Sol.
Initial moles of N2 = 0.180 mol
Volume of vessel = 4 L
So , Initial concentration of N2
= 0.180 mol / 4 L = 0.045 M
Also , Initial moles of O2 = 0.650 mol
Initial concentration of O2
= 0.650 mol / 4 L = 0.1625 M
Reaction : N2(g) + O2(g) <----> 2NO(g)
Initial 0.045 0.1625 0
change - x - x + 2x
equilibrium 0.045 - x 0.1625 - x 2x
So , Kc = [NO]2 / ( [N2] × [O2] )
1.70 × 10-3 = (2x)2 / ( ( 0.045 - x ) ( 0.1625 - x ) )
Assuming that x is very small , so ,
0.045 - x = 0.045
and 0.1625 - x = 0.1625
Therefore ,
1.70 × 10-3 = 4x2 / ( 0.045 × 0.1625 )
x2 = 0.0000031078
x = 0.00176
Equilibrium Concentrations are :
[N2] = 0.045 - x = 0.045 - 0.00176 = 0.04324 M
[O2] = 0.1625 - x = 0.1625 - 0.00176 = 0.16074 M
[NO] = 2x = 2 × 0.00176 = 0.00352 M
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