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:

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

where,

k = rate constant = $2.42\times 10^{-2}\text{ hr}^{-1}$

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

$t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}$

$t=17.0\text{ hr}$

Therefore, the time taken will be, 17.0 hr

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