For the reaction: N2O5 (g) / longrightarrow& longrightarrow; 2NO2 (g) + 1/2O2 (g)

If N2O5 is decomposing with an instantaneous rate of 7.81 mol/L·s, what is the instantaneous rate of formation of NO2?

I don't know where to begin

The given equation from the problem above is already balance,

N2O5 - - - > 2NO2 + 0.5O2

Since, in every mole of N2O5 consumed, 2 moles of NO2 are formed, we can answer the problem by multiplying the given rate, 7.81 mol/L. s with the ratio.

(7.81 mol/L. s) x (2 moles NO2 formed / 1 mole of N2O5 consumed)

= 15.62 mol/L. s

The answer is the rate of formation of NO2 is approximately 15.62 mol/L. s.