Respuesta :
Answer:
Explanation:
For first order reaction the equation is
[tex]k = \frac{1}{t} ln\frac{a}{a_t}[/tex]
k is rate constant , t is time , a is initial concentration and a_t is concentration after time t .
Putting the values
[tex]6.82\times 10^{-3} = \frac{1}{2.5\times 60} ln\frac{.02}{a_t}\\[/tex]
[tex]ln\frac{.02}{a_t}[/tex] = 1.023
.02 / a_t = 2.78
a_t = 7.2 x 10⁻³ moles
b ) Let after t time the mole becomes .005
again putting the values in the equation
[tex]6.82\times 10^{-3} = \frac{1}{t\times 60} ln\frac{.02}{.005}\\[/tex]
409.2 t x 10⁻³ = [tex]ln\frac{.2 }{.005\\}[/tex] = 3.689
t = 9.01 min
c ) for half life a_t = a / 2
putting the values in the equation
6.82 x 10⁻³ x 60 x t = ln 2
t = 1.7 min.
a). The number of moles remaining post 2.5 minutes would be:
[tex]= 7.2 [/tex] × [tex]10^{-3} moles[/tex]
b). The number of minutes required for [tex]N_{2} O_{5}[/tex]'s quantity to fall to 0.005 moles would be:
[tex]= 9.01 min[/tex]
c). The half-life of [tex]N_{2} O_{5}[/tex] at 70°C would be:
[tex]= 1.7 minutes[/tex]
a). Given that,
The first-order rate of reaction for [tex]N_{2} O_{5}[/tex] decomposition [tex]= 6.82[/tex] × [tex]10^{-3} s^{-1} [/tex]
Amount of [tex]N_{2} O_{5}[/tex] [tex]= 0.0200 mol[/tex]
Volume of solution [tex]= 2.5 liters[/tex]
by using
[tex]k = (\frac{1}{t})In(\frac{a}{a_{1} })[/tex]
After applying the values to it, we get
[tex]= 6.82[/tex] × [tex]10^{-3} s^{-1} [/tex] [tex]= [1/(2.5 * 60)]In(.02/a_{t}) [/tex]
⇒ [tex].02 / a_t = 2.78[/tex]
∵ [tex]a_{t} [/tex] [tex]= 7.2 [/tex] × [tex]10^{-3} moles[/tex]
b). The number of minutes required for [tex]N_{2} O_{5}[/tex] quantity to fall to 0.005 moles would be:
Let t as the time,
[tex]= 6.82[/tex] × [tex]10^{-3} s^{-1} [/tex] [tex]= (1 /(t * 60) In (.02/.005)[/tex]
∵ [tex]t = 9.01 min[/tex]
c). The half-life of [tex]N_{2} O_{5}[/tex] at 70°C would be:
[tex]a_t = a / 2[/tex]
applying the values in the above equation
[tex]6.82 [/tex] ×[tex]10^{-3} [/tex] × [tex]60 [/tex] × [tex]t = In.2[/tex]
∵ [tex]t = 1.7 min[/tex]
Learn more about "Decomposition" here:
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