For a reaction, N₂(g) + 3H₂(g) → 2NH₃(g). The rate of formation of NH₃ was found to be 2 x 10^-2 mol/dm³/s. What is the rate of consumption of N₂? 

1 x 10^-2 mol/dm³/s

N₂(g) + 3H₂(g) → 2NH₃(g)

Rate = -\(\frac{1}{1}\)\(\frac{d[N_2]}{dt}\) = -\(\frac{1}{3}\)\(\frac{d[H_2]}{dt}\) = +\(\frac{1}{2}\)\(\frac{d[NH_3]}{dt}\)

\(\frac{d[NH_3]}{dt}\) = 2 x 10^-2 mol/dm³/s

-\(\frac{1}{1}\)\(\frac{d[N_2]}{dt}\) = +\(\frac{1}{2}\)\(\frac{d[NH_3]}{dt}\) = \(\frac{1}{2}\) x 2 x 10^-2 mol/dm³/s

-\(\frac{d[N_2]}{dt}\) = 1 x 10^-2 mol/dm³/s