0.6 mole of `NH_(3)` in a reaction vessel of `2dm^(3)` capacity was brought to equilibrium. The vessel was then found to contain 0.15 mole of `H_(2)` formed by the reaction `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)`. Which of the following statements is true?

Options

A. 0.15 mole of the original `NH_3` had dissociated at equilibrium
B. 0.55 mole of ammonia is left in the vessel
C. At equilibrium the vessel contained 0.45 mole of `N_2`
D. The concentration of `NH_3` at equilibrium is 0.25 mole per `dm^3`

Answer

D

Explanation

`2NH_3 (g) hArr N_(2)(g) +3H_2(g)`
`{:((0.6)/(2)"mole/litre",0,0,"at start"),(((0.6)/2-(2x)/3),(x/2),((3x)/2),"at equilibrium"):}`
`(3x)/2=(0.15)/2`
(a) Dissociated moles of `NH_3 = 2x = 0.10`
(b) Left ammonia = 0.6 - 2x = 0.50 moles
(c) Moles of `N_2 = x= 0.05`
(d) Concentration of `NH_(3) = (0.6)/2 -(2x)/2 = (0.25"mole")/(dm^3)`