Question

2. A chemist studies the reaction below. 2NO(g) Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and measures the initial reaction rates. Experiment [Cl2] (mol/L) [NO] (mol/L) Initial Rate ((mol/L)/s) 1 0.3 0.4 0.02 2 0.3 0.8 0.08 3 0.6 0.8 0.16 A. Which two experiments should you compare to find the rate order of Cl2

Answer 1

Answer : The rate order of
`Cl_2` is, 1

Explanation :

The given balanced reaction is,

`2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)`

As we know that rate is an experimentally determined value. So, we can not determine the rate law by looking at the reaction.

The general rate law expression will be,

`R=k[NO]^x[Cl_2]^y`

where,

R = rate

k = rate constant

`[NO]` and
`[Cl_2]` = concentration of NO and
`Cl_2` reactant

x and y are the order of the reaction of NO and
`Cl_2` reactant respectively.

To calculate the order with respect to
`Cl_2`, we can compare the rates of two reactions in which the concentration of NO remains the same.

Now we have to compare the rates 2 and 3 experiment because in this NO concentration remains same and we get:

`(R_2)/(R_3)=(k[NO]^x[Cl_2]^y)/(k[NO]^x[Cl_2]^y)`

`(0.08)/(0.16)=(k[0.8]^x[0.3]^y)/(k[0.8]^x[0.6]^y)`

`(1)/(2)=((0.3)^y)/((0.6)^y)`

`(1)/(2)=((1)/(2))^y`

`y=1`

Therefore, the rate order of
`Cl_2` is, 1