Respuesta :
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
Answer : The time taken to decompose is, 0.0206 seconds.
Explanation :
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]76\text{s}^{-1}[/tex]
t = time passed by the sample = ?
a = initial amount of the reactant = 1.63
a - x = amount left after decay process = [tex][1.63-79\% \times (1.630)]=[1.63-\frac{79}{100}\times (1.63)]=0.342[/tex]
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{76}\log\frac{1.63}{0.342}[/tex]
[tex]t=0.0206s[/tex]
Therefore, the time taken to decompose is, 0.0206 seconds.
