# For the oxidation of ammonia it was found that the rate of formation of N2 was 0.27 mol L–1 s–1. 4NH3+3O2→2N2+6H2O At what rate was water being formed?

Apr 17, 2015

Water was being formed at a rate of ${\text{0.81 mol L"^(-1)"s}}^{- 1}$.

The key to answering this question lies in the balanced chemical equation, more specifically in the mole ratios that exist between the species involved in the reaction.

$4 N {H}_{3} + 3 {O}_{2} \to \textcolor{red}{2} {N}_{2} + \textcolor{b l u e}{6} {H}_{2} O$

Notice that you have a $\textcolor{red}{1} : \textcolor{b l u e}{3}$ mole ratio between nitrogen and water, which tells you that, for every mole of nitrogen produced by the reaction, three times more moles of water will come along for the ride.

The rates of formation of the products are proportional to the stoichiometric coefficients each product has in the balacned chemical equation

$\frac{1}{\textcolor{red}{2}} \cdot {\underbrace{\frac{\Delta \left[{N}_{2}\right]}{\Delta t}}}_{\text{rate of formation") = 1/(color(blue)(6)) * underbrace((Delta[H_2O])/(Deltat))_("rate of formation}}$

This is equivalent to

$\frac{\Delta \left[{N}_{2}\right]}{\Delta t} = \frac{1}{3} \cdot \frac{\Delta \left[{H}_{2} O\right]}{\Delta t} \implies \frac{\Delta \left[{H}_{2} O\right]}{\Delta t} = 3 \cdot \frac{\Delta \left[{N}_{2}\right]}{\Delta t}$

The calculation will yield

(Delta[H_2O])/(Deltat) = 3 * "0.27 mol L"^(-1)"s"^(-1) = color(green)("0.81 mol L"^(-1)"s"^(-1))