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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