Question

question 5 of 32
what is the equilibrium expression for the reaction below?
n₂(g) + 3h₂(g) ⇌ 2nh₃(g)
a. \\(\frac{\text{nh}_3^2}{\text{n}_2\text{h}_2^3}\\)
b. \\(\frac{\text{nh}_3}{\text{n}_2\text{h}_2}\\)
c. \\(\frac{\text{n}_2 + 3\text{h}_2}{2\text{nh}_3}\\)
d. \\(\frac{3\text{n}_2\text{h}_2}{2\text{nh}_3}\\)

Explanation:

Step1: Recall Equilibrium Expression Rule

For a reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant expression \( K \) is \( K=\frac{[C]^c[D]^d}{[A]^a[B]^b} \), where concentrations are raised to their stoichiometric coefficients.

Step2: Apply to Given Reaction

The reaction is \( \ce{N2(g) + 3H2(g)
ightleftharpoons 2NH3(g)} \). Here, \( a = 1 \) (for \( \ce{N2} \)), \( b = 3 \) (for \( \ce{H2} \)), \( c = 2 \) (for \( \ce{NH3} \)). So the equilibrium expression is \( \frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} \).

Answer:

A. \(\frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}\)