Respuesta :
Answer:
A. [tex]r=k[NO]^2[H_2][/tex]
B. [tex]k=0.121\frac{L^2}{mol^2*s}[/tex]
C. Second-order with respect to NO and first-order with respect to H₂
Explanation:
Hello,
In this case, for the reaction:
[tex]H_2O(g) + 2NO(g) \rightarrow N_2O + H_2O(g)[/tex]
The rate law is determined by writing the following hypothetical rate laws:
[tex]3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n[/tex]
Whereas we can compute m as follows:
[tex]\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2[/tex]
Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:
[tex]\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1[/tex]
A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:
[tex]r=k[NO]^2[H_2][/tex]
B) Rate constant is computed from one kinetic data:
[tex]k=\frac{1.529x10^{-2}\frac{mol}{L*s} }{(0.6\frac{mol}{L} )^2(0.35\frac{mol}{L})}\\\\k=0.121\frac{L^2}{mol^2*s}[/tex]
C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.
Best regards.
The order with respect to NO has been 2, while the order with respect to water has been 1.
The rate law has been the representation of the chemical concentration responsible for determining the rate of the reaction. The balanced chemical equation has been:
[tex]\rm H_2O\;+\;2\;NO\;\rightarrow\;N_2O\;+\;H_2O[/tex]
A. The rate equation for the reaction has been:
Rate = Rate constant [tex]\rm [H_2O]^m\;[NO]^n[/tex]
Where, m and n are the rate of the respective reactants.
The rate with respect to NO from the given data can be given as:
The ratio of Rate 1 to rate 2 with concentration of NO
[tex]\rm \dfrac{3.822\;\times\;10^-^3}{1.529\;\times\;10^-^2}[/tex] = [tex]\rm \dfrac{[0.3]^m\;[0.35]^n}{[0.6]^m\;[0.35]^n}[/tex]
m = 2
The rate of reaction with respect of NO has been 2.
The rate of reaction with water concentration has been:
[tex]\rm \dfrac{1.529\;\times\;10^-^2}{3.058\;\times\;10^-^2}[/tex] = [tex]\rm \dfrac{[0.6]^m\;[0.35]^n}{[0.6]^m\;[0.7]^n}[/tex]
n = 1
The rate of reaction with respect of water has been 1.
The rate law for reaction has been:
Rate = k [tex]\rm [H_2O]^1\;[NO]^2[/tex]
B. The rate constant can be given as:
Rate = 3.822 [tex]\rm \times\;10^-^3[/tex]
Concentration of NO = 0.6
Concentration of water = 0.35
The rate constant from the rate law can be given as:
3.822 [tex]\rm \times\;10^-^3[/tex] = k [tex]\rm [0.6]^2\;[0.35][/tex]
k = 0.121 [tex]\rm mol^-^2\;L^2\;s^-^1[/tex]
C. The order with respect to NO has been 2, while the order with respect to water has been 1.
For more information about the rate of reaction, refer to the link:
https://brainly.com/question/14221385
