Question

2N2O5 ---> 4NO2 + O2.
If the rate of formation of O2 is = 9.0×10^-6.
1. What is the rate of formation of Nitrogen dioxide
2. The rate of decomposition of dinitrogen pentoxide
3. Develop the overall rate relationship

Answer 1

Answer:

Solution ::

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Explanation:

$\huge\underline\bold{Answer}$

-1/2d(N205)/dt=1/4d(No2)/dt=d(02)/dt

$\underline{\underline{{\purple{\sf Given}}}}$

the rate of formation of o2 =9.0x10*-6

Answer(1)

$\longrightarrow$The rate of formation of nitrogen dioxide is ::

-1/2d(N2o5)/dt=1/4d(No2)=9.0x10*-6

-1/2d(N2o5)/dt=

1/4d(No2)/dt=9.0x10*-6

d(N02)/dt=36.0x10*-6

Answer(2)

$\longrightarrow$ The rate of decomposition of dinitrogen pentoxide

is

-1/2d(N205)/dt=9.0x10*-6

d(n205)/dt=-18.0x10*-6

Answer (3)

$\longrightarrow$ Develop the overall rate relationship ::

-1/2d(N205)/dt=1/4d(N02)/dt=d(02)/dt