The following concentrations were observed at 500 K for the formation of $\mathrm{NH}_{3}$ from $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$. At equilibrium : $[\mathrm{N}_{2}]=2 \times 10^{-2} \mathrm{M},[\mathrm{H}_{2}]=3 \times 10^{-2} \mathrm{M}$ and $[\mathrm{NH}_{3}]=1.5 \times 10^{-2} \mathrm{M}$. Equilibrium constant for the reaction is ____.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Equilibrium

The following concentrations were observed at 500 K for the formation of

NH_{3}

from

N_{2}

and

H_{2}

. At equilibrium :

[N_{2}] = 2 \times 1 0^{- 2} M, [H_{2}] = 3 \times 1 0^{- 2} M

and

[NH_{3}] = 1.5 \times 1 0^{- 2} M

. Equilibrium constant for the reaction is ____.

Answer: 417

For

N_{2} (g) + 3 H_{2} (g) ⇌ 2N H_{3} (g)

K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} = \frac{( 1.5 \times 1 0 ^{- 2} ) ^{2}}{( 2 \times 1 0 ^{- 2} ) ( 3 \times 1 0 ^{- 2} ) ^{3}}

The numerator is

2.25 \times 1 0^{- 4}

; the denominator is

(2 \times 1 0^{- 2}) (2.7 \times 1 0^{- 5}) = 5.4 \times 1 0^{- 7}

K_{c} = \frac{2.25 \times 1 0 ^{- 4}}{5.4 \times 1 0 ^{- 7}} = 417

The equilibrium constant is 417.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer