Chemical Equilibrium: Common Mistakes and Fixes (5)

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Tags Equilibrium

Question

For the reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), the equilibrium constant Kc=0.5K_c = 0.5 at 400400 K. Initially, a 11 L vessel contains 22 mol N2N_2, 33 mol H2H_2, and 11 mol NH3NH_3. Predict the direction of the reaction.

Solution — Step by Step

QcQ_c has the same form as KcK_c but uses initial concentrations:

Qc=[NH3]2[N2][H2]3=(1)2(2)(3)3=154≈0.0185Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(1)^2}{(2)(3)^3} = \frac{1}{54} \approx 0.0185

Q=0.0185<K=0.5Q = 0.0185 < K = 0.5. So the system has too little product (or too much reactant) compared to equilibrium.

To increase QQ toward KK, the system must produce more NH3NH_3 and consume N2+H2N_2 + H_2. So the reaction proceeds in the forward direction.
  • Q<KQ < K: forward reaction
  • Q>KQ > K: reverse reaction
  • Q=KQ = K: at equilibrium, no net change

Final answer: forward direction (toward NH3NH_3).

Why This Works

The reaction quotient QQ tells you where the system is right now in concentration space; KK tells you where equilibrium is. The system always moves toward equilibrium, which means Q→KQ \to K.

This is one of the most-asked types of question in JEE Main and NEET — easy 4 marks if you write the formula correctly.

Alternative Method

Conceptual: at the start, products are low and reactants are high. Le Chatelier says the system shifts to consume reactants and form products — forward direction. Same conclusion without arithmetic.

The classic mistake: writing QQ with reactants on top and products on bottom. The convention is products in the numerator, reactants in the denominator — same as KK. If you flip them, you’ll predict the opposite direction.

For equilibrium problems involving KcK_c vs KpK_p: Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n} where Δn\Delta n = (moles of gaseous products) − (moles of gaseous reactants). For ammonia synthesis, Δn=2−4=−2\Delta n = 2 - 4 = -2, so Kp≠KcK_p \ne K_c unless temperature is very specific.

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