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

Answer 2

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:

$t = 37.1 s$

First we have to use the reaction equation given as:

$2 N_2O_5(g) \rightarrow 4 NO_2 + O_2$

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

$0.110 - 2X + 4X + X = 0.150 atm\\3X = 0.150 - 0.110\\X = 0.0133 atm$

Pressure in reactant side, will be:

$0.110 - 2X \rightarrow 0.110 - 2 (0.0133) = 0.0834 atm$

The integral rate law expression is given as, knowing that the rate constant is  $7.48*10^{-3}*s^{-1}$

$\int\limits^{0.0834}_{0.11} {(7.48*10^{-3}) t} \, dt = t = 37.1 s$

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