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

41.3 minutes

Step-by-step 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)= (0.693)/(K)`

So, fraction of original pressure =
`(1)/(2)^2`

n here is number of half life

therefore,
`(1)/(8)= (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=(2.303)/(t)\log(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 =
`(1)/(8)* (X)=(X)/(8)`

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

`k=(2.303)/(124s)\log(X)/(((X)/(8)))}`

`k=0.0167s^(-1)`

Now we have to calculate the half-life.

`k=(0.693)/(t_(1/2))`

`t_(1/2)=(0.693)/(0.0167s^(-1))`

`t_(1/2)=41.5s`

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