The rate law depicts the effect of concentration on reaction rate. Second mechanism 2NO(g) ⇄ N₂O₂(g) [fast], N₂O₂(g) + O₂(g) → 2NO₂(g) [slow] is most reasonable. Thus, option b is correct.
<h3>What is rate law?</h3>
Rate law and equation give the rate at which the reaction takes place under the influence of the concentration of the reactants. The balanced chemical reaction is given as,
2NO(g) + O₂(g) → 2NO₂(g)
The rate of the equation is given as,
rate = k [NO]² [O₂]
In a multi-step chemical reaction, the slowest step is the rate-determining step. The second mechanism is given as,
2NO (g) → N₂O₂ (g) [fast]
N₂O₂(g) +O₂(g) → 2NO₂ (g) [slow]
Rate is given as,
rate = k [N₂O₂] [O₂]
Therefore, option b. the second mechanism is the most reasonable.
Learn more about rate law, here:
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Answer:
Energy lost is 7.63×10⁻²⁰J
Explanation:
Hello,
I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5
E = hc/λ(1/n₂² - 1/n₁²)
n₁ = 15
n₂ = 5
hc/λ = 2.18×10⁻¹⁸J (according to the data)
E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)
E = 2.18×10⁻¹⁸ (1/15² - 1/5²)
E = 2.18×10⁻¹⁸ ×(-0.035)
E = -7.63×10⁻²⁰J
The energy lost is 7.63×10⁻²⁰J
Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level