Question

The reaction ch3-n ≡ c → ch3-c ≡ n is a first-order reaction. at 230.3°c, k = 6.29 ⋅ 10-4s-1. if [ch3 = n≡c] is 1.00 ⋅ 10-3 initially, [ch3 = n≡c] is ________ after 1.000 ⋅ 103 s.

Answer 1

Answer:Concentration of [
`CH_3=N\equiv C] after [tex]1.000* 10^3 s` is
`2.38* 10^(-8) mol/L`.

Step-by-step explanation:

Rate constant of the reaction = k =
`6.29* 10^(-4) s^(-1)`

Initial concentration of
`[A_o]=1.00* 10^(-3) mol/L`

Concentration after time t= [A]

`t = 1.000* 10^3 s`

`log[A]=2.303* \log[A_o]-kt`

`log[A]=2.303* \log[1.00* 10^(-3)]-6.29* 10^(-4) s^(-1)* 1.000* 10^3 s`

`[A]=2.38* 10^(-8) mol/L`

Concentration of [
`CH_3=N\equiv C] after [tex]1.000* 10^3 s` is
`2.38* 10^(-8) mol/L`.

Answer 2

The answer is 5.33 x 10^-4

to find the answer follow these steps:

the reaction is first order reaction

here k = 6.29 x 10^-4 s^-1

initial concentration = 1.00 x 10^-3

After time t = 1.000 x 10^3 s concentration is = ?

So, - 6.29 x 10^-4 x 1.00 x 10^3 + ln (1.00 x 10^-3) = - 7.54

C = e ^ - 7.54= 5.33 x 10^-4 M