The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

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

where,

k = rate constant  =

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

a - x = concentration left = 0.053 M

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

Therefore, the time taken will be, 17.0 hr

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.