Question

The isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 minutes at OC. The total time it takes for the partial pressure of cyclopropane to decrease from 1.0 atmosphere to 0.125 atmosphere at 500 °C is closest to:_______

(A) 38 minutes
(B) 28.5 38
(C) 76 minutes
(D) 152 minutes

Answer 1

Answer : The total time it takes is, 57 min

Explanation :

Half-life = 19 min

First we have to calculate the rate constant, we use the formula :

`k=\frac{0.693}{19\text{ min}}`

`k=0.0365\text{ min}^(-1)`

Now we have to calculate the time taken for decay.

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

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant

t = time taken top decay = ?

a = initial pressure of the reactant = 1.0 atm

a - x = pressure left after decay process = 0.125 atm

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

`t=(2.303)/(0.0365)\log(1.0)/(0.125)`

`t=57min`

Therefore, the total time it takes is, 57 min