Identify the reaction for which Kp ≠ Kc :

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ChemistryChemical Equilibrium

Options

H₂(g) + I₂(g) ⇌ 2HI(g)

N₂(g) + O₂(g) ⇌ 2NO(g)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)

Correct Answer

Option 3 : N₂ + 3H₂ ⇌ 2NH₃

Solution

Relation: Kp = Kc(RT)^Δnᵍ

Kp ≠ Kc when Δnᵍ ≠ 0

Δnᵍ = moles of gaseous products − moles of gaseous reactants

Calculate Δnᵍ for each:
Option 1: H₂+I₂⇌2HI → Δnᵍ = 2−(1+1) = 0 → Kp=Kc
Option 2: N₂+O₂⇌2NO → Δnᵍ = 2−(1+1) = 0 → Kp=Kc
Option 3: N₂+3H₂⇌2NH₃ → Δnᵍ = 2−(1+3) = 2−4 = −2 → Kp≠Kc ✅
Option 4: H₂O+CO⇌H₂+CO₂ → Δnᵍ = (1+1)−(1+1) = 0 → Kp=Kc

Kp = Kc(RT)^Δnᵍ → Kp ≠ Kc only when Δnᵍ ≠ 0
N₂+3H₂⇌2NH₃: Δnᵍ = −2 → Kp ≠ Kc

Theory: Chemical Equilibrium

1. Kp vs Kc Relationship

Kp = Kc × (RT)^Δnᵍ, where Δnᵍ = Δn(gas) = moles of gaseous products − moles of gaseous reactants. R = 0·0821 L·atm/mol/K. When Δnᵍ = 0: Kp = Kc (they are equal). When Δnᵍ > 0: Kp > Kc. When Δnᵍ < 0: Kp < Kc. For reactions involving only solids/liquids: Kp = Kc if no gases. Only gas-phase moles count for Δnᵍ.

2. Equilibrium Constants Kc and Kp

Kc = [products]^coeff / [reactants]^coeff (concentration in mol/L). Kp = (partial pressures of products)^coeff / (partial pressures of reactants)^coeff (pressure in atm). Both are dimensionless in the thermodynamic sense (activities used). Kc is used when concentrations are measured; Kp when pressures are measured. Both constants depend only on temperature.

3. Le Chatelier's Principle

If equilibrium is disturbed, system adjusts to counteract the change. Adding reactant → equilibrium shifts forward (more products). Removing product → equilibrium shifts forward. Increasing pressure → equilibrium shifts to side with fewer gas moles. Decreasing pressure → shifts to side with more gas moles. For N₂+3H₂⇌2NH₃: high pressure favours NH₃ (Δnᵍ=−2, fewer moles). For H₂+I₂⇌2HI: pressure has NO effect (Δnᵍ=0).

4. Haber Process and Equilibrium

N₂+3H₂⇌2NH₃, ΔH=−92kJ/mol. For high yield: high pressure (Δnᵍ=−2, fewer moles), low temperature (exothermic, Le Chatelier). But low T → slow rate. Compromise: 400-500°C, 200-300 atm, Fe catalyst. Kp for this reaction: Kp = Kc(RT)^(−2). At 500°C(773K): if Kc known, Kp = Kc × (0·0821 × 773)^(−2) = Kc/4019. So Kp much smaller than Kc.

5. Equilibrium Constant Expression

Kc = [C]^c[D]^d / [A]^a[B]^b for aA + bB ⇌ cC + dD. Solids and pure liquids NOT included (activity = 1). If reaction is multiplied by n: new Kc = (Kc)^n. If reaction reversed: new Kc = 1/Kc. If reactions added: Kc(overall) = Kc₁ × Kc₂. These relationships are extremely important for NEET calculations.

6. Reaction Quotient Q vs K

Q is calculated same way as K but with ANY concentrations (not necessarily equilibrium). If Q < K: reaction proceeds forward (more products needed). If Q > K: reaction proceeds reverse. If Q = K: system at equilibrium. Q is used to predict direction of reaction when not at equilibrium.

7. Degree of Dissociation and K

For A ⇌ 2B, starting with 1 mol A, degree of dissociation α: at equilibrium, moles A = 1−α, moles B = 2α, total = 1+α. Kp = (2α)²P/((1−α)(1+α)) = 4α²P/(1−α²). If α<<1: Kp ≈ 4α²P. Can solve for α given Kp and P. This type of calculation appears frequently in NEET.

