Question

A possible mechanism for the overall reaction, ОН + NO HNO in the presence of N, is: (HO... NO,) (HO...NO2) OH NO NO2 ОН (HO...NO,) N2 HNO N Apply the steady state approximation to the mechanism, what is the rate of the production of HNO,?

Answer 1

Answer :

According to STEADY STATE APPROXIMATION ( SSA) , the concentration of the reaction intermediate is remains constant throughout the reaction i.e. the rate of change in the concentration of the intermediate is zero.

For the given reaction, there is three steps and the mechanism involved the formatiin of a intermediate which is ( HO..... NO₂)^$\neq$​​​​​​

Let consider, INT = (OH.... NO₂)^$\neq$

By considering all the three steps given in the question, Rate of change of the intermediate is,

d[ INT ]/dt = k₁[OH][NO₂] - k_-1 [ INT ] - k₂[ INT ][N₂]

Now according to Steady state approximation (SSA),

d[INT] /dt = 0

Or,   k₁[OH][NO₂] - k_-1 [ INT ] - k₂[ INT ][N₂] = 0

Or, k₁​​​​​​[OH] [NO₂] = k_-1[ INT ] + k₂ [ INT ].[N₂]

Or, k_-1[ INT ] + k₂ [ INT ] [N₂] = k₁ [OH][NO₂​​​​​​]

Or, { k_-1 + k₂ [ N₂ ] } [ INT ] = k₁[OH][NO₂]

Or, [ INT ] = { k₁[ OH ][ NO₂ ] }/ { k_-1 + k₂ [ N₂ ] }

Now from the last step i.e. the 3^rd step, the rate of formation of the HNO₃ is,

d[ HNO₃ ]/dt = k₂ [ INT ].[ N₂]

Now putting the expression of [ INT ] from the SSA calculation,

d[HNO₃ ] /dt = k₂.[ N₂ ].{ k₁[OH][ NO₂ ] }/{ k_-1 + k₂ [ N₂ ] }

Or, d[ HNO₃ ] /dt = k₁.k₂.[ OH ][ NO₂ ][ N₂ ] / { k_-1 + k₂ [ N₂ ] }

So, this is the expression for rate of the formation of HNO₃

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