Question

Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise to 0.150 atm .

Answer 1

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

                  2 N2O5(g)  --> 4 NO2 + O2

Initial:          0.110                  -             -

change:        -2x                  +4x        +x

Final:          0.110 - 2x           +4x        +x

But final = 0.150atm;

0.110 - 2x    +  4x   +  x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k =  rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3)  t

upon solving, t = 37.1 s