The decomposition of dinitrogen pentoxide is a first order reaction with a rate constant of 5.10 x 10-4 s-1 at 318K :
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
If the initial concentration of N2O5 was 0.250 M, what is the concentration after 3.20 mins?
a. 0.249 M
b. 0.227 M
c. 0.275 M
d. 0.199 M
e. 1.29 M
Chemistry - Rate Law question?
2016-05-31 11:59 am
回答 (1)
2016-05-31 12:14 pm
The rate equation for the first order reaction :
Rate = k [N₂O₅]
The integrated-form rate equation :
[N₂O₅] = [N₂O₅]ₒ e^(-kt)
k = 5.10 × 10⁻⁴ s⁻¹ at 318 K
t = 3.20 min = 3.20 × 60 s
[N₂O₅]ₒ = 0.250 M
Substitute into the integrated-form rate equation :
[N₂O₅] = 0.250 e^[-(5.10 × 10⁻⁴) × (3.20 × 60)] M
Concentration after 3.20 min, [N₂O₅] = 0.227 M
...... The answer is : b. 0.227 M
Rate = k [N₂O₅]
The integrated-form rate equation :
[N₂O₅] = [N₂O₅]ₒ e^(-kt)
k = 5.10 × 10⁻⁴ s⁻¹ at 318 K
t = 3.20 min = 3.20 × 60 s
[N₂O₅]ₒ = 0.250 M
Substitute into the integrated-form rate equation :
[N₂O₅] = 0.250 e^[-(5.10 × 10⁻⁴) × (3.20 × 60)] M
Concentration after 3.20 min, [N₂O₅] = 0.227 M
...... The answer is : b. 0.227 M
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