When 50 mL of 2M solution of `N_2 O_5` was heated, 0.28 L of `O_2` at NTP was formed after 30 minutes. Calculate the concentration of unreacted `N_2 O_5` at that time and also find the average rate of reaction.

Answer

`1.5 M;0.000278 Ms^(-)`

Explanation

`22.4 L N_2 O_5 at` NTP = 1 mol
` 0.56 L N_2 O_5` at `NTP= (1)/(22.4 ) xx 0.56 = 0.025 mol`
` therefore ` No. of moles of `N_2 O_3` decomposed = 0.025 mol
Total number of moles of `N_2 O_5` in 50 mL 2M `N_2 O_5`
`= ( 2)/(1000) xx 50 = 0.100 mol`
Number of moles of `N_2 O_5` left undecomposed =0.100 -0.025 =0.075 mol
Molarity of unreacted `N_2 O_5 =(0.075 )/( 50 ) xx 1000 = 1.5 M`
Average rate ` = (2M - 1.5 M)/( 30 xx 60 sec -0 ) = ( 0.5 M )/( 1800 s)`
`=2.78 xx 10^(-4) Ms^(-1)`