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
