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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

At 450 K450\ K, Kp=2.0×1010/barK_p= 2.0 × 10^{10}/bar for the given reaction at equilibrium.
2SO2(g)+O2(g)⇋2SO3(g)2SO_2(g) + O_2(g) ⇋ 2SO_3 (g)
What is KcK_c at this temperature ?

Answer

For the given reaction, Δn=2−3=−1Δn = 2 - 3 = -1
T=450 KT = 450 \ K
R=0.0831 bar L bar K−1 mol−1R = 0.0831\ bar\ L \ bar\ K^{-1}\ mol^{-1}
Kp=2.0×1010 bar−1K_p = 2.0 × 10^{10} \ bar^{-1}
We know that, Kp=Kc(RT)ΔnK_p = K_c (RT)^{Δn}
⇒ 2.0×1010bar−1=Kc(0.0831 L bar K−1 mol−1×450 K)−12.0 × 10^{10} bar^{-1} = K_c (0.0831 \ L\ bar \ K^{-1}\ mol^{-1} × 450 \ K)^{-1}

⇒ Kc=2.0×1010 bar−1(0.0831 Lbar K−1 mol−1×450 K)−1K_c = \frac {2.0 × 10^{10}\ bar^{- 1}}{ (0.0831 \ L bar\ K^{-1}\ mol^{-1} × 450 \ K)^{-1}}
⇒ Kc = (2.0 × 1010 bar-1)(0.0831 L bar K-1mol-1 × 450 K)
⇒ Kc = 74.79 × 1010 L mol-1
⇒ Kc = 7.48 × 1011 L mol-1
⇒ Kc = 7.48 × 1011 M-1