How many hours will it take for the concentration of methyl isonitrile to drop to 14.0 %% of its initial value?
1 answer:
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:
where,
k = rate constant =
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
Therefore, the time taken will be, 10.7 hours
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<h3>Further explanation</h3>Given
⁴²K (potassium -42)
Required
The number of hours
Solution
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually, radioactive elements have an unstable atomic nucleus.
Based on Table N(attached), the half-life for ⁴²K is 12.4 hours, which means half of a sample of ⁴²K will decay in 12.4 hours
For three half-life periods :
