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svet-max [94.6K]
4 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]4 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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3 years ago
There are 34 qeustions on a test John answers 22 of them correctly what is his percent error
Aleks [24]

Answer:

35.29%

Explanation:

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saw5 [17]

Given parameters:

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Unknown:

Volume in cubic meter and cubic decimeter = ?

Solution:

The ideal gas law is given as;

       PV  = nRT

where P is the pressure on the gas

          V is the volume of the gas

         n is the number of moles

         R is the gas constant

         T is the temperature of the gas

Let us convert to appropriate units;

      1000°C to K;

       K  = 1000 + 273  = 1273K

100000Pa to atm;

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           100000Pa = 0.99atm

 Input the parameters and solve for the unknown;

      0.99 x V  = 2 x 0.082 x 1273

                   V  = 210.9dm³

to m³;

        1000dm³   = 1m³

        210.9dm³ = \frac{210.9}{1000}   = 0.21m³

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Explanation:

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