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In a reversible reaction $\text{ 2N}{{\text{O}}_{\text{2}}}\text{ }\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ the rate of disappearance of $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$ is equal to:


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Last updated date: 14th Jun 2024
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Answer
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Hint: in a reversible reaction, the net disappearance of the reactant A species is equal to the difference in the rate of disappearance of reactant in forwarding reaction and rate of formation of reactant in the reverse reaction.it is represented as,
$\text{ }-\dfrac{d\left[ \text{A} \right]}{dt}\text{ = }{{\text{k}}_{f}}\left[ \text{A} \right]\text{ }-{{\text{k}}_{r}}\left[ \text{A} \right]\text{ }$
Where A is a reactant,$\text{ }{{\text{k}}_{f}}\text{ }$ is the rate constant for the forward reaction and $\text{ }{{\text{k}}_{r}}\text{ }$ is the rate constant for the backward reaction.

Complete step by step solution:
The reversible reaction follows a simple relation between the rate constant of forwarding and backward reaction. To understand the relation we will study the rate of reaction for forwarding and backward reaction.
Consider a reversible reaction of the nitrogen dioxide to give nitrogen trioxide. The reaction of $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$ and $\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ is given as follows,
$\text{ 2N}{{\text{O}}_{\text{2}}}\text{ }\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$
Where$\text{ }{{\text{K}}_{\text{1}}}\text{ }$ is the rate constant for forwards reaction and $\text{ }{{\text{K}}_{2}}\text{ }$is the rate constant for the backward reaction.
Forward reaction: two molecules of nitrogen dioxide react to form dinitrogen tetroxide. Thus the rate of forwarding reaction would be proportional to the concentration of$\text{ N}{{\text{O}}_{\text{2}}}\text{ }$. Thus rate for the forward reaction is written as,
$\text{ }-\dfrac{d\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}{dt}\text{ = Rat}{{\text{e}}_{\text{forward}}}\text{ = 2}{{\text{k}}_{1}}{{\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}^{\text{2}}}\text{ }$
Backward or reverse reaction: one molecule of dinitrogen tetroxide decomposes to form two molecules of nitrogen dioxide. Thus the rate of backward reaction would be proportional to the concentration of dinitrogen tetroxide. Thus rate for the reverse reaction is written as,
$\text{ }\dfrac{d\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}{dt}\text{ = Rat}{{\text{e}}_{\text{backward}}}\text{ = }{{\text{k}}_{2}}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{4}} \right]\text{ }$
But the rate of reverse reaction can be expressed as the rate of formation of nitrogen dioxide.
We are interested to determine the rate of disappearance of the nitrogen dioxide in the reversible reaction. Rate of the disappearance of $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$ in the reversible reaction would be equal to the difference between the rate of disappearance of $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$ in the forward reaction and rate of formation of nitrogen dioxide$\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$. Thus the rate of disappearance is as follows,
$\text{ }-\dfrac{d\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}{dt}\text{ = 2}{{\text{k}}_{1}}{{\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}^{\text{2}}}-{{\text{k}}_{2}}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{4}} \right]\text{ }$
Therefore in a reversible reaction of nitrogen dioxide and $\text{ }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ the rate of disappearance of $\text{ N}{{\text{O}}_{\text{2}}}\text{ }$ is equal to,
$\text{ Rate of disappearance of N}{{\text{O}}_{\text{2}}}=\text{ 2}{{\text{k}}_{1}}{{\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}^{\text{2}}}-{{\text{k}}_{2}}\left[ {{\text{N}}_{\text{2}}}{{\text{O}}_{4}} \right]\text{ }$

Note: Note that, in reversible reaction conversion of reactant to product and product to reactant takes place simultaneously. Thus the rate of formation or disappearance of a reactant or product are always related to each other. If the system is at equilibrium. Therefore, the rate of forwarding reaction is equal to the rate of backward reaction. Therefore at equilibrium, we have,
$\text{ Rat}{{\text{e}}_{\text{forward}}}\text{ = Rat}{{\text{e}}_{\text{backward}}}\text{ }$