The rate constant for the first-order decomposition of cyclobutane at 500 °C is 9.2 x 10−3 s−1. How long will it take (in seconds) for 80.0% of a sample of C4H8 to decompose?

Respuesta :

The time taken to complete the reaction is 179 seconds.

Rate of reaction?

Rate of reaction is the time taken for a reaction to attain completion. From the data presented, we have;

Initial concentration  = No

Final concentration = No - (0.8No) = 0.2No

rate constant = 9.2 x 10−3 s−1

time taken = t

Given that;

N =Noe^-kt

0.2No =Noe^-( 9.2 x 10−3 t)

0.2 = e^-( 9.2 x 10−3 t)

ln(0.2) = 9.2 x 10−3 t

t = ln(0.2)/-9.2 x 10−3

t = -1.609/-0.009

t= 179 seconds

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