Most Important Selected Qs for JEE AdvancedChemistryChemical Equilibrium
Paragraph : 10 moles of NH ₃ is heated at 15 atm from 27^ C to 347^ C assuming volume constant. The pressure at equilibrium is found to be 50 atm . The equilibrium constant for the dissociation of NH ₃2 NH ₃( ~g ) N ₂( ~g )+3 H ₂( ~g ), H =91.94 ~kJ Can be written as K _ P = P _ N ₂ ( P _ H ₂ )^3 ( P _ NH ₃ )^2 ( ~atm )^2 Answer, the following questions: Question : The percentage of dissociation of NH ₃ is
Options
- A61.2 %
- B20 %
- C48 %
- D30 %
Correct answer
A. 61.2 %
Step-by-step solution
10-2 x 2 NH ₃ x ~N ₂ + 3 x 3 H ₂ Pressure increases due to increase in temperature as well as due to increase in moles. 15 300 P 620 P = 620 300 15=31 ~atm at 10 moles of NH ₃ at 620 K . Now, NH ₃ is dissociated to attain 50 atm at 620 K aligned & P n or 10 31 & (10+2 x ) 50 & 10+2 x 10 = 50 31 =1.61,10+2 x =16.12 & 2 x =6.12 & = 2 x 10 100 & = 6.12 100 10 =6.12 aligned