In 20 s, the concentration of NO2 decreases from 450 mM to 320 mM in the reaction
2 NO2 --> 2 NO + O2
(a) Determine the rate of reaction of NO2
(b) Determine the rate of formation of O2
(c) What is the unique rate of the reaction?
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rate = 1/2*d(NO2)/dT
d(NO2) = 450-320 = 130
dT = 20
rate formation O2 is 2x rate NO2 decomposition
To determine the rate of reaction and the rate of formation of O2 in this reaction, we'll need to use the given concentration changes and the stoichiometry of the balanced equation.
(a) To determine the rate of reaction of NO2, we'll use the equation:
Rate of reaction = Δ[NO2] / Δt
Where Δ[NO2] represents the change in concentration of NO2, and Δt represents the change in time.
Given: Δ[NO2] = 450 mM - 320 mM = 130 mM
Δt = 20 s
Plugging in these values into the equation, we have:
Rate of reaction = 130 mM / 20 s = 6.5 mM/s
Therefore, the rate of reaction of NO2 is 6.5 mM/s.
(b) To determine the rate of formation of O2, we'll use the stoichiometry of the balanced equation:
2 NO2 --> 2 NO + O2
According to the balanced equation, for every 2 moles of NO2 that react, 1 mole of O2 is formed. Therefore, the molar ratio between NO2 and O2 is 2:1.
Since the given rate is for the reaction of NO2, we'll need to divide it by 2 to find the rate of formation of O2:
Rate of formation of O2 = 6.5 mM/s / 2 = 3.25 mM/s
Therefore, the rate of formation of O2 is 3.25 mM/s.
(c) The unique rate of the reaction is determined by the coefficient of the balanced equation. In this case, since the balanced equation shows that 2 moles of NO2 react to give 2 moles of NO and 1 mole of O2, the unique rate of the reaction would be equal to the rate of reaction of NO2 divided by 2:
Unique rate of the reaction = 6.5 mM/s / 2 = 3.25 mM/s
Therefore, the unique rate of the reaction is 3.25 mM/s.