Chemical Equilibrium: PYQ Walkthrough (6)

hard • 3 min read
Tags Equilibrium

Question

(JEE Main style.) For the equilibrium N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) at a certain temperature, Kp=1.6×104K_p = 1.6 \times 10^4 atm−2^{-2}. A mixture has partial pressures PN2=0.5P_{N_2} = 0.5 atm, PH2=0.5P_{H_2} = 0.5 atm, PNH3=1.0P_{NH_3} = 1.0 atm. Predict the direction of reaction and find KcK_c. Take T=400T = 400 K, R=0.0821R = 0.0821 L·atm·K−1^{-1}·mol−1^{-1}.

Solution — Step by Step

Qp=(PNH3)2(PN2)(PH2)3=(1.0)2(0.5)(0.5)3=10.0625=16 atm−2Q_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} = \frac{(1.0)^2}{(0.5)(0.5)^3} = \frac{1}{0.0625} = 16 \text{ atm}^{-2}

Qp=16Q_p = 16, Kp=1.6×104K_p = 1.6 \times 10^4. Since Q<KQ < K, the reaction proceeds forward (in the direction of products) to reach equilibrium.

The relation: Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n}, where Δn=\Delta n = moles of gaseous products − moles of gaseous reactants.

For N2+3H2→2NH3N_2 + 3H_2 \to 2NH_3: Δn=2−4=−2\Delta n = 2 - 4 = -2.

So Kc=Kp⋅(RT)2=(1.6×104)(0.0821×400)2K_c = K_p \cdot (RT)^{2} = (1.6 \times 10^4)(0.0821 \times 400)^2.

RT=0.0821×400=32.84RT = 0.0821 \times 400 = 32.84. (RT)2=32.842≈1078.5(RT)^2 = 32.84^2 \approx 1078.5.

Kc=1.6×104×1078.5≈1.73×107K_c = 1.6 \times 10^4 \times 1078.5 \approx 1.73 \times 10^7 M−2^{-2}.

Why This Works

Comparing QQ with KK tells us whether we have too few products (Q<KQ < K, reaction goes forward) or too many (Q>KQ > K, reaction goes backward). At Q=KQ = K, system is at equilibrium and stays put.

The KpK_p–KcK_c relation comes from the ideal gas law: P=(n/V)RT=cRTP = (n/V)RT = cRT. Substituting concentrations as c=P/(RT)c = P/(RT) in the equilibrium expression generates the (RT)Δn(RT)^{\Delta n} factor.

Memorize: for any equilibrium, sign of Δn\Delta n determines the relationship.

  • Δn=0\Delta n = 0: Kp=KcK_p = K_c (no conversion needed).
  • Δn>0\Delta n > 0: Kp>KcK_p > K_c.
  • Δn<0\Delta n < 0: Kp<KcK_p < K_c.

For ammonia synthesis, Δn=−2\Delta n = -2, so Kp≪KcK_p \ll K_c. Always sanity-check this.

Alternative Method

Compute Q in terms of concentrations directly. Using ideal gas law: ci=Pi/(RT)c_i = P_i/(RT).

cN2=0.5/32.84≈0.0152c_{N_2} = 0.5/32.84 \approx 0.0152 M, cH2≈0.0152c_{H_2} \approx 0.0152 M, cNH3=1.0/32.84≈0.0305c_{NH_3} = 1.0/32.84 \approx 0.0305 M.

Qc=(0.0305)2(0.0152)(0.0152)3=9.3×10−45.27×10−8≈1.77×104≈Qp⋅(RT)2Q_c = \dfrac{(0.0305)^2}{(0.0152)(0.0152)^3} = \dfrac{9.3 \times 10^{-4}}{5.27 \times 10^{-8}} \approx 1.77 \times 10^4 \approx Q_p \cdot (RT)^2.

Same conclusion: Q<KQ < K, forward reaction.

Students forget that Δn\Delta n is calculated only for gaseous species. If liquid or solid species appear in the equation, they don’t contribute to Δn\Delta n. For ammonia synthesis, all species are gases, so Δn=2−(1+3)=−2\Delta n = 2 - (1+3) = -2.

Final answer: Q<KQ < K, reaction proceeds forward. Kc≈1.73×107K_c \approx 1.73 \times 10^7 M−2^{-2}.

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