At 450 K, Kp = 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2leftharpoons 2SO3(g). What is Kc at this temperature?

Question

At 450 K, Kp = 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2leftharpoons 2SO3(g). What is Kc at this temperature? (A) 8 × 10-2 M-1  (B) 5.47 × 10-11 M-1  (C) 3 × 10-5 M-1  (D) 7.47 × 1011 M-1

Tardigrade Admin · Accepted Answer

Kp =Kc(RT)Δ n or Kc = (Kp/(RT)Δ n) Δ n = 2 - (2 + 1) = -1 , T = 450 K, R = 0.083 bar K-l mol-1 ∴ (Kp/(RT)-1) =KP(RT) =2.0 × 1010 bar -1 × 0.0831 L bar K -1 mol -1 × 450 K =74.8 × 1010 mol -1=7.48 × 1011 L mol -1

Q. At $450 K, K_{p} = 2.0 \times 1 0^{10} / ba r$ for the given reaction at equilibrium. $2 S O_{2} (g) + O_{2} ⇌ 2 S O_{3} (g) .$ What is $K_{c}$ at this temperature?

$8 \times 1 0^{- 2} M^{- 1}$

$5.47 \times 1 0^{- 11} M^{- 1}$

$3 \times 1 0^{- 5} M^{- 1}$

$7.47 \times 1 0^{11} M^{- 1}$

Q. At 450K,Kp​=2.0×1010/bar for the given reaction at equilibrium. 2SO2​(g)+O2​⇌2SO3​(g). What is Kc​ at this temperature?

8×10−2M−1

5.47×10−11M−1

3×10−5M−1

7.47×1011M−1

Kp​=Kc​(RT)ΔnorKc​=(RT)ΔnKp​​ Δn=2−(2+1)=−1,T=450K, R=0.083barK−lmol−1 ∴(RT)−1Kp​​=KP​(RT) =2.0×1010bar−1×0.0831LbarK−1mol−1×450K =74.8×1010mol−1=7.48×1011Lmol−1

Q. At $450 K, K_{p} = 2.0 \times 1 0^{10} / ba r$ for the given reaction at equilibrium. $2 S O_{2} (g) + O_{2} ⇌ 2 S O_{3} (g) .$ What is $K_{c}$ at this temperature?

$8 \times 1 0^{- 2} M^{- 1}$

$5.47 \times 1 0^{- 11} M^{- 1}$

$3 \times 1 0^{- 5} M^{- 1}$

$7.47 \times 1 0^{11} M^{- 1}$