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SashulF [63]
3 years ago
10

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 .
Chemistry
1 answer:
vodka [1.7K]3 years ago
5 0

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

     

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