The reaction of ch3 ch2 no2 + oh- --> h2 o + ch3 chno2 - is a second order and k at 273 k is 39.1 min-1mol-1 an aqueous solution is 0.005 molar in nitroethane and 0.006 molar in naoh. how long will it take for 90% of the nitroethane to react?

t = 23,4 min

Explanation:

The integrated rate law for a reaction of the form A + B → Products is:

(1)

For the reaction:

CH₃CH₂NO₂ + OH⁻ → H₂O + CH₃CHNO₂

The [A]₀ is [CH₃CH₂NO₂]₀ = 0,005M and [B]₀ = [OH⁻]₀ = 0,006M

90% of [A]₀ is 0,005M×0,9 = 0,0045M consumed

That means [A] = 0,005M - 0,0045M = 0,0005M

And [B] = 0,006M - 0,0045M = 0,0015M

Replacing in (1) knowing k = 39,1min⁻¹mol⁻¹:

ln 2,5 = (39,1min⁻¹mol⁻¹× 0,001M) t

t = 23,4 min

I hope it helps!

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