Explanation
In a multi-step reaction, the elementary step having the slowest-rate is always the rate determining step of the overall reaction.
Therefore, a step labeled slow will be the rate determining step.
Correct Answer: Option D
Explanation:
1. 2HI \rightarrow H_{2} + I_{2} \dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{1}\left [ HI \right ]^{2} Therefore, it’s a second order reaction.
2. 2NO_{2} \rightarrow 2NO + O_{2} \dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{2}\left [ NO_{2} \right ]^{2} Therefore, it’s a second order reaction.
3. 2NO + O_{2} \rightarrow 2NO_{2}\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{3}\left [ NO \right ]^{2} \left [ O_{2} \right ]^{1} Therefore, it’s a third order reaction.
4. NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O\dfrac{\mathrm{d} r}{\mathrm{d} t} = K_{4} \left [ NH_{4}NO_{2} \right ]^{1} Therefore, it’s a first order reaction.
Hence, NH_{4}NO_{2} \rightarrow N_{2} + 2H_{2}O is a first order reaction.
Since rate constant of zero order reaction is [A]=[A]_o−kt \implies t = \frac{[A]_o−[A]}{k} \implies t = \frac{a - 0}{k} \implies t =\frac{a}{k}
When zero order kinetic rate law is followed, let [A] is the current concentration, [A]_0 is the initial concentration, and k is the reaction constant and t is time Then relation is given by [A]=[A]_0−k
In order to find the half life we need to isolate t on its own, and divide it by 2. We would end up with a formula as such depict how long it takes for the initial concentration to dwindle by half given by -: \implies t_{1/2}=\frac{[A]_0}{2k}
Since the rate constant of a zero-order reaction is Rate = K.
The unit of rate constant will be mol . L^{-1} min^{-1}.
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