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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`2N_2O_5(g) -gt4NO_2(g) + O_2(g) `
From the stoichiometry of the reaction:
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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