Question

At 1000°C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 76 1/s, to two molecules of ethylene (C2H4). The initial cyclobutane concentration is 1.63. How long (in seconds) will it take for 79% of the cyclobutane to decompose? Enter to 4 decimal places.

Answer 1

Answer : The time taken to decompose is, 0.0206 seconds.

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  = $76\text{s}^{-1}$

t = time passed by the sample  = ?

a = initial amount of the reactant  = 1.63

a - x = amount left after decay process = $[1.63-79\% \times (1.630)]=[1.63-\frac{79}{100}\times (1.63)]=0.342$

Now put all the given values in above equation, we get

$t=\frac{2.303}{76}\log\frac{1.63}{0.342}$

$t=0.0206s$

Therefore, the time taken to decompose is, 0.0206 seconds.

Answer 2

Answer:

It will take 0.0205s for 79% of cyclobutane to decompose

Explanation:

Step 1: Data given

Temperature = 1000 °C

Reaction is first order

rate constant = 76*1/s

The initial cyclobutane concentration is 1.63M

How long (in seconds) will it take for 79% of the cyclobutane to decompose?

When 79% decomposes, there will remain 21 %

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

There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]0, [A], and k, and need to find t.

⇒ with [A]0 = the original amount = 100 % =1

⇒ with [A]t = the amount after it's decomposed = 21 % =0.21

⇒ k = the rate constant = 76/s

⇒ t= the time needed = ?

ln (100/21)  = 76t

t = 0.0205 s

It will take 0.0205s for 79% of cyclobutane to decompose