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The first-order decomposition of n2o at 1000 k has a rate constant of 0.76 s-1. if the initial concentration of n2o is 10.9 m, what is the concentration of n2o after 9.6 s

Respuesta :

Erythrosin17
For a first-order reaction, the rate law would be expressed as:

r = dC / dt = -kC

Integrating it from time zero and the initial concentration, Co, to time, t, and the final concentration, C. We will obtain the first-order integrated law as follows:

ln C/Co= -kt

To determine the concentration of N2O in the system at a certain time, we simply substitute the given values from the problem statement as follows:

ln C / Co = -kt
ln C / 10.9 = -0.76 (9.6)
e^ln C / 10.9 = e^-0.76 (9.6)
C / 10.9 = 6.78 x 10^-4
C= 7.39 x 10^-3 m

Therefore, the concentration of N2O in the system after 9.6 s would be 7.39 x10^-3 m.
okpalawalter8

The concentration of [tex]N_20[/tex] is mathematically given as

{N_2O}_t=7.4*10^{-3}M

The concentration of N2O

Question Parameters:

The first-order decomposition of N20 at 1000 k has a rate constant of 0.76 s-1.

if the initial concentration of N20 is 10.9 m,

Generally the equation for the First order Reaction is mathematically given as

At=Ae(-kt)

Therefore

[tex]\frac{At}{A}=exp(-kt)\\\\\frac{N_20_t}{N_20}=e(-kt)\\\\\frac{N_20_t}{N_20}=e(-0.76^{-1}*9.6)[/tex]

{N_2O}_t=7.4*10^{-3}M

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