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!
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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
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