Consider the following reaction equilibrium: N ₂(g)+3 H ₂(g) 2 NH ₃(g) Initially, 1 mole of N ₂ and 3 moles of H ₂ are taken in a 2 ~L flask. At equilibrium state if, the number of moles of N ₂ is 0.6 , what is the total number of moles of all gases present in the flask?

Step-by-step solution

N ₂(g)+3 H ₂(g) 2 NH ₃(g) array ll 1 mole 3 mole & 0 moles (2 L) (initially) 0.6 or (3-3 x) & 2 x moles (1-x) moles moles & (at equilibrium) array Number of moles of N ₂=0.6=1-x x=1-0.6=0.4 So, 3-3 x=3-3 0.4=1.8 , and 2 x=2 0.4=0.8 Therefore, the total number of moles at equilibrium. =(1+x)+(3-3 x)+2 x=0.6+1.8+0.8=3.2 ~mol