Question

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?

Answer 1

t = 23,4 min

Step-by-step explanation:

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

`ln([B][A]_0)/([B]_(0)[A]) = k ([B]_(0) - [A]_0)t` (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([0,0015M][0,005M])/([0,006M][0,0005M]) = k ([0,006M] - [0,005M])t`

ln 2,5 = (39,1min⁻¹mol⁻¹× 0,001M) t

t = 23,4 min

I hope it helps!