The given equilibrium is
N2(g)+3H2(g)⇌2NH3(g)At t=04160At equilibrium4−x16−3x 2x
Total gaseous moles at equilibrium =4−x+16−3x+2x=(20−2x)
Since, pressure has fallen to 9/10 of its original value, hence no. of mole will also fall up to the same extent.
(20−2x)=910×20=18
x=1
[N2]=4−x1=4−11=3mole/litre
[NH3]=2x1=2 mole/litre