You visited us 1 times! Enjoying our articles? Unlock Full Access!

Industrial Preparation of Ammonia

For the react...

Question

For the reaction, $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ , the equilibrium expression may be written as:
$K = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}$ .

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A True
The equilibrium constant is the ratio of the molar concentrations of the products to the molar concentrations of the reactants. The molar concentrations are raised to appropriate powers which are the stoichiometric coefficients.
Thus the equilibrium constant can be written as $\frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}$ .

Suggest Corrections

Similar questions

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

K \frac{{[N H_{3}]}_{3}^{2}}{[N_{2}] {[H_{2}]}_{2}^{3}}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{6}

2.37 \times 10^{- 3}

[N_{2}] = 2. M [H_{2}] = 3 M

N H_{3}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}, K_{c}

K

2.37 \times 10^{- 3}

[N_{2}] = 2 M, [H_{2}] = 3 M

N H_{3}

N_{2} + 3 H_{2} \to 2 N H_{3}

\frac{d [H_{2}]}{d t}

\frac{d [N H_{3}]}{d t}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}; K_{3}

N_{2} + O_{2} ⇌ 2 N O; K_{2}

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O; K_{1}

2 N H_{3} + 5 / 2 O_{2} K ⇌ 2 N O + 3 H_{2} O

Join BYJU'S Learning Program