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In: Chemistry

NO2 (g) + OH• (g)à HNO3 (g) The rate constant for this reaction is 1.2x10-11 cm3...

NO₂ (g) + OH• (g)à HNO₃ (g)

The rate constant for this reaction is 1.2x10^-11 cm³ molecule^-1s^-1.

Write the rate law for this reaction.
hydroxyl radical concentration in atmosphere is 2.0 x10⁶ molecules cm^-3 at 25 ^oC, find the atmospheric lifetime of NO₂ at 25 ^oC (in hours)
Use the information from first second , determine the concentration of hydroxyl radicals in ppbv at 25 ^oC and 1.0 atm.

Expert Solution

NO₂ (g) + OH^•(g) ==> HNO₃ (g)

The rate constant for this reaction is 1.2x10^-11 cm³ molecule^-1s^-1.

the rate law for this reaction : it is 2nd order reaction ( as also evident from unit of rate constant)

rate = K₂ [NO₂ ][OH^•]

[OH^•] in atmosphere = 2.0 x10⁶ molecules cm^-3 at 25 ^oC,

find the atmospheric lifetime of NO₂ at 25 ^oC:

from 2nd order kinetics , (^T(NO₂)) lifetime of NO₂ can be calculated :

^T(NO₂) = 1/K₂ [NO₂ ][OH^•] = 1/ ((1.2x10^-11 cm³ molecule^-1s^-1)(2.0 x10⁶ molecule cm^-3 ))

^T(NO₂) = 41666.7 sec = 11.57 Hrs.

concentration of hydroxyl radicals in ppbv at 25 ^oC and 1.0 atm:

[OH^•] in atmosphere = 2.0 x10⁶ molecules cm^-3 at 25 ^oC

or in molar terms : 2.0 x10⁶ molecules cm^-3 / 6.023*10²³ molecule/mol = 3.32*10^-18 moles cm^-3

amount of OH^• = 17 g/mol *3.32*10^-18 moles cm^-3 = 5.65 *10^-17 g cm^-3 = 5.65 *10^-11 ug cm^-3

( 1g = 10⁶ microgram (ug))

since 1 ppbv = 1 ug/L ; 1 cm3 = 0.001 L

thus

concentration of hydroxyl radicals in ppbv : 5.65 *10^-8 (ppbv)

queen_honey_blossom answered 1 year ago

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