Answer:
A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
(4x^2)x
Kc= -----------
(A-2x)^2
PV=nRT
n/v = P/RT = .745/(0.0821)(334+273) = .01495
To Find the initial molarity of NO2
(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L
Thus:
46(A-2x) + 2x(30) + 32x = .515 g/L
46A-92x+60x+32x = .515
46A=.515
A=.01120 M
Using the total molarity found
(A-2x)+2x+x = .01495 M
A+x=.01495
Plug in A found into the above equation:
.01120+x = .01495
x=.00375
Now Plug A and x into the original Equilibrium Constant Expression:
(4x^2)x
Kc= -----------
(A-2x)^2
Kc = 0.000014
Explanation:
