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:

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

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

$t=\frac{2.303}{0.0365}\log\frac{1.0}{0.125}$

$t=57min$

Therefore, the total time it takes is, 57 min