For ammonia formation from constituent elements, the expression for $K_c$ is

Question

For the reaction $ \text{N}_2(g) + 3 \, \text{H_2(g) \rightleftharpoons 2 \, \text{NH}_3(g) $, the equilibrium constant $ K_c $ at a certain temperature is $ 0.5 $. If the initial concentrations of $ \text{N}_2 $, $ \text{H}_2 $, and $ \text{NH}_3 $ are $ 1.0 \, \text{mol/L} $, $ 1.0 \, \text{mol/L} $, and $ 0 \, \text{mol/L} $ respectively, calculate the equilibrium concentrations of all species.}

Answer 1

Step 1: Write the balanced equation for ammonia formation.
The Haber process reaction (formation from constituent elements): \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] It is crucial to write the balanced equation first because the equilibrium expression depends on stoichiometric coefficients.
Step 2: Recall the law of mass action.
For a general reaction $ aA + bB \rightleftharpoons cC + dD $, the equilibrium constant is: \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] Products go in the numerator, reactants in the denominator, each raised to the power of its stoichiometric coefficient.
Step 3: Apply to the ammonia equilibrium.
Here: products = $ NH_3 $ (coefficient 2), reactants = $ N_2 $ (coefficient 1) and $ H_2 $ (coefficient 3). Therefore: \[ K_c = \frac{[NH_3]^2}{[N_2]^1[H_2]^3} = \frac{[NH_3]^2}{[N_2][H_2]^3} \]
Step 4: Check the other options are wrong.
Option (1) uses wrong exponents (cubed for both products and reactants). Option (3) ignores stoichiometry entirely. Option (4) omits reactants completely.
Step 5: Physical interpretation of Kc.
A large $ K_c $ value means the equilibrium lies to the right (more products), favoring ammonia formation. In practice, the Haber process uses high pressure and a catalyst to achieve reasonable yield despite moderate $ K_c $.
Step 6: Final answer.
\[ \boxed{K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}} \]

For ammonia formation from constituent elements, the expression for \(K_c\) is

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The Correct Option is B

Solution and Explanation

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