A sample of pure NO2NO2 is heated to 335 ∘C∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g)2NO2(g)⇌2NO(g)+ O2(g) At equilibrium the density of the gas mixture is 0.525 g/Lg/L at 0.750 atmatm .Calculate Kc for the reaction

1. 1 × 10⁻²

Explanation:

Given that:

Temperature = 335 ° C = (335+ 273)K = 608

Pressure = 0.750 atm

Volume = 1 Litre

number of moles of NO2 = ???

Rate Constant =0.0821 L atm /K/mol

Using the Ideal gas equation

PV = nRT

n =

n =

n = 0.015

n = 1.5 × 10⁻² mole

Density = 0.525 g/L

The equation for the reaction can be illustrated as:

                     2NO2(g)         ⇌          2NO(g)         +         O2(g)

For the ICE table; we have:

Initial                 x                                   0                              0

Change            -2y                               + 2y                          +y

Equilibrium        (x - 2y)                        2y                             y

Total moles at equilibrium = (x-2y)+2y+y

= x + y moles

However,

1.5 × 10⁻² mole of the mixture has a mass of 0.525 g

i.e x + y moles = 1.5 × 10⁻² mole

Now, molar mass of 1 mole of NO2 = 46g/mol

Since number of moles =

mass of (x-2y) moles = 46 × (x-2y) g

Molar mass of NO = 30 g/mol

Also, mass of NO = 2y × 30 = 60y

Molar mass of O2 = 32 g/mol

Mass of O2 = y × 32 = 32y

Total mass = ( 46x - 90y)+60y+32y = 0.525

46x = 0.525

x =

x = 0.0114

x = 1.14 × 10⁻²

x + y moles = 1.5 × 10⁻²

y =  1.5 × 10⁻² -  1.14 × 10⁻²

y = 0.0036

y = 3.6 × 10⁻³

At equilibrium

[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M

[NO] = 2 ( 3.6 × 10⁻³)  = 7.2 × 10⁻³ M

[O2] = 3.6 × 10⁻³ M

0.011

1. 1 × 10⁻²

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