Question

How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?

Answer 1

Complete question:

The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M. How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?

Answer:

The time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours

Step-by-step explanation:

The initial concentration of CH3NC is 3.00 X 10⁻²M =0.03M

The rate constant K= 5.11 X 10⁻⁵s⁻¹

If the concentration of methyl isonitrile to drop to 15.0 %;

The new concentration of methyl isonitrile becomes 0.15 X 0.03 = 0.0045 M

The time taken to drop to 0.0045 M, can be calculated as follows:

`t = -ln[((CH_3NC))/((CH_3NC)_0)]/K`

`t = (-ln[(0.0045)/(0.03)]/5.11 X 10^(-5))X((1 min)/(60 s)) = 618.8 mins`

→ 618.8 mins X 1hr/60mins = 10.3 hours

Therefore, the time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours

Answer 2

The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Step-by-step explanation:

As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

`ln ((C)/(C_0))=-kt`

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as
`5.11 * 10^(-5) \, s^(-1)`

So time t is given as

`ln ((C)/(C_0))=-kt\\ln(0.15)=-5.11 * 10^(-5) * t\\t=(ln(0.15))/(-5.11 * 10^(-5) )\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs`

So the concentration will become 15% of the initial value after 10.31 hrs.