Question

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

Answer 1

The answer is 0.405 M/s

- (1/3) d[O2]/dt = 1/2 d[N2]/dt

- d[O2]/dt = 3/2 d[N2]/dt

- d[O2]/dt = 3/2 × 0.27

- d[O2]/dt = 0.405 mol L^(-1) s^(-1)