Question:
In the reversible reaction,
$2 NO _{2} \ce{->[{k_{1}}][{K_{2}}]} N _{2} O _{4}$
the rate of disappearance of $N O_{2}$ is equal to
In the reversible reaction,
$2 NO _{2} \ce{->[{k_{1}}][{K_{2}}]} N _{2} O _{4}$
the rate of disappearance of $N O_{2}$ is equal to
Updated On: Apr 26, 2024
- $\frac{2k_{1}}{k_{2}}\left[NO_{2}\right]^{2}$
- $2k_{1}\left[NO_{2}\right]^{2} - 2k_{2}\left[N_{2}O_{4}\right]$
- $2k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$
- $\left(2k_{1} - k_{2}\right) \left[NO_{2}\right]$
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The Correct Option is C
Solution and Explanation
$2NO_{2} \ce{->[{k_1}][{k_2}]} N_{2}O_{4}$
Rate = $-\frac{1}{2} \frac{d\left[NO_{2}\right]}{dt} = k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$
$\therefore$ Rate of disappearance of $NO_{2}$ i.e.,
$-\frac{d\left[NO_{2}\right]}{dt} = 2k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$
Rate = $-\frac{1}{2} \frac{d\left[NO_{2}\right]}{dt} = k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$
$\therefore$ Rate of disappearance of $NO_{2}$ i.e.,
$-\frac{d\left[NO_{2}\right]}{dt} = 2k_{1}\left[NO_{2}\right]^{2} - k_{2}\left[N_{2}O_{4}\right]$
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Concepts Used:
Rate of a Chemical Reaction
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or products per unit time.
Consider the reaction A → B,
Rate of the reaction is given by,
Rate = −d[A]/ dt=+d[B]/ dt
Where, [A] → concentration of reactant A
[B] → concentration of product B
(-) A negative sign indicates a decrease in the concentration of A with time.
(+) A positive sign indicates an increase in the concentration of B with time.
Factors Determining the Rate of a Reaction:
There are certain factors that determine the rate of a reaction:
- Temperature
- Catalyst
- Reactant Concentration
- Chemical nature of Reactant
- Reactant Subdivision rate