The equilibrium constant of the reaction SO2(g) + 1/2 O2(g) leftharpoons SO3(g) is 4 × 10-3 atm-1/2. The equilibrium constant of the reaction 2SO3(g) leftharpoons 2SO2(g) + O2(g) would be:

Question

The equilibrium constant of the reaction SO2(g) + 1/2 O2(g) leftharpoons SO3(g) is 4 × 10-3 atm-1/2. The equilibrium constant of the reaction 2SO3(g) leftharpoons 2SO2(g) + O2(g) would be: (A) 250 atm (B) 4 × 103 atm (C) 0.25 × 104 atm (D) 6.25 × 104 atm

Tardigrade Admin · Accepted Answer

SO2(g)+(1/2) O2(g) → SO3(g) Kp=4 × 10-3 SO3 → SO2 (g) + (1/2) O2 (g) K'p = (1/Kp) 2SO3 → 2SO2 +O2 (g) K'p =( (1/4 × 10-3)) K''p =(K'p)2 =[(1/4 × 10-3)] 2 =[ (1000/4)] =6250 =625 × 102 =6.25 × 104 atm

Q. The equilibrium constant of the reaction
$S O_{2} (g) + 1/2 O_{2} (g) ⇌ S O_{3} (g)$ is $4 \times 1 0^{- 3} a t m^{- 1/2}$ . The equilibrium constant of the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ would be:

$250 a t m$

$4 \times 1 0^{3} a t m$

$0.25 \times 1 0^{4} a t m$

$6.25 \times 1 0^{4} a t m$

Q. The equilibrium constant of the reaction SO2​(g)+1/2O2​(g)⇌SO3​(g) is 4×10−3atm−1/2. The equilibrium constant of the reaction 2SO3​(g)⇌2SO2​(g)+O2​(g) would be:

250atm

4×103atm

0.25×104atm

6.25×104atm

SO2​(g)+21​O2​(g)→SO3​(g)Kp​=4×10−3 SO3​→SO2​(g)+21​O2​(g)Kp′​=Kp​1​ 2SO3​→2SO2​+O2​(g)Kp′​=(4×10−31​) Kp′′​=(Kp′​)2​=[4×10−31​]2=[41000​] =6250=625×102=6.25×104 atm

Q. The equilibrium constant of the reaction
$S O_{2} (g) + 1/2 O_{2} (g) ⇌ S O_{3} (g)$ is $4 \times 1 0^{- 3} a t m^{- 1/2}$ . The equilibrium constant of the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ would be:

$250 a t m$

$4 \times 1 0^{3} a t m$

$0.25 \times 1 0^{4} a t m$

$6.25 \times 1 0^{4} a t m$