Question

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1k=3.36×10−5 s−1 at 720 KK. If the initial cyclopropane concentration is 0.0445 MM, what will the cyclopropane concentration be after 235.0 min?

Answer 1

Step-by-step explanation:

r =−dA/dt

= k[A]

Where,

k = rate constant

= 3.36×10^-5 s^-1

Initial concentration, Ao = 0.0445 M

Change in concentration, dA = (0.0445 - y), y is the concentration at time, t (At)

Time,t = 235 min

= 235 × 60

= 14,100 s

3.36×10^-5 × y = (0.0445 - y)/14,100

3.36×10^-5y × 14,100 = 0.0445 - y

0.4738 × y = 0.0445 - y

0.4738y + y = 0.0445

1.4738y = 0.0445

y = 0.0445/1.4738

= 0.0302 M

Answer 2

The concentration would be 0.0302 M.

Step-by-step explanation:

The reaction follows a first-order in cyclopropane.

Rate = kC = change in concentration/time

Let the concentration of cyclopropane after 235 minutes be y

k is rate constant = 3.36×10^-5 s^-1

Initial concentration = 0.0445 M

Change in concentration = (0.0445 - y)

Time = 235 minutes = 235 × 60 = 14,100 s

3.36×10^-5y = (0.0445 - y)/14,100

3.36×10^-5y × 14,100 = 0.0445 - y

0.47376y = 0.0445 - y

0.47376y + y = 0.0445

1.47376y = 0.0445

y = 0.0445/1.47376 = 0.0302 M