Question

Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressure of cyclobutene in the reaction vessel drops to one-eighth its original value in 124 seconds. What is the half-life for this reaction at this temperature?

Answer 1

Answer:

41.3 minutes

Explanation:

Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

$t_{1/2}= \frac{0.693}{K}$

So, fraction of original pressure = $\frac{1}{2}^2$

n here is number of half life

therefore, $\frac{1}{8}= \frac{1}{2}^3$

⇒ n= 3

it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.

Answer 2

Answer : The half-life of this reaction at this temperature is, 41.5 seconds.

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant  = ?

t = time passed by the sample  = 124 s

a = let initial amount of the reactant  = X

a - x = amount left after decay process = $\frac{1}{8}\times (X)=\frac{X}{8}$

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

$k=\frac{2.303}{124s}\log\frac{X}{(\frac{X}{8})}}$

$k=0.0167s^{-1}$

Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$t_{1/2}=\frac{0.693}{0.0167s^{-1}}$

$t_{1/2}=41.5s$

Therefore, the half-life of this reaction at this temperature is, 41.5 seconds.