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svet-max [94.6K]
3 years ago
15

When the system is at equilibrium, it contains NO 2 at a pressure of 0.817 atm , and N 2 O 4 at a pressure of 0.0667 atm . The v

olume of the container is then reduced to half its original volume. What is the pressure of each gas after equilibrium is reestablished
Chemistry
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

Pressure of NO2 = 1.533 atm

Pressure of N2O4 = 0.2344

Explanation:

Step 1: Data given

Pressure of NO2 = 0.817 atm

Pressure of N2O4 = 0.0667 atm

Step 2: The balanced equation

N2 O4  ⇌ 2NO2

Step 3: calculate the total pressure

Total pressure = 0.817 atm + 0.0667 atm

Total pressure = 0.8837 atm

Step 4: Calculate the value of Kp

Kp = p(NO2)² / p(N2O4)

Kp = 0.817²/0.0667

Kp = 10.0

Step 5: Calculate

For the second equilibrium, since the volume of the container is halved, the total pressure will be doubled: 2*0.8837 = 1.7674 atm

The partial pressures of N2O4 and NO2 at the equilibrium will be at equilibrium will be 1.7674−x atm and x atm respectively.

Kp = 10.0 =p(NO2)² / p(N2O4)

10.0 = X² / (1.7674 - X)

X = 1.533

Pressure of NO2 = 1.533 atm

Pressure of N2O4 = 0.2344

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babunello [35]

Answer:

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

Explanation:

                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

Where S = Solubility(mole/lit) and Ksp = Solubility product

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⇒ 27S⁴ = 2.3x10−4

⇒ S = 0.05 mol/lit

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0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole

0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)    \frac{0.15}{3}                \frac{0.24}{1}

                                   = 0.05            = 0.24

Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

                       

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Explanation:

Transfer of mass A into stagnant film B depends on the availability of driving force.

Whereas driving force is the pressure difference at the surface of A and the bulk.

As,       N_{A} \propto (P_{A1} - P_{A2})

           N_{A} = K_{G} \times (P_{A1} - P_{A2})

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Thus, we can conclude that the flux of A from a surface into a mixture of A and B is 0.132 mol/h.ft^{2}

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