The equilibrium constant K-c the reaction-SO-3-g- rightleftharpoons SO-2-g- - dfrac-1-2-O-2-g- is 4-9 times 10-2- The value of K-c the reaction- will be-2SO-2-g- - O-2-g- rightleftharpoons 2SO-3-g-2-40 times 10-3-4164-9 times 10-2-9-8 times 10-2-

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Toppr Admin · Accepted Answer

SO3-g-x21CC-SO2-g-12O2-g-KC-4-9-xD7-10-x2212-2-x2234-SO2-g-12O2-g-x21CC-SO3-g-KC-14-9-xD7-102-x2234-SO2-g-O2-g-x21CC-2SO3-g- -The whole reaction is multiplied by 2-x2234-KC will also be squared-x2234-KC-1004-9-2-416

The equilibrium constant Kc for the reaction,SO3(g)⇌SO_{2(g}}+12O2(g) is 4.9×10−2 The value of Kc for the reaction, will be:2SO2(g)+O2(g)⇌2SO3(g)2.40×10−34164.9×10−29.8×10−2

SO3(g)⇌SO2(g)+12O2(g);KC=4.9×10−2∴SO2(g)+12O2(g)⇌SO3(g);KC=14.9×102∴SO2(g)+O2(g)⇌2SO3(g) (The whole reaction is multiplied by 2)∴KC will also be squared.∴KC=(1004.9)2=416

The equilibrium constant $K_{c}$ for the reaction,
$S O_{3 (g)} ⇌$ SO_{2(g}} $+ \frac{1}{2} O_{2} (g)$ is $4.9 \times 10^{- 2}$
The value of $K_{c}$ for the reaction, will be:
$2 S O_{2 (g)} + O_{2} (g) ⇌ 2 S O_{3} (g)$
$2.40 \times 10^{- 3}$
416
$4.9 \times 10^{- 2}$
$9.8 \times 10^{- 2}$