Respuesta :
Answer : The rate of NO formation is, 1 M/s
Explanation :
The general rate of reaction is,
[tex]aA+bB\rightarrow cC+dD[/tex]
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
[tex]\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}[/tex]
[tex]\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}[/tex]
[tex]\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}[/tex]
[tex]\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
[tex]Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
The given balanced chemical reaction is:
[tex]N_2(g)+O_2(g)\rightarrow 2NO(g)[/tex]
The rate law expression for this reaction will be:
[tex]Rate=-\frac{d[N_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[NO]}{dt}[/tex]
Given:
Initial concentration of [tex]N_2[/tex] = 0.500 M
Final concentration of [tex]N_2[/tex] = 0.450 M
Time taken = 0.100 s
Now we have to calculate the rate of disappearance of [tex]N_2[/tex]
[tex]Rate=\frac{d[N_2]}{dt}[/tex]
or,
[tex]Rate=\frac{(C_{final}-C_{initial})}{t}[/tex]
[tex]Rate=\frac{(0.500-0.450)M}{0.100s}=0.5M/s[/tex]
Now we have to calculate the rate of formation of [tex]NO[/tex]
The relation between rate of [tex]N_2[/tex] and [tex]NO[/tex] is:
[tex]Rate=-\frac{d[N_2]}{dt}=+\frac{1}{2}\frac{d[NO]}{dt}[/tex]
[tex]2\times \text{Rate of }N_2=\frac{d[NO]}{dt}[/tex]
[tex]2\times 0.5M/s=\frac{d[NO]}{dt}[/tex]
[tex]1M/s=\frac{d[NO]}{dt}[/tex]
or,
[tex]\frac{d[NO]}{dt}=1M/s[/tex]
Thus, the rate of NO formation is, 1 M/s
Based on the data provided and rate of disappearance of N_{2}, the rate of formation of NO is 1.0 M/s.
What is the rate of a chemical reaction?
The rate of a chemical reaction is the rate at which reactants disappear or the rate at which products are formed.
- Reaction rate = change in concentration of reactant/product/time
The equation of the given reaction is as follows:
- N2(g) + O2(g) –> 2NO(g)
Rate of disappearance of N_{2} = (0.500 - 0.450) M /0.1 s
Rate of disappearance of N_{2} = 0.5 M/s
From the equation of reaction;
Rate of formation of NO = 2 × Rate of disappearance of N_{2}
Rate of formation of NO = 2 × 0.5 M/s = 1.0 M/s.
Therefore, the rate of formation of NO is 1.0 M/s.
Learn more about rate of reaction at: https://brainly.com/question/24795637
