This is an incomplete question, here is a complete question.
The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?
Answer : The time taken will be, 17.0 hr
Explanation :
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]2.42\times 10^{-2}\text{ hr}^{-1}[/tex]
t = time passed by the sample = ?
a = initial concentration of the reactant = 0.080 M
a - x = concentration left = 0.053 M
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}[/tex]
[tex]t=17.0\text{ hr}[/tex]
Therefore, the time taken will be, 17.0 hr
