For the reaction : 2 NO_{2}^{-} + 4H+ + 2I– → I2 + 2NO + 2H2O The rate of reaction can be written as : \frac{- d \left[\right. NO_{2}^{-} \left]\right.}{dt} = k_{1} \left[\right. NO_{2}^{-} \left]\right. \left(\left[\right. H^{+} \left]\right.\right)^{2} \left[\right. I^{-} \left]\right. \frac{d \left[\right. H^{+} \left]\right.}{dt} = - k_{2} \left[\right. NO_{2}^{-} \left]\right. \left(\left[\right. H^{+} \left]\right.\right)^{2} \left[\right. I^{-} \left]\right. \frac{d \left[\right. NO \left]\right.}{dt} = k_{3} \left[\right. NO_{2}^{-} \left]\right. \left(\left[\right. H^{+} \left]\right.\right)^{2} \left[\right. I^{-} \left]\right.

Engineering

Chemistry

Rate of Chemical Reaction

Order of Reaction and Law of Mass Action

Question

For the reaction :

$2 {NO}_{2}^{-}$ + 4H⁺ + 2I^– → I₂ + 2NO + 2H₂O
The rate of reaction can be written as :

$\frac{- d [{NO}_{2}^{-}]}{dt} = k_{1} [{NO}_{2}^{-}] {[H^{+}]}^{2} [I^{-}]$

$\frac{d [H^{+}]}{dt} = - k_{2} [{NO}_{2}^{-}] {[H^{+}]}^{2} [I^{-}]$

$\frac{d [NO]}{dt} = k_{3} [{NO}_{2}^{-}] {[H^{+}]}^{2} [I^{-}]$

k₁ = k₂ = k₃

$k_{1} = \frac{k_{2}}{2} = k_{3}$

k₁ = 2k₂ = k₃

2k₁ = k₂ = 3k₃

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$- \frac{1}{2} \frac{{dNO}_{2}^{-}}{dt} = \frac{1}{4} \frac{d [H^{+}]}{dt} = \frac{1}{2} \frac{dNO}{dt} = k [{NO}_{2}^{-}] [H^{+}] [I ¯]$

$\Rightarrow \frac{{dNO}_{2}}{dt} = 2 k_{0} X = k_{1} [X]$

$\frac{d [H^{+}]}{dt} = 4 k_{0} X = k_{2} X$

$\frac{dNO}{dt} = 2 k_{0} X = k_{3} [X]$

$K_{0} = \frac{k_{1}}{2} = \frac{k_{2}}{4} = \frac{k_{3}}{2}$

$k_{1} = \frac{k_{1}}{2} = k_{3}$