Question

Using the given data, calculate the rate constant of this reaction: A+B ---> C+D    

Trial              [A] (M)         [B} (M)           Rate (M/s)

1                    0.230           0.330             0.0170

2                    0.230           0.792             0.0979

3                    0.276           0.330             0.0204

Answer 1

Let the rate law for the reaction be,

rate = k[A]^x.[B]^y

with,

k being the rate constant

x and y be order with respect to [A] and [B]

compare trial 1 and 2, [A] is constant and gets cancelled out,

rate1/rate2 = 0.0170/0.0979 = (0.33/0.792)^y

y = 2

compare trial 1 and 3, [B] is same and gets cancelled out,

rate1/rate3 = (0.0170/0.0204) = (0.230/0.276)^x

x = 1

So the rate law becomes,

rate = k[A].[B]^2

Now take trial 1,

k = 0.0170/(0.230) x (0.330)^2 = 0.68 1/M^2.s

Is the rate constant.