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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

Find out the value of KcK_c for each of the following equilibria from the value of KpK_p:

  1. 2NOCl(g)⇋2NO(g)+Cl2(g); Kp=1.8×10–2 at 500 K2NOCl (g) ⇋ 2NO (g) + Cl_2 (g); \ K_p= 1.8 × 10^{–2}\ at \ 500\ K
  2. CaCO3(s)⇋CaO(s)+CO2(g); Kp=167 at 1073 KCaCO_3 (s) ⇋ CaO(s) + CO_2(g); \ K_p= 167 \ at \ 1073\ K
Answer

The relation between KpK_p and KcK_c is given as:
Kp=Kc(RT)ΔnK_p = K_c (RT)^{Δn}
(a) Here, Δn=3−2=1Δn = 3 - 2 = 1
R=0.0831 bar L mol−1K−1R = 0.0831 \ bar \ L \ mol^{-1}K^{-1}
T=500 KT = 500 \ K
Kp=1.8×10−2K_p = 1.8 \times 10^{-2}
Now, Kp=Kc(RT)ΔnK_p = K_c(RT)^{Δn}
⇒ 1.8×10−2=Kc(0.0831×500)11.8 × 10^{-2} = K_c(0.0831 × 500)^1
⇒ Kc=1.8×10−20.0831×500K_c= \frac {1.8 × 10^{- 2}}{0.0831 × 500 }
Kc=4.33×10−4K_c = 4.33 × 10^{-4} (approximately)

(b) Here, Δn=2−1=1Δn = 2 - 1 = 1
R=0.0831 bar L mol−1K−1R = 0.0831 \ bar\ L\ mol^{-1}K^{-1}
T=1073KT = 1073 K
Kp=167K_p= 167
Now, Kp=Kc(RT)ΔnKp = Kc (RT)^{Δn}
⇒ 167=Kc(0.0831×1073)Δn167 = K_c (0.0831 × 1073)^{Δn}

⇒ Kc=1670.0831×1073K_c = \frac {167}{0.0831 × 1073}
⇒ Kc=1.87K_c=1.87 (approximately)