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$\frac{1}{2}N_{2}(g)+O_{2}\rightarrow NO_{2}(g)$ .............(K₁)

$2NO_{2}(g)\rightarrow N_{2}O_{4}(g)$ ...............(K₂)

Given that above reactions have equilibrium constants K1 and K2 respectively. What would be the expression for the equilibrium constant K for the following reaction in terms of K₁ and K₂?
Option: 1
$K_{1}K_{2}$

Option: 2
$\frac{1}{K_{1}(K_{2})^{2}}$

Option: 3
$\frac{1}{K_{2}(K_{1})^{2}}$

Option: 4
$\frac{1}{K_{1}K_{2}}$

Answers (1)

best_answer

Rate of reaction -

The rate of the reaction during the courses of a reaction in any instant of time is the change in concentration of reacting species. Rate is a positive quantity.

Unit = $mol \:L^{-} \:sec^{-}=\:[concentration][time]^{-}$

- wherein

$R\rightarrow P$

t = t₁

$R_1\rightarrow P_1$

t = t₂

$R_2\rightarrow P_2$

$\Delta t=t_2-t_1 \:\Delta R=[R_2]-[R_1]$

$\DeltaP=[P_2]-[P_1]$

Rates in presence of stoichiometry of reactants/products -

When stoichiometry coefficients of reactants/ products are not equal to one, the rate of disappearance of & the rate of appearance of products is divided by their respective stoichiometric coefficients

- wherein

$e.g.\:2HI(g)\rightarrow H_{2}(g)+I_{2}(g)$

$r=\frac{-1}{2}.\frac{d}{dt}[HI]$

$=\frac{+d}{dt}[H2]=\frac{+ d}{dt}[I_{2}]$

Instantaneous Rate -

The rate of a reaction calculated at a particular instant of time is called Instantaneous Rate

- wherein

$r_{av}=\frac{-\Delta R}{\Delta t}=\frac{+\Delta P}{\Delta t}$

$\lim_{\Delta\rightarrow 0}r_{av} = r_{inst}$

Formula = $\frac{-d[R]}{dt}= \frac{-d[P]}{dt}$

$N_{2}O_{4}$ ----------> $2NO_{2}$ $\frac{1}{k_{2}}$

$2NO_{2}$ -----------> $N_{2}+2O_{2}$ $\frac{1}{k_{1}^{2}}$

Posted by

sudhir kumar

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