Question

Cyclopropane is converted into propene via the reaction below: H₂C CH₂ CH₂ --> H H C = C H CH3. At 750K, the k value is 1.8 x10⁻⁴ sec-¹. If the initial concentration is 0.15 M, how much will remain after 55 seconds?

Answer 1

The amount of cyclopropane remaining after 55 seconds can be determined using the first-order rate equation. With an initial concentration of 0.15 M and a rate constant of 1.8 x 10^-4 sec^-1, approximately 0.128 M of cyclopropane will remain after 55 seconds.

The given reaction is:

H2C CH2 CH2 → H H C = C H CH3

The rate constant (k) for this reaction is given as 1.8 x 10-4 sec-1 at 750K.

To determine the amount of cyclopropane remaining after 55 seconds, we need to use the first-order rate equation:

[A] = [A]0 * e-kt

Where [A] is the concentration of the reactant at time 't', [A]0 is the initial concentration, 'k' is the rate constant, and 't' is the time.

Let's substitute the given values:

[A]0 = 0.15 M

k = 1.8 x 10-4 sec-1

t = 55 seconds

Now, let's calculate the concentration of cyclopropane ([A]) after 55 seconds:

[A] = 0.15 * e-(1.8 x 10-4) * 55

[A] ≈ 0.128 M

Therefore, approximately 0.128 M of cyclopropane will remain after 55 seconds.