Question

Determine the rate law and the value of k for the following reaction using the data provided.2 NO(g) + O2(g) → 2 NO2(g)[NO]i (M)[O2]i (M)Initial Rate (M-2s-1)0.0300.00558.55 x 10-30.0300.01101.71 x 10-20.0600.00553.42 x 10-2

Answer 1

Answer: Rate law=
`Rate=k[NO]^2[O_2]^1`

Rate law constant is
`1727.3L^2mol^(-2)s^(-1)`

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`2NO(g)+O_2(g)\rightarrow 2NO_2(g)`

`Rate=k[NO]^x[O_2]^y`

k= rate constant

x = order with respect to NO

y = order with respect to
`O_2`

n = x+y = Total order

a) From trial 1:
`8.55 x 10^(-3)=k[0.030]^x[0.0055]^y` (1)

From trial 2:
`1.71 x 10^(-2)=k[0.030]^x[0.0110]^y` (2)

Dividing 2 by 1 :
`(1.71* 10^(-2))/(8.55* 10^(-3))=(k[0.030]^x[0.0110]^y)/(k[0.030]^x[0.0055]^y)`

`2=2^y,2^1=2^y` therefore y=1.

b) From trial 1 :
`8.55 x 10^(-3)=k[0.030]^x[0.0055]^y` (3)

From trial 3:
`3.42* 10^(-2)=k[0.060]^x[0.0055]^y` (4)

Dividing 4 by 3:
`(3.42* 10^(-2))/(8.55* 10^(-3))=(k[0.060]^x[0.0055]^y)/(k[0.030]^x[0.0055]^y)`

`4=2^x,2^2=2^x`, x=2Thus rate law is
`Rate=k[NO]^2[O_2]^1`

Thus order with respect to
`NO` is 2 , order with respect to
`O_2` is 1 and total order is 1+2=3.

Rate law is
`Rate=k[NO]^2[O_2]^1`

b) For calculating k:

Using trial 1:
`8.55* 10^(-3)=k[0.030]^2[0.0055]^1`

`k=1727.3L^2mol^(-2)s^(-1)`

The value of rate constant is
`1727.3L^2mol^(-2)s^(-1)`