# 2N2O5(g) ->4NO2(g) + O2(g) The rate of disappearance of N2O5(g) at a certain temperature is 0.016 mol.L-1min-1. What is the rate of formation of NO2(g) (in mol.L-1.min-1) at this...

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

The rate of disappearance of N2O5(g) at a certain temperature is 0.016 mol.L-1min-1.

What is the rate of formation of NO2(g) (in mol.L-1.min-1) at this temperature?

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### 1 Answer

`2N_2O_5(g) -gt4NO_2(g) + O_2(g) `

From the stoichiometry of the reaction:

`(-d[N_2O_5])/(dt)=1/2*(d[NO_2])/(dt)`

`rArr (d[NO_2])/(dt)=2*((-d[N_2O_5])/(dt))`

where `(dx)/dt` stands for rate of appearance or disappearance of species `x` with respect to time.

Here, the rate of disappearance of `N_2O_5` (g) at a certain temperature is 0.016 mol.L-1min-1.

Therefore, the the rate of formation of `NO_2` (g) at this temperature

= 2*0.016 =**0.032** mol.L-1min-1.

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