Question

N2+3H2=2NH3 IN ONE LITRE FLASK N2=3M H2= 9M NH3=4M AT EQULIBRIUM THEN INTIAL CONC OF N2 AND H2 ARE

Answer 1

Given:

• The volume of the flask (v) = 1L
• The equilibrium concentration of N₂ = 3M
• The equilibrium concentration of H₂ = 9M
• The equilibrium concentration of NH₃ = 4M

To find:

• The initial concentration of N₂
• The initial concentration of H₂

Solution:

• Since the volume of the flask is 1L , all the concentration will be equal to the number of moles of the substance.
• Two mole of NH₃ is produced from one mole of N₂. Hence, four mole of NH₃ will be produced from two moles of N₂.
• So, initial concentration of N₂ = 3 + 2 = 5M

• Two mole of NH₃ is produced from three mole of H₂. Hence, four mole of NH₃ will be produced from six moles of H₂.

• So, initial concentration of N₂ = 9 + 6 = 15M

Answer:

• The initial concentration of N₂ = 5M

• The initial concentration of H₂ = 15M