Exactly 1.0 mol n2o4 is placed in an empty 1.0-l container and is allowed to reach equilibriumdescribed by the equation n2o4(g)↔ 2no2(g). if at equilibrium then2o4is 40.% dissociated, what is the value of the equilibriumconstant (in units of moles per liter) for the reaction under theseconditions?
a. 0.20
b. 0.84
c. 1.1
d. 1.5
e. 2.0

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration .

The given equilibrium reaction is,

Initially                      c                 0

At equilibrium             

The expression of will be,

where,

= degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

Therefore, the value of equilibrium constant for this reaction is, 1.1

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally