The equilibrium constant for the reaction SO3(g) leftharpoons SO2(g)+(1/2)O2(g) is Kc=4.9× 10-2 . The value of Kc or the reaction 2SO2(g)+O2 (g) leftharpoons 2SO3(g) will be

Question

The equilibrium constant for the reaction SO3(g) leftharpoons SO2(g)+(1/2)O2(g) is Kc=4.9× 10-2 . The value of Kc or the reaction 2SO2(g)+O2 (g) leftharpoons 2SO3(g) will be (A)  416  (B)  2.40× 10-3  (C)  9.8× 10-2  (D)  4.9× 10-2

Tardigrade Admin · Accepted Answer

Equilibrium constant for the reaction, SO2(g)+(1/2)O2(g) leftharpoons SO3(g) Kc=(1/4.9× 10-2) and for 2SO2(g)+O2 leftharpoons (g)2SO3(g) Kc=( (1/4.9× 10-2) )2 =(104/(4.9)2)=416.49

Q. The equilibrium constant for the reaction
$S O_{3} (g) ⇌ S O_{2} (g) + \frac{1}{2} O_{2} (g)$
is $K_{c} = 4.9 \times 10^{- 2}$ . The value of $K_{c}$ or the reaction
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ will be

$416$

$2.40 \times 10^{- 3}$

$9.8 \times 10^{- 2}$

$4.9 \times 10^{- 2}$

Equilibrium constant for the reaction,
$S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)$
$K_{c} = \frac{1}{4.9 \times 10 ^{- 2}}$
and for $2 S O_{2} (g) + O_{2} ⇌ (g) 2 S O_{3} (g)$
$K_{c} = (\frac{1}{4.9 \times 10 ^{- 2}})^{2}$
$= \frac{10 ^{4}}{( 4.9 ) ^{2}} = 416.49$

Q. The equilibrium constant for the reaction SO3​(g)⇌SO2​(g)+21​O2​(g) is Kc​=4.9×10−2 . The value of Kc​ or the reaction 2SO2​(g)+O2​(g)⇌2SO3​(g) will be

416

2.40×10−3

9.8×10−2

4.9×10−2