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Anastaziya [24]

3 years ago

5

The evaporation rate of chloroform (CHCl3) in air has been experimentally measured in an Arnold diffusion cell at 25°C and 1.00

atm. Over a period of 10 hr, the chloroform dropped 0.44 cm from the initial surface 7.40 cm from the top of the tube. The density of liquid chloroform is 1.485 g/cm3 and its vapor pressure at 25°C is 200 mm Hg. 1. Assuming air is an ideal gas, determine the molar concentration of air in mol/cm3 . 2. Determine the mole fraction of chloroform in air at the liquid-air interface. 3. Determine the flux of chloroform through the air column in mol/(cm2 s) 4. Determine the diffusion coefficient of chloroform in air in cm2 /s.

1 answer:

Zanzabum3 years ago

5 0

Answer:

1. C = 0.041 mol/L, 2. Mole Fraction = 0.2632, 3. N(A) = 9.322 x 10⁻⁵ mol/cm².s, 4. D(AB) = 5.688 x 10⁻⁵ cm²/s

Explanation:

Use the ideal gas equation

PV = nRT

Can be re4arranged as

P = nRT/V, where n/V = C

Hence

P = CRT

1. Concentration of air

C = P/RT

C = 101325/(8.314 x 298)

C = 40.89mol/m³ = 0.041 mol/L

2. Mole Fraction of chloroform in air at the liquid- air interface

Mole fraction = vapor pressure(chloroform)/vapor pressure(total)

Mole Fraction = 200/760 = 0.2632

3. Flux of chloroform

N(A) = (D(AB)//Z) . (P(T)/RT) .(P(A1) – P(A2))/P(BM)

P(BM) = (760 – 560)/ln(760/560) = 655 mm Hg

D(AB) = RTP(BM) (Z²₁ – Z²₀)/{2PM(A)(P(A1) – P(A2))}

D(AB) = 8.314 x 298 x 655 (0.0758² – 0.074²)/(2 x 101325 x 0.1195 x 200 x 36000)

D(AB) = 5.688 x 10⁴ = 5.688 x 10⁻⁵cm²/s (multiplying by 10⁴ to convert into cm2)

N(A) = (5.688 x 10⁻⁵)/(7.62) x (101325/8 .314 x 298) x (200/655)

N(A) = 9.322 x 10⁻⁵ mol/cm².s

4. D(AB) has already been calculated in the solution of 3

D(AB) = 5.688 x 10⁻⁵ cm²/s

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As we can see, C appears only on two comopunds, CO and CO₂, and since both have 1 C each, their coefficients have to be the same for C to be balanced. However, CO has 1 O and CO₂ has 2, so there is a difference of 1 O betwee them.

The other source of O is Fe₂O₃, that has 3 O. So, we must choose a coefficient for CO and CO₂ such that the difference between the numbers of O is a multiple of 3, that way we can fix this difference with the O from Fe₂O₃. So, we can put coefficients of 3 on both of them:

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That way, we maintained C balanced (3 on each side) and now we have 3 + 3 O on the left side and 6 O on the right side, so the same amount.

Now, we just have to calance Fe, but it is easy since we have it alone in Fe. Since we have 2 on the left side, it is enough to put a coefficient of 2 on Fe to get the balanced reaction:

$Fe_2O_3+3CO\longrightarrow2Fe+3CO_2$

Now, to convert from mass to number of moles, we need the molar masses of the reactants, which we can calculate from the atomic weights of the elemnts in each of them:

$M_{Fe_2O_3}=2\cdot M_{Fe}+3\cdot M_O=(2\cdot55.845+3\cdot15.9994)g/mol=159.6882g/mol$ $M_{CO}=1\cdot M_C+1\cdot M_O=(1\cdot12.0107+1\cdot15.9994)g/mol=28.0101g/mol$

Now, we can convert their masses to number of moles:

$\begin{gathered} M_{Fe_{2}O_{3}}=\frac{m_{Fe_2O_3}}{n_{Fe_{2}O_{3}}} \\ n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_{2}O_{3}}}=\frac{167g}{159.6882g/mol}=1.045787\ldots mol \end{gathered}$ $\begin{gathered} M_{CO}=\frac{m_{CO}}{n_{CO}} \\ n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{85.8g}{28.0101g/mol}=3.063180\ldots mol \end{gathered}$

Now, to determine the limiting reactant, we need to divide both the number of mole by their coefficients on the balanced reaction, so we can see how many we need per reaction of each:

$\begin{gathered} Fe_2O_3\colon\frac{n_{Fe_2O_3}}{1}=\frac{1.045787\ldots mol}{1}=1.045787\ldots mol \\ CO\colon\frac{n_{CO}}{3}=\frac{3.063180\ldots mol}{3}=1.021060\ldots mol \end{gathered}$

Now, the limiting reactant is the one we have less number of moles per reaction. We can see that we have less CO than Fe₂O₃, so the limiting reactant is CO.

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