The equilibrium of the reaction SO2g-12O2g-SO3g is 4-10-3atm-1-3- The equilibrium constant of the reaction 2SO3g- 2SO2g-O2g would be-

The correct option is D 6.25×10^4atm SO_2(g)+1/2 O_2(g)⇌ SO_3 (g) K=4× 10^ 3 Reverse the reaction SO_3⇌ SO_2(g)+1/2 O_2(g)⇒ K=14× 10^ 3 #160; ...

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Standard X

Chemistry

Industrial Preparation of Ammonia

The equilibri...

Question

The equilibrium of the reaction ${S O}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g)$ is $4 \times 10^{- 3} {a t m}^{- 1 / 3}$ . 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

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4 \times 10^{3} a t m

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0.25 \times 10^{4} a t m

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6.25 \times 10^{4} a t m

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Solution

The correct option is D $6.25 \times 10^{4} a t m$
$S O_{2} (g) + 1 / 2 O_{2} (g) ⇌ S O_{3} (g) K = 4 \times 10^{- 3}$
Reverse the reaction $S O_{3} ⇌ S O_{2} (g) + 1 / 2 O_{2} (g) \Rightarrow K = \frac{1}{4 \times 10^{- 3}}$
$2 S O_{3} ⇌ 2 S O_{2} (g) + O_{2} (g) \Rightarrow {(\frac{1}{4 \times 10^{- 3}})}^{2}$
$K = \frac{1}{16 \times 10^{- 6}}$
$k = 0.0625 \times 10^{6} a t m$
$(D) K = 6.25 \times 10^{4} a t m$

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The equilibrium of the reaction SO2(g)+12O2(g)⇌SO3(g) is 4×10−3atm−1/3. The equilibrium constant of the reaction 2SO3(g)⇌2SO2(g)+O2(g) would be:

The correct option is D 6.25×104atmSO2(g)+1/2O2(g)⇌SO3(g) K=4×10−3Reverse the reaction SO3⇌SO2(g)+1/2O2(g)⇒K=14×10−3 2SO3⇌2SO2(g)+O2(g)⇒(14×10−3)2 K=116×10−6k=0.0625×106 atm (D)K=6.25×104 atm

The equilibrium of the reaction ${S O}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {S O}_{3} (g)$ is $4 \times 10^{- 3} {a t m}^{- 1 / 3}$ . The equilibrium constant of the reaction $2 {S O}_{3} (g) ⇌ 2 {S O}_{2} (g) + O_{2} (g)$ would be: