Question

Find the rate law expression and the rate law constant for the reaction whose data is in the table below.

Trial	[X]	[Y]	Rate
1	2.0	2.0	2.5 x 10—6
2	2.0	6.0	2.25 x 10—5
3	4.0	2.0	2.0 x 10—5

b. Calculate the rate of the reaction described in part a) when [X] = 0.25 mol/L and [Y] = 6.0 mol/L.

Answer 1

By seeing the first 2 rows of the table, on increasing the concentration of Y by 3 times increased the rate by 8.79 or approximately 9 times (obtained by taking the values of rate in row 2 to 1)

That means reaction is second order.

Since rate = k [A]^x

now A became 3A and rate became 9 rate

thus 3^x = 8.79 or

x log 3 = log 8.79

x = 1.99 which is approximately equal to 2

order is 2 with respect to reactant y

similarly, from the row 1 and 3, the concentration of X is doubled change in rate = 0.8 or approximately

2^x = 0.8

or x = -0.32

rate law expression is

rate = k [X]^-0.3[Y]²

b) For this purpose we have to calculte k first

k = rate/[X]^-0.3[Y]² substitute value from one row of the table

2.5 x 10^-6 / [2]^-0.3 2³

k = 0.39 x 10^-6

Now substitute this k and use new concentrations for finding rate

rate = k [X]^-0.3[Y]²

rate = 0.39 x 10^-6 x (0.25)^-0.3 x 6² = 14.04 x 10^{-6 units of time}

^{Note: if u have provide unit of time for the value of rate in the table use that units here}