Respuesta :

Answer:

A) Rate = -k [N₂O]²

B) Rate = 0.0031 M/s

C) Rate of change in the concentration of N₂O = -0.0062 M/s

Explanation:

2N₂O(g)-->2N₂(g)+O₂(g)

A) To first guess what the rate of reaction light be

Rate = -k [N₂O]² (minus sign because the concentration of reactant is reducing)

B) Rate of reaction = (Change in concentration of reactants)/time

It is also equal to (Change in concentration of products too)/time.

So,

Rate of reaction = (Change in concentration of O₂)/time

Change in concentration of O₂ = (0.019/0.47) - 0 = 0.0404 M

Time = 13.0 s

Rate of reaction = (0.0404/13) = 0.00311 M/s = 0.0031 M/s to 2 s.f.

C) Rate of change in the concentration of N2O over this time interval.

(1/2) -[Δ(N₂O)/Δt] = (Δ(O₂)/Δt)

[Δ(N₂O)/Δt] = -2 × (Δ(O₂)/Δt)

[Δ(N₂O)/Δt] = -2 × 0.0031 = -0.0062 M/s

Hope this Helps!!!

The solution to the three given questions for rate of reaction, the average rate of reaction over a time interval and prediction of rate of change of 2N₂O  over the given time interval respectively are;

A)  [tex]-\frac{1}{2} \frac{d[N_{2}O]}{dt} = \frac{1}{2} \frac{d[N_{2}]}{dt} = \frac{1}{2} \frac{d[O_{2}]}{dt}[/tex]

B) 0.00311 M/s

C) -0.00622 M/s

The complete question is given in the comment section under the question.

From the question, we have the reaction;

2N₂O (g) → 2N₂ (g) + O₂ (g)

We are given;

  • Moles of O₂ = 1.9 × 10⁻² mol
  • Volume of vessel = 0.470 L
  • Change in time = 13 s

A) The expression for the rate of reaction in terms of change in concentration of each reactant and products is;

Rate = [tex]-\frac{1}{2} \frac{d[N_{2}O]}{dt} = \frac{1}{2} \frac{d[N_{2}]}{dt} = \frac{1}{2} \frac{d[O_{2}]}{dt}[/tex]

B) At t = 0 s, concentration of O₂ is;

M = number of moles of O₂/Volume of vessel

M = 1.9 × 10⁻²/0.47

M = 0.0404 mol/L

  • The average rate of formation of  O₂  from t = 0 to 13 s is gotten from the formula:

Average rate of reaction = Δ[O₂]/Δt

Average rate of reaction = 0.0404/13

Average rate of reaction = 0.00311 M/s

C) The rate of decomposition for N₂O for t = 0 s to 13 s is;

Rate = [tex]-\frac{1}{2} \frac{d[N_{2}O]}{dt} = \frac{d[O_{2}]}{dt}[/tex]

Since  Δ[O₂]/Δt = 0.00311 M/s

Then;

[tex]-\frac{1}{2} \frac{d[N_{2}O]}{dt} = 0.00311 M/s[/tex]

Thus;

[tex]\frac{d[N_{2}O]}{dt} = -2 * 0.00311 M/s[/tex]

[tex]\frac{d[N_{2}O]}{dt} = -0.00622 M/s[/tex]

Thus, the concentration of N₂O will decrease with time

Read more about Change in concentration at; https://brainly.com/question/13524327

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