Question

The conversion of cyclopropane to propene occurs with a first-order rate constant of . 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 1

To find the time it takes for the concentration of cyclopropane to decrease from 0.080 mol/L to 0.053 mol/L in a first-order reaction, you can use the equation t = -(ln([A]t / [A]0)) / k.

Step-by-step explanation:

The reaction described is the conversion of cyclopropane to propene, and it is a first-order reaction.

First-order reactions follow an exponential decay pattern. The equation for a first-order reaction is:

ln([A]t / [A]0) = -kt

Where [A]t is the final concentration, [A]0 is the initial concentration, k is the rate constant, and t is the time.

To find the time it takes for the concentration of cyclopropane to decrease from 0.080 mol/L to 0.053 mol/L, we can use the equation. Rearranging the equation to solve for t:

t = -(ln([A]t / [A]0)) / k

Plugging in the values, we get:

t = -(ln(0.053 / 0.080)) / 5.95 x 10-4

Solving this equation gives us the time it takes for the concentration to decrease from 0.080 mol/L to 0.053 mol/L.

Answer 2

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:

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

where,

k = rate constant =
`2.42* 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=(2.303)/(2.42* 10^(-2))\log(0.080)/(0.053)`

`t=17.0\text{ hr}`

Therefore, the time taken will be, 17.0 hr