Question

I cannot seem to answer this question correctly. Could you show me step by step how to work this?

Using the given data, calculate the rate constant of this reaction.

A+B ------ C+D                     Trial          A(M)            (B)M        (C)

                                              1              0.330           0.370      0.0135

                                               2              0.330          0.999       0.0984

                                               3               0.495          0.370       0.0203

                K=      

Answer 1

Given reaction is A + B -----------> C + D

rate = k [A]^m [B]ⁿ    k = rate constant

From trial 1,

0.0135 = k [0.33]^m [0.37]ⁿ ------- Eq (1)

From trial 2,

0.0984 = k [0.33]^m [0.999]ⁿ --------Eq (2)

From trial 3,

0.0203 = k [0.495]^m [0.37]ⁿ --------Eq (3)

Do Eq(2)/Eq(1),

(0.0984/0.0135) = (k/k) (0.33/0.33)^m (0.999/0.37)ⁿ

7.28 = 1^m x 2.7ⁿ  

7.28= 1 x 2.7ⁿ

Take log on both sides,

log 7.28 = n log 2.7

n = log 7.28/ log 2.7

n =2

Therefore, n =2

Do Eq(3)/Eq(1),

(0.0203/0.0135) = (k/k) (0.495/0.33)^m (0.37/0.37)n

1.5 = (1.5)^m x (1)ⁿ

1.5 = (1.5)^m x 1

So,

m = 1

Therefore, m= 1, n= 2

Hence,

rate = k [A]¹ [B]²

i.e. rate = k [A] [B]²

Use trail 1, to get the value of k

0.0135 M/s = k [0.33 M] [0.37 M]²

k = 0.298 M^-2 s^-1

Therefore, rate constant for given reaction = 0.298 M^-2 s^-1