Kc for the reaction N2O4 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?

p (N2O4) = 0.318 atm

p (NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2] = 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 - x) M

[NO2] = 2x

Step 6: Calculate Kc

Kc = 0.619 = [NO2]² / [N2O4]

0.619 = (2x) ² / (0.259-x)

0.619 = 4x² / (0.259 - x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 - x) M = 0.1217 M

[NO2] = 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT) / V

p (N2O4) = (0.0256 * 0.08206 * 318) / 2.10

p (N2O4) = 0.318 atm

p (NO2) = (0.577 * 0.08206 * 318) / 2.10

p (NO2) = 7.17 atm