8. Effect of Temperature on K

Van't Hoff equation: d(ln K)/dT = ΔH°/RT². For exothermic (ΔH<0): K decreases with T (equilibrium shifts left). For endothermic (ΔH>0): K increases with T. For N₂+3H₂⇌2NH₃ (ΔH=−92kJ): K decreases at higher T → less NH₃ at equilibrium at high T. This is why Haber process uses 400-500°C (compromise between reasonable K and reasonable rate).

Frequently Asked Questions

1. When is Kp = Kc? ⌄

Kp = Kc when Δnᵍ = 0 (no change in moles of gas). Examples: H₂+I₂⇌2HI (Δnᵍ=0), N₂+O₂⇌2NO (Δnᵍ=0), H₂O+CO⇌H₂+CO₂ (Δnᵍ=0). For these reactions, pressure has no effect on equilibrium yield. For all other gas-phase reactions where moles change: Kp ≠ Kc.

2. What is Δnᵍ for N₂+3H₂⇌2NH₃? ⌄

Gaseous products: 2 mol NH₃. Gaseous reactants: 1 mol N₂ + 3 mol H₂ = 4 mol. Δnᵍ = 2 − 4 = −2. So Kp = Kc × (RT)^(−2) = Kc/(RT)². Since RT > 1 (at any temperature above ~12 K), (RT)^(−2) < 1 → Kp < Kc for this reaction. At 500°C(773K): RT = 0·0821 × 773 = 63·5, so Kp = Kc/63·5² = Kc/4032.

3. Why does pressure affect N₂+3H₂⇌2NH₃ but not H₂+I₂⇌2HI? ⌄

N₂+3H₂⇌2NH₃: Δnᵍ=−2. Increasing pressure → system shifts to side with fewer moles (right → more NH₃). High pressure increases NH₃ yield. H₂+I₂⇌2HI: Δnᵍ=0. Equal moles on both sides. Pressure change does NOT affect equilibrium position — yield of HI unchanged by pressure. Le Chatelier: pressure only affects equilibrium if Δnᵍ ≠ 0.

4. What is Kp for the decomposition of PCl₅? ⌄

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Δnᵍ = (1+1) − 1 = +1. Kp = Kc × RT. At 250°C(523K): Kp = Kc × 0·0821 × 523 = 42·9 × Kc. So Kp > Kc (because Δnᵍ = +1 > 0). High pressure reduces PCl₅ dissociation (Le Chatelier — shifting left to reduce moles).

5. What is the water-gas shift reaction equilibrium? ⌄

H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g). Δnᵍ = (1+1) − (1+1) = 0 → Kp = Kc. This reaction is used industrially to convert CO to CO₂ and produce more H₂. Since Δnᵍ=0, pressure has no effect. Temperature controls the equilibrium (exothermic → lower T gives more H₂ and CO₂ — this is why it's done at 150-350°C).

6. How to calculate Kc from Kp? ⌄

Kc = Kp / (RT)^Δnᵍ = Kp × (RT)^(−Δnᵍ). Example: For N₂+3H₂⇌2NH₃, Kp = 6·8×10⁻² at 500°C. Δnᵍ = −2. Kc = Kp/(RT)^(−2) = Kp × (RT)². At 500°C(773K): RT = 0·0821×773 = 63·5. Kc = 6·8×10⁻² × 63·5² = 6·8×10⁻² × 4032 = 274. So Kc >> Kp for this reaction.

7. What is the effect of catalyst on equilibrium? ⌄

Catalyst DOES NOT affect Kc or Kp. Catalyst speeds up both forward and backward reactions equally. It decreases the time to reach equilibrium but does NOT change the position of equilibrium (same Kc). The Fe catalyst in Haber process allows the reaction to reach equilibrium faster at 400-500°C — without it, the rate would be too slow at any useful temperature. Catalyst = kinetic effect, not thermodynamic.

8. What are the units of Kp and Kc? ⌄

Strictly, both Kp and Kc are dimensionless (thermodynamic equilibrium constants use activities: aᵢ = Pᵢ/P° or [Cᵢ]/C°, where P°=1 atm and C°=1 mol/L). However, in NEET, they are often treated as having units: Kp has units of (atm)^Δnᵍ and Kc has units of (mol/L)^Δnᵍ. When Δnᵍ=0, both are dimensionless. The Kp=Kc(RT)^Δnᵍ equation uses R=0·0821 L·atm/mol/K when Kp is in atm and Kc is in mol/L.

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