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:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $5.11\times 10^{-5}s^{-1}$

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

$t=\frac{2.303}{5.11\times 10^{-5}}\log\frac{100}{14}$

$t=38482.72s=\frac{38482.72}{3600}=10.7hr$

Therefore, the time taken will be, 10.7 hours