At 450 K , Kp=2.0 × 1010 / bar for the given reaction at equilibrium, 2 SO 2(g)+ O 2(g) leftharpoons 2 SO 3(g) The value of KC at this temperature will be

Question

At 450 K , Kp=2.0 × 1010 / bar for the given reaction at equilibrium, 2 SO 2(g)+ O 2(g) leftharpoons 2 SO 3(g) The value of KC at this temperature will be (A) 6.4 × 1012 L mol -1 (B) 7.479 × 1011 L mol -1 (C) 7.00 × 10-11 L mol -1 (D) 5.66 × 106 L mol -1

Tardigrade Admin · Accepted Answer

2 SO 2(g)+ O 2(g) leftharpoons 2 SO 3(g) Kp=KC(R T)Δ ng ∵ Δ ng=2-3=-1 Given, Kp=2.0 × 1010 textbar -1 R=0.0831 L bar K -1 mol -1 and T=450 K ∴ KC=(Kp/(R T)Δ ng)=(Kp/(R T)-1) or KC=Kp × R T =2.0 × 1010 textbar -1 × 0.0831 L bar K -1 mol -1 × 450 K KC=7.479 × 1011 L mol -1

Q. At $450 K, K_{p} = 2.0 \times 1 0^{10}$ / bar for the given reaction at equilibrium,
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
The value of $K_{C}$ at this temperature will be

$6.4 \times 1 0^{12} L m o l^{- 1}$

$7.479 \times 1 0^{11} L m o l^{- 1}$

$7.00 \times 1 0^{- 11} L m o l^{- 1}$

$5.66 \times 1 0^{6} L m o l^{- 1}$

Q. At 450K,Kp​=2.0×1010/ bar for the given reaction at equilibrium, 2SO2​(g)+O2​(g)⇌2SO3​(g) The value of KC​ at this temperature will be

6.4×1012Lmol−1

7.479×1011Lmol−1

7.00×10−11Lmol−1

5.66×106Lmol−1

2SO2​(g)+O2​(g)⇌2SO3​(g) Kp​=KC​(RT)Δng​ ∵Δng​=2−3=−1 Given, Kp​=2.0×1010bar−1 R=0.0831L bar K−1mol−1 and T=450K ∴KC​=(RT)Δng​Kp​​=(RT)−1Kp​​ or KC​=Kp​×RT =2.0×1010bar−1×0.0831L bar K−1mol−1×450K KC​=7.479×1011Lmol−1

Q. At $450 K, K_{p} = 2.0 \times 1 0^{10}$ / bar for the given reaction at equilibrium,
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
The value of $K_{C}$ at this temperature will be

$6.4 \times 1 0^{12} L m o l^{- 1}$

$7.479 \times 1 0^{11} L m o l^{- 1}$

$7.00 \times 1 0^{- 11} L m o l^{- 1}$

$5.66 \times 1 0^{6} L m o l^{- 1}$