Question

\[\ce{N2 + 3H2 <=> 2NH3 }\]

1 mole N₂ and 3 mole H₂ are present at start in 1L flask. At equilibrium NH₃ formed required 100 mL of 5M HCl for neutralisation hence K_C is ______.

Options

• `(0.5)^2/((0.75)(2.25)^3)`

• `(0.5)^2/((0.5)(2.5)^3)`

• `(0.5 " L")/((0.75)(2.5)^3)`

• None of these

Answer 1

1 mole N₂ and 3 mole H₂ are present at start in 1L flask. At equilibrium NH₃ formed required 100 mL of 5M HCl for neutralisation hence K_C is `underline((0.5)^2/((0.75)(2.25)^3))`.

Explanation:

NH₃ formed = `5 xx 100/1000` = 0.5 mole

\[\ce{\underset{1 - 0.25}{N2} + \underset{3 - 0.75}{3H2} -> 2NH3}\]

`therefore "K"_"C" = 0.5^2/(0.75 xx (2.25)^3)`