Answer:
The order of the reaction is 4.
The rate law of the reaction will be :
[tex]R=k[NO]3[H_2]^1[/tex]
Rate constant of the reaction: k =[tex] 6\times 10^5 M^{-3}s^{-1}[/tex]
Explanation:
[tex]2NO(g) +2H_2 (g)\rightarrow N_2(g) + 2H_2O(g)[/tex]
Let the stoichiometric coefficient of the NO and [tex]H_2[/tex] in rate law be x and y .
Rate of the reaction is given by :
[tex]R=k[NO]^[H_2]^y[/tex]
1) When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]
R = 0.00600 M/s
[tex]0.00600 M/s=k[0.0100 M]^x[0.0100 M]^y[/tex]..[1]
2) When , [tex][NO]=0.0200 M, [H_2]=0.0300 M[/tex]
R = 0.144 M/s
[tex]0.144 M/s=k[0.0200 M]^x[0.0300 M]^y[/tex]..[2]
3) When , [tex][NO]=0.0100 M, [H_2]=0.0200 M[/tex]
R = 0.0120 M/s
[tex]0.0120 M/s=k[0.0100 M]^x[0.0200 M]^y[/tex]..[3]
Dividing [1] and [3]
[tex]\frac{0.00600 M/s}{0.0120 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^y}{k[0.0100 M]^x\times [0.0200 M]^y}[/tex]
y = 1
Dividing [1] and [2]
[tex]\frac{0.00600 M/s}{0.144 M/s}=\frac{k[0.0100 M]^x\times [0.0100 M]^1}{k[0.0200 M]^x\times [0.0300 M]^1}[/tex]
x = 3
The order of the reaction = x + y = 3 + 1 = 4
The rate law of the reaction will be :
[tex]R=k[NO]3[H_2]^1[/tex]
Rate constant of the reaction: k
When , [tex][NO]=0.0100 M, [H_2]=0.0100 M[/tex]
R = 0.00600 M/s
[tex]0.00600 M/s=k[0.0100 M]^3[0.0100 M]^1[/tex]..[1]
[tex]k=\frac{0.00600 M/s}{[0.0100 M]^3[0.0100 M]^1}=6\times 10^5 M^{-3}s^{-1}[/tex]
