Question

Hydrogen gas and oxygen gas react to form water. The balanced equation for the reaction is the following. 2 H2(g) + O2(g) → 2 H2O(g) Write equations to describe the rate of disappearance of H2(g) and the rate of disappearance of O2(g). Then write a third equation to show how the rates are related. Rate of disappearance of H2(g): rate of disappearance of H2 = − Δ[H2] Δt rate of disappearance of H2 = Δ[H2] Δt rate of disappearance of H2 = Δt Δ[H2] rate of disappearance of H2 = − Δt Δ[H2] Rate of disappearance of O2(g): rate of disappearance of O2 = Δ[O2] Δt rate of disappearance of O2 = Δt Δ[O2] rate of disappearance of O2 = − Δ[O2] Δt rate of disappearance of O2 = − Δt Δ[O2] Relationship between the rates: rate of disappearance of H2 = − 2 mol H2 1 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = 2 mol H2 1 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = 1 mol H2 2 mol O2 × rate of disappearance of O2 rate of disappearance of H2 = − 1 mol H2 2 mol O2 × rate of disappearance of O2

Answer 1

Answer: Thus rate of disappearance of
`H_2=2* {\text {rate of disappearance of}O_2}`

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`2H_2(g)+O_2(g)\rightarrow 2H_2O(g)`

Given: Order with respect to
`H_2` = 2

Order with respect to
`O_2` = 1

Thus rate law is:

`Rate=k[H_2]^2[O_2]^1`

k= rate constant

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

`Rate=-(1d[H_2])/(2dt)=-(1d[O_2])/(dt)`

`Rate=-(1d[H_2])/(dt)=2* -(1d[O_2])/(dt)`

Thus rate of disappearance of
`H_2=2* {\text {rate of disappearance of}O_2}`