Question

The gas no reacts with h2, forming n2 and h2o: $$2no(g)+2h2​(g) 2h2​o(g)+n2​(g) if δ[no]/ δt = –24.0 m/s under a given set of conditions, what are the rates of change of [n2] and [h2o]?

Answer 1

The rate of change of [N₂] is –12.0 m/s and the rate of change of [H₂O] is –24.0 m/s, based on the stoichiometry of the reaction and the given rate of NO consumption.

Step-by-step explanation:

The rate at which nitrogen, N₂, and water, H₂O, are formed in the reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) can be deduced from the given rate of consumption of NO. Given the stoichiometry of the reaction, for every 2 moles of NO consumed, 1 mole of N₂ and 2 moles of H₂O are produced.

Since the rate of disappearance of NO (δ[NO]/δt) is –24.0 m/s, we apply stoichiometry to determine the rates of formation for N₂ and H₂O. The rate of formation of N₂ (δ[N₂]/δt) would be half of NO's rate, thus –12.0 m/s. Meanwhile, the rate of formation of H₂O is the same as the rate of disappearance of NO since they have a 1:1 stoichiometric ratio in moles, therefore, δ[H₂O]/δt = –24.0 m/s.

Answer 2

`2NO + 2H_(2)`→
`2H_(2)O+N_(2)`. For this reaction, rate of reaction is decrease in the concentration of NO with respect to time or decrease in the concentration of H₂ with respect to time or increase in the concentration of H₂O with respect to time or increase in the concentration of N₂ with respect to time.

`-(1)/(2) (d[NO])/(dt)= -(1)/(2) (d[H_(2) ])/(dt)=+(1)/(2) (d[H_(2) O])/(dt)= +(d[N_(2) ])/(dt)`

Given,
`(d[NO])/(dt)=24.0 m/s`. So, rates of change of [N₂]=
`-(1)/(2) (d[NO])/(dt)`=
`(1)/(2)`×24.0 m/s=12 m/s. Rate of change of [H₂O] =
`+(1)/(2) (d[H_(2) O])/(dt)=-(1)/(2) (d[NO])/(dt)`= 24.0 m/s.