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:

$\frac{R_2}{R_3}=\frac{k[NO]^x[Cl_2]^y}{k[NO]^x[Cl_2]^y}$

$\frac{0.08}{0.16}=\frac{k[0.8]^x[0.3]^y}{k[0.8]^x[0.6]^y}$

$\frac{1}{2}=\frac{(0.3)^y}{(0.6)^y}$

$\frac{1}{2}=(\frac{1}{2})^y$

$y=1$

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