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zalisa [80]
3 years ago
10

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

g. The distribution coefficient of benzoic acid in diethyl ether and water is approximately 10. Calculate the amount of acid that would be left in the water solution after four 20-mL extractions with ether.
Chemistry
1 answer:
dimaraw [331]3 years ago
3 0

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

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3 0
4 years ago
Consider the equilibrium
vladimir1956 [14]

Answer:

Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}

\Delta _rG=1.01x10^5J/mol

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar

Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}

Finally, the Gibbs free energy for the reaction at 298.15K is:

\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol

Best regards.

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4 years ago
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xxTIMURxx [149]
Data Given:

                  Initial Volume = V₁ =  36.7 L

                  Initial Pressure = P₁ =  145 kPa

                  Initial Temperature = T₁ =  65 °C + 273 = 338 K

                  Final Volume = V₂ = ?

                  Final Pressure = P₂ = 101.325 kPa           (Standard Pressure)

                  Final Temperature = T₂ = 273 K                (Standard Temperature)

Formula used:
                      As number of moles are constant, so Ideal Gas equation in following form is used,
                                        P₁ V₁ / T₁  =  P₂ V₂ / T₂
Solving for V₂,
                                    V₂  = P₁ V₁ T₂ / T₁ P₂
Putting Values,

                  V₂ = (145 kPa × 36.7 L × 338 K) ÷ (273 K × 101.325 kPa)

                  V₂ = 1798667 ÷ 27661.25

                  V₂ = 65.02 L
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