Question

A sample of pure NO2 is heated to 335 ∘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 0.525 g/L at 0.750 atm .

Answer 1

The equilibrium constant for the reaction is 0.00662

Explanation:

The balanced chemical equation is :

2NO2(g)⇌2NO(g)+O2(g

At t=t  1-2x ⇔ 2x + x moles

The ideal gas law equation will be used here

PV=nRT

here n= $\frac{w}{W}$ = $\frac{w}{V}$= density

P = $\frac{density RT}{M}$       density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm

putting the values in reaction

0.75 = $\frac{0.525 x 0.0821 x 608.15 }{M}$

  M    = 34.61

         

to calculate the Kc

Kc=$\frac{ [NO] [O2]}{NO2}$

  $\frac{1-2x}{1+x}$ x M NO2 + $\frac{2x}{1+x}$ M NO+ $\frac{x}{1+x}$ M O2

Putting the values as molecular weight of NO2, NO,O2

$\frac{46(1-2x) +30(2x)+32x}{1+x}$

34.61= $\frac{46}{1+x}$

x= 0.33

Kc= $\frac{4x^2)x}{1-2x^2}$

    putting the values in the above equation

Kc = 0.00662