Consider the following reaction in a 1 ~L closed vessel. N ₂+3 H ₂ 2 NH ₃ If all the species; N ₂, H ₂ and NH ₃ are in 1 ~mol in the beginning of the reaction and equilibrium is attained after unreacted N ₂ is 0.7 ~mol . What is the value of equilibrium constant?

Step-by-step solution

For the given reaction: N ₂+3 H ₂ 2 NH ₃ array llll Initial moles & 1 & 1 & 1 array At equilibrium 1-x 1-3 x 2 x Given, 1-x=0.7 ~mol x=0.3 ~mol Therefore, concentration of N ₂, H ₂ and NH ₃ at equilibrium will be aligned [ N ₂ ] & =[0.7] [ H ₂ ] & =1-(3 0.3)=[0.1] [ NH ₃ ] & =1+2 x=1+(2 0.3)=[1.6] aligned According to law of equilibrium constant (K_C ) aligned K_C & = [ NH ₃ ]^2 [ ~N ^2 ], [ H ₂ ]^3 = [ l .6]^2 [0.7][0.1]^3 K_C & = (256) 0.007 =3657.14 aligned Hence, equilibrium constant (K_C )=3657.14 and (b) is t