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Elena L [17]
3 years ago
11

How to set up the rate expressions for the following mechanism?

Chemistry
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

Explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law = \dfrac{d[D]}{dt}=k_3[B][C] --- (1)

\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]

Using steady-state approximation;

\dfrac{d[B]}{dt}=0

k_1[A]-k_2[B]-k_3[B][C] = 0

[B] = \dfrac{k_1[A]}{k_2+k_3[C]}

From equation (1), we have:

\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}

when the pressure is high;

k₂ << k₃[C]

\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \  \text{first order}

k₂ >> k₃[C]

\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \  \text{second order}

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Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

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c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

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Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

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ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

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