Chemistry, 06.11.2019 21:31 katherineweightman

If a 100 kg mol mixture of n-hexane and n-octane containing 50 mole% n-hexane is to be separated by flash distillation @ 1 atm to obtain 40 kg mol in distillated (ie. v=40 kg mol) a) whatare the compositions of the liquid and vapor phases when v = 40 kg mol? b) what is the flash drum temperature?

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Answer from: amandasantiago2001

(a) $x_1=0.34\\x_2=0.66\\y_1=0.74\\y_2=0.24$

(b) 95.24^0C

Explanation:

Hello,

The Rachford-Rice equation should be used in this case:

$\frac{z_1(K_1-1)}{1+V/F(K_1-1)} +\frac{z_2(K_2-1)}{1+V/F(K_2-1)} =0\\\frac{0.5(\frac{P_{1,sat}}{P}-1)}{1+V/F(\frac{P_{1,sat}}{P}-1)} +\frac{z_2(\frac{P_{2,sat}}{P}-1)}{1+V/F(\frac{P_{2,sat}}{P}-1)} =0$

The distribution coefficients could be approximated via Raoult's law and the Antoine equation for the vapor pressure:

$K_i=\frac{P_{i,sat}}{P}=\frac{10^{A-\frac{B}{C+T} }}{P}$

For n-hexane: A=6,9895, B=1216,92 and C=227,451

For n-octane: A=7,14462, B=1498,96 and C=225,874

*Those values and the Antoine law are used in °C and mmHg.

Replacing each value in the Rachford-Rice equation and solving for T, one gets the flash drum temperature as 95.24°C (attached picture), thus, the compositions in both the liquid and the vapor phases are:

$x_1=\frac{z_1}{1+V/F(K_1-1)}=\frac{0.5}{1+0.4*(2.175-1)} =0.34\\x_2=1-x_1=0.66\\y1=K_1*x_1=2.175*0.34=0.74\\y_2=1-y_1=0.24$

Best regards.

If a 100 kg mol mixture of n-hexane and n-octane containing 50 mole% n-hexane is to be separated by

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

Electrons are transferred from atoms of to atoms of this transfer makes the sodium atoms and the

Answer from: Quest

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

$\delta h=e_p-e_r$

$\delta h$ = enthaply

$e_p$ = energy of products

$e_r$ = energy of reactants

as $e_p< e_r$ , thus $\delta h$ comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

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If a 100 kg mol mixture of n-hexane and n-octane containing 50 mole% n-hexane is to be separated by...

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