ------------------> (2)

5 x10^-2 =k [0.1]^x [0.1]^y [0.01]^z -------------> (3)

6.25x10^-3 =k [0.05]^x [0.05]^y [0.02]^z ----------------> (4)

????    =k [0.2]^x [0.3]^y [0.3]^z ----------------> (5)

by solving 1 and 2

x = 1

by solving 2 and 3

y = 2

by solving 3 and 4

z = 0

rate law = k [A]¹ [B]² [C]⁰

overall order = 1+ 2+0 =3

from any of above equations we can solve rate constant k value

5 x10^-2 =k [0.1]^x [0.1]^y [0.01]^z -------------> (3)

5 x10^-2 =k [0.1]¹ [0.1]² [0.01]⁰ -------------> (3)

k = 50 mol^-2 lit^2 sec^1

for unknown rate

????    =k [0.2]^x [0.3]^y [0.3]^z ----------------> (5)

rate = 50 x [0.2] [0.3]² [0.3]⁰ ----------------> (5)

rate        = 0.9 M/s