Respuesta :
Answer:
a. 1,157x10⁻⁵M/s
b. 5,787x10⁻⁶M/s
Explanation:
For the reaction:
2H₂O₂(aq) → 2H₂O(l) + O₂(g).
a. The rate law of descomposition is:
[tex]rate=-\frac{1}{2} \frac{d[H_{2}O_{2}]}{dt}=\frac{d[O_{2}]}{dt}[/tex]
Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:
[tex]rate=-\frac{1}{2} \frac{0,500M}{-2,16x10^4s}[/tex]
[tex]rate=1,157x10^{-5}M/s[/tex]
As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s
b. Between 2,16x10⁴s and 4,32x10⁴s, rate law is:
[tex]rate=-\frac{1}{2} \frac{0,500M-0,250M}{2,16x10^4s-4,32x10^4s}[/tex]
[tex]rate=5,787x10^{-6}M/s[/tex]
The rates are 5,787x10⁻⁶M/s
I hope it helps!
The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.
Thus the rate is,
[tex]rate=1157\times10^{-5}[/tex]
What is rate law?
The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.
The balanced reaction is:-
[tex]2H_2O_2(aq)----->2H_2O(l)+o_2(g)[/tex]
The rate law of decomposition:-
[tex]rate=-\dfrac{1}{2}\dfrac{d[H_2O_2]}{dt}[/tex]
Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:
[tex]rate=-\dfrac{1}{2}\dfrac{0.5}{2.16\times10^{4}}[/tex]
[tex]rate=1157\times10^{-5}[/tex]
As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s
For more information about the rate of reaction, refer to the link:-
brainly.com/question/16759172