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 .

In this exercise we have to use the knowledge of chemistry to calculate the necessary time that the pressure can hold the system, in this way we can say that this time corresponds to:

First we have to use the reaction equation given as:

This equation can be rewritten in terms of pressure, so it will be;              

Pressure in reactant side, will be:

The integral rate law expression is given as, knowing that the rate constant is  

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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

a. hope i'm right

explanation:

d.) circle motion requires a force to give it a centripetal acceleration.