Question

At 700 K, the rate constant for the following reaction is 6.2x 10-4 min-1. how many minutes are requiredfor 20 percent of a sample of cyclopropane to isomerize topropane?

C3H6(cyclopropane)-->C3H6(propane)
A. 1,120 min
B. 360 min
C. 3710 min
D. 1.4 x 10-4 min
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Answer 1

Answer: 360 minutes are required for 20 percent of a sample of cyclopropane to isomerize topropane.

Step-by-step explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

As the units of rate constant is
`min^(-1)` , the reaction follows first order kinetics.

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

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

where,

k = rate constant =
`6.2* 10^(-4)min^(-1)`

t = time for decomposition = ?

a = let initial amount of the reactant = 100

x = amount decayed =
`(20)/(100)* 100=20`

a - x = amount left after decay process = 100-20= 80

`t=(2.303)/(6.2* 10^(-4)min^(-1))\log(100)/(100-20)`

`t=360min`

Thus 360 minutes are requiredfor 20 percent of a sample of cyclopropane to isomerize topropane.