Respuesta :
Answer : The rate of appearance of [tex]O_2[/tex] related to the rate of disappearance of [tex]KO_2[/tex] is,
[tex]\frac{d[O_2]}{dt}=-\frac{3}{4}\frac{d[KO_2]}{dt}[/tex]
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 rate of reaction is,
[tex]4KO_2+2H_2O\rightarrow 4KOH+3O_2[/tex]
The expression for rate of reaction :
[tex]\text{Rate of disappearance of }KO_2=-\frac{1}{4}\frac{d[KO_2]}{dt}[/tex]
[tex]\text{Rate of disappearance of }H_2O=-\frac{1}{2}\frac{d[H_2O]}{dt}[/tex]
[tex]\text{Rate of formation of }KOH=+\frac{1}{4}\frac{d[KOH]}{dt}[/tex]
[tex]\text{Rate of formation of }O_2=+\frac{1}{3}\frac{d[O_2]}{dt}[/tex]
The overall rate expression is:
[tex]\text{Rate of reaction}=-\frac{1}{4}\frac{d[KO_2]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{1}{4}\frac{d[KOH]}{dt}=+\frac{1}{3}\frac{d[O_2]}{dt}[/tex]
Now we have to determine the rate of appearance of [tex]O_2[/tex] related to the rate of disappearance of [tex]KO_2[/tex].
[tex]-\frac{1}{4}\frac{d[KO_2]}{dt}=+\frac{1}{3}\frac{d[O_2]}{dt}[/tex]
or,
[tex]+\frac{1}{3}\frac{d[O_2]}{dt}=-\frac{1}{4}\frac{d[KO_2]}{dt}[/tex]
[tex]\frac{d[O_2]}{dt}=-\frac{3}{4}\frac{d[KO_2]}{dt}[/tex]
Thus, the rate of appearance of [tex]O_2[/tex] related to the rate of disappearance of [tex]KO_2[/tex] is,
[tex]\frac{d[O_2]}{dt}=-\frac{3}{4}\frac{d[KO_2]}{dt}[/tex]
The appearance of Oxygen molecule is directly related or dependent to the rate of disappearance of potassium superoxide (KO2).
The rate of appearance of O2 is directly related to the rate of disappearance of KO2 because potassium superoxide (KO2) has oxygen molecule which is attached to the potassium so when the potassium reacts with water molecules, the potassium hydroxide (KOH ) and oxygen molecules is formed.
If the chemical reaction occurs then we get oxygen molecules so we can conclude that appearance of Oxygen molecule is directly related or dependent to the rate of disappearance of potassium superoxide (KO2).
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