Question

Cyclopropane, C3H6, is converted to its isomer propylene, CH2=CHCH3, when heated. The rate law is first order in cyclopropane, and the rate constant is 6.0 × 10−4/s at 500°C. If the initial concentration of cy- clopropane is 0.0226 mol/L, what is the concentration after 525 s?

Answer 1

Answer: 0.011 M

Step-by-step explanation:

`C_3H_6\rightarrrow CH_2=CH-CH_3`

`Rate=k[C_3H_6]^1`

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

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

where,

k = rate constant =
`6.0* 10^(-4)\text{s}^(-1)`

t = time taken = 525 sec

a = initial amount of the reactant = 0.0226 mol/L

a - x = amount left = ?

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

`525=(2.303)/(6.0* 10^(-4))\log(0.0226)/((a-x))`

`(a-x)=0.011mol/L`

Thus the concentration after 525 s is 0.011 mol/L.