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)?

1 Answer
Nov 23, 2015

P_(NO_2)=0.33atm

P_(N_2O_4)=0.67atm

K_P=0.163

Explanation:

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

From the equilibrium number of moles, we can find the mole fraction of each gas:

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