Answer:
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
