For the reaction, K=0.538 at 313 K. N2O4(g)⇌2NO2(g)
If a reaction vessel initially contains an N2O4 concentration of 0.0650 molL−1 at 313 K, what are the equilibrium concentrations of N2O4 and NO2 at 313 K? (Hint: Ensure use of proper units for the equilibrium constant.)
Sol.
Reaction : N2O4(g) <---> 2NO2(g)
initial 0.0650 0
change - x + 2x
equilibrium 0.0650 - x 2x
As Equilibrium constant = K = [NO2]2 / [N2O4]
0.538 = (2x)2 / ( 0.0650 - x )
4x2 = 0.538 ( 0.0650 - x ) = 0.03497 - 0.538x
4x2 + 0.538x - 0.03497 = 0
x2 + 0.1345x - 0.0087 = 0
Solving this quadratic equation ,
x = ( - 0.1345 +- ( 0.1345*0.1345 - 4*1*(-0.0087) )1/2 ) / 2
x = (- 0.1345 +- 0.2299 ) / 2
So , x = ( - 0.1345 + 0.2299 ) / 2 = 0.0477
or , x = ( - 0.1345 - 0.2299 ) / 2 = - 0.1822
But x cannot be negative
So , x = 0.0477
Therefore , Equilibrium Concentrations are :
[NO2] = 2x = 2*0.0477 = 0.0954 M
[N2O4] = 0.0650 - x = 0.0650 - 0.0477 = 0.0173 M
For the reaction, K=0.538 at 313 K. N2O4(g)⇌2NO2(g) If a reaction vessel initially contains an N2O4...
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