Question

0.1 mole of N2O4 was sealed in a tube under one atmosphere conditions at 25 degree celcius.Calculate the number of moles of NO2 present,if the equilibrium N2O4-------2NO2(KP=0.14) is reached?PLS HELP ME FRNDS!

Answer 1

Given details:

Pressure P = 1 atm.

No. of moles of N2O4 = 0.1

At equilibrium (when Partial pressure Kp=0.14), the chemical reaction that occurs is:

N2O4  <-----> 2NO2

Solution:

Before reaching equilibrium, no. moles of N2O4  and NO2 are  0.1,  0.

So, total no. of moles = 0.1 + 0 = 0.1

After reaching equilibrium, no. moles of N2O4  and NO2 are 0.1 – x, 2x.

So, total no. of moles = 0.1 - x + 2x = 0.1 + x

Using Dalton's law of partial pressure,

Kp = Pp(NO2) ^2 */ Pp(N2O4)

Kp   =     [P (2x) /(0.1 + x)]^2  / [P * (0.1-x)/(0.1 + x)]

On solving,

Kp = P * 4x^2  / (0.01 -x)

Value of x is negligible, so let's ignore it in denominator.

0.14 = 1 * 4x^2 / 0.01

x^2 = 0.00035

x   =  0.017  

No. of moles of NO2 = 2x  = 0.017×2  = 0.034 mole