At 27°C ,1atm, 20% of N2O4 dissociated into NO2. Determine the partial pressure of each gas at equilibrium and the equilibrium constant Kp at this temperature(assume that the number of mole of N2O4 before dissociation is 1 mole)?

Question

Peyton Adams · Accepted Answer

#P_(NO_2)=0.33atm#

#P_(N_2O_4)=0.67atm#

#K_P=0.163# The dissociation reaction of #N_2O_4# is: #N_2O_4(g)->2NO_2(g)# For this question, the temperature and pressure are constant. If 20% dissociates of #1mol# of #N_2O_4#, therefore, #0.2 mol# will be consumed. Using the #ICE# table: #" " " " " " " " " "N_2O_4(g)->2NO_2(g)# #Initial" " " " " "1mol" " " "0mol# #"Change" " " " "-0.2mol" " " "+0.4mol# #"Equilibrium" " " "0.8mol" " " " " " "0.4mol# Finding the mole fraction of each gas is possible by using the equilibrium number of moles: #chi_(NO_2)=n_(NO_2)/n_("total")=(0.4cancel(mol))/(1.2cancel(mol))=0.33# #chi_(N_2O_4)=n_(N_2O_4)/n_("total")=(0.8cancel(mol))/(1.2cancel(mol))=0.67# or #chi_(N_2O_4)=1-chi_(NO_2) = 0.67# Since the pressure is constant #P_("total")=1atm# #chi_(NO_2)=(P_(NO_2))/P_("total")=>P_(NO_2)=chi_(NO_2)xxP_("total")=0.33xx1atm=0.33atm# #P_(N_2O_4)=chi_(N_2O_4)xxP_("total")=0.67xx1atm=0.67atm# The equilibrium constant #K_P# expression can be written as: #K_P=(P_(NO_2)^2)/P_(N_2O_4)=(0.33)^2/0.67=0.163#

Jeremiah Adams · Answer

undefined At equilibrium, the partial pressures are: $ P_{N_2O_4} = 0.8 \, atm $ and $ P_{NO_2} = 0.2 \, atm $. The equilibrium constant $ K_p $ at this temperature is $ K_p = 0.16 \, atm^2 $.