Question

the gas phase decompostion of no2 into no and o2 is second order and has a rate constant o 1.1 if the intial concentrtion of no2 is 1.9*10^-2 m, how long will it take for 75% of the no2 to decompose/ what is the new concentration of no2 after 75% has decomposed ?

Answer 1

For 2^nd order reaction, rate law can be written as

1/C_A -1/C_A0=-kt

Here, C_A0=1.9*10^-2 M

k=1.1 M^-1 s^-1

Fraction of NO₂ decomposed, x=75%=0.75

Concentration of NO₂ reamining=(1-x)*C_A0

                                                                =(1-0.75)*1.9*10^-2=0.48*10^-2 M

Putting these values in 2^nd order rate equation

(1/1.9*10^-2) –(1/0.48*10^-2)=-1.1*t

t=141.5 sec