Question

Help with Rate Constant Calculations?

The isomerization reaction CH3NC → CH3CN obeys the first-order rate law,
rate = k[CH3NC],
in the presence of an excess of argon. Measurement at 500. K reveals that in 485 seconds, the concentration of CH3NC has decreased to 73% of its original value. Calculate the rate constant (k) of the reaction at 500. K.

___s−1

(The integrated form for the first-order rate law can be written in the general terms
ln[A]t − ln[A]0 = −kt, where
[A]0
is the initial concentration of reactant A,

[A]t
is the concentration of A at time t, and k is the rate constant.)

Answer 1

`k = 0.00065 s^-1`

Explanation:

`ln[A]_t − ln[A]_0` can be written as `ln([A]_t/[A]_0)`

The problem says that at t = 485 s, `[A]_t` = 73% `[A]_0`
Therefore `[A]_t = 0.73 [A]_0`

`ln([A]_t/[A]_0) = -kt`
`ln(0.73 [A]_0/[A]_0) = -k(485)`
`ln (0.73) = -k(485)`
`-0.3147 = -k(485)`

`k = 0.00065 s^-1`

Answer 2

Consider,

`CH_3NC to CH_3CN` where `R(t) = k[CH_3NC]`

The nitrile group is much more stable than the odd configuration nitrogen is in the first isomer of that molecule (it even has a formal charge).

Temperature in this case is merely descriptive and perhaps to illustrate that reaction rates are generally proportional to temperature.

This is a first order reaction. Recall,

`ln[A]_t = -kt + ln[A]_0`
`=> (ln[A]_t)/(ln[A]_0) = -kt`

Hence,

`ln(0.73M)/(1.0M) = -k * 485s`
`therefore k approx 6.5*10^-4s^-1`

is the rate constant of this reaction given these data.