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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Respuesta :

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

Explanation:

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

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

As the units of rate constant is [tex]min^{-1}[/tex] , the reaction follows first order kinetics.

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

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  =[tex]6.2\times 10^{-4}min^{-1}[/tex]

t = time for decomposition = ?

a = let initial amount of the reactant = 100

x = amount decayed = [tex]\frac{20}{100}\times 100=20[/tex]

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

[tex]t=\frac{2.303}{6.2\times 10^{-4}min^{-1}}\log\frac{100}{100-20}[/tex]

[tex]t=360min[/tex]

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

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