Question

How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 %% of its initial value?

Answer 1

This is an incomplete question, here is a complete question.

The rearrangement of methyl isonitrile (CH₃NC) to acetonitrile (CH₃NC) is a first-order reaction and has a rate constant of 5.11 × 10⁻⁵ s⁻¹ at 472 K. If the initial concentration of CH₃NC is 3.00 × 10⁻² M :

How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 % of its initial value?

Answer : The time taken will be, 10.7 hours

Explanation :

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`5.11* 10^(-5)s^(-1)`

t = time passed by the sample = ?

a = let initial amount of the reactant = 100

a - x = amount left after decay process = 14 % of 100 = 14

Now put all the given values in above equation, we get

`t=(2.303)/(5.11* 10^(-5))\log(100)/(14)`

`t=38482.72s=(38482.72)/(3600)=10.7hr`

Therefore, the time taken will be, 10.7 hours