Question

The first-order rate constant for the decomposition of N2O5, given below, at 70°C is 6.82 10-3 s-1. Suppose we start with 0.0550 mol of N2O5(g) in a volume of 2.5 L. 2 N2O5(g) → 4 NO2(g) + O2(g) (a) How many moles of N2O5 will remain after 2.5 min? mol (b) How many minutes will it take for the quantity of N2O5 to drop to 0.005 mol? min (c) What is the half-life of N2O5 at 70°C? min

Answer 1

Part A

first order rate equation

k = 2.33 / t log [A0/A]

where k = rate constant 6.82 10-3 s-1

t = time 3 min convert in to sec = 2.5x 60 = 150 sec

A0 = intial concentration = 0.055 mol

A =concentration after 3 min

put all these in above equation

6.82 10-3 s-1 = 2.303 / 150 log[0.055 / A]

log[0.055 / A] = 0.445

[0.055 / A] = 10 ^0.445

[0.055 / A] = 2.79

A = 0.055 / 2.79

A = 0.02 M

Part B

use same formula

we need to find t

A0 = 0.055 mol

A = 0.005 mol

6.82 10-3 s-1 = 2.303 / t log[0.055 / 0.005]

6.82 10-3 s-1 = 2.303 / t  x 1.0414

t = 351.66 s

t = 5.86 min

part C

k = 0.693 /t12

t1/2 = 0.693 / 6.82 10-3 s-1

= 101.613 s

= 1.7 min