Skip to content
Exercises · 6.5

Q.Find out the value of Kc for each of the following equilibria from the value of Kp:

(i) 2NOCl
(g) ⇌ 2NO
(g) + Cl2 (g); Kp= 1.8 × 10⁻² at 500 K
(ii) CaCO3 (s) ⇌ CaO(s) + CO2(g); Kp= 167 at 1073 K
Jharkhand JacTextbookSubjective· 2mImportance★★★★★est
21% · 33/155 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Using Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n} with R=0.0831 L bar K−1mol−1R = 0.0831\ \text{L bar K}^{-1}\text{mol}^{-1}: reaction (i) gives Kc=4.33×10−4K_c = 4.33 \times 10^{-4} at 500 K; reaction (ii) gives Kc=1.87K_c = 1.87 at 1073 K.

Formula

Kp=Kc(RT)Δn,Kc=Kp(RT)ΔnK_p = K_c(RT)^{\Delta n}, \qquad K_c = \frac{K_p}{(RT)^{\Delta n}}

where Δn\Delta n counts only gaseous species (products −- reactants). Solids do not contribute.

(i) 2NOCl(g)⇌2NO(g)+Cl2(g)2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)} + \text{Cl}_2\text{(g)}

Δn=(2+1)−2=1\Delta n = (2 + 1) - 2 = 1.

RT=0.0831×500=41.55RT = 0.0831 \times 500 = 41.55

Kc=1.8×10−2(41.55)1=4.33×10−4K_c = \frac{1.8 \times 10^{-2}}{(41.55)^1} = 4.33 \times 10^{-4} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.