Question

When 50 ml of 2M solution of N2O5 was heated, 0.28L of O2 at NTP was fgormed after 30 min. Cal the conc of unreacted N2O5 at that time and find rate of reaction.

Answer 1

Reaction will take place like this

2 N2O5 -> 4 NO + 3 O2

Now Calculate the amount of O2 formed

1 mole = 22.4 L of O2

1L = 1/22.4 Mole of O2

0.28 L at NTP = 0.28 / 22.4 = 0.0125 mol

3 mole oxygen production will use 2 mole N2O5

1 mole oxygen production will use 2/3 mole N2O5

0.0125 mole oxygen will use 0.0125* 2/3 = 0.00833 mol N2O5

Now Calculate the initial and final amount of N2O5

50mL of 2M solution

50 mL = 0.050 L

Molarity = Number of mole / Volume

Number of mole of N2O5 initially present = 0.050 L * 2 (mol/L) = 0.100 mol N2O5 available Initially.

After 30 minutes 0.100 - 0.00833 = 0.0917 mol is left.

The concentration is then 0.0917 mol / 0.050 L = 1.83 mol/Liter

Reaction rate = Mole consumed / Time Taken

Time = 30 min = 30*60 = 1800 sec

Reaction rate = 0.00833 mol / 30 min = 4.63 * 10^-6 mol/sec