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 .

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:

im not exactly sure but i think its b!