Question

The decomposition of N2O5 in solution in carbon tetrachloride proceeds via the reaction 2 N2O5(soln) → 4 NO2(soln) + O2(soln) The reaction is first order and has a rate constant of 4.82 × 10-3 s-1 at 64°C. If the reaction is initiated with 0.085 mol in a 1.00-L vessel, how many moles remain after 151 s?

Answer 1

Answer: The amount remained after 151 seconds are 0.041 moles

Step-by-step explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant =
`4.82* 10^(-3)s^(-1)`

t = time taken for decay process = 151 sec

`[A_o]` = initial amount of the reactant = 0.085 moles

[A] = amount left after decay process = ?

Putting values in above equation, we get:

`4.82* 10^(-3)=(2.303)/(151)\log(0.085)/([A])`

`[A]=0.041moles`

Hence, the amount remained after 151 seconds are 0.041 moles