MCQ
The equilibrium constant for the reaction

$SO_3(g) \rightleftharpoons  SO_2(g)+ \frac{1}{2} O_2(g)$

is $K_c= 4.9 \times 10^{-2}.$ The value of $K_c$ for the reaction

$2SO_2(g) + O_2(g) \rightleftharpoons  2SO_3(g)$

will be

  • A
    $9.8 \times 10^{-2}$
  • B
    $4.9 \times 10^{-2}$
  • $416$
  • D
    $2.40  \times 10^{-3}$

Answer

Correct option: C.
$416$
c
$S O_{3}(g) \rightleftharpoons S O_{2}(g)+\frac{1}{2} O_{2}(g)$

$K_{c}=\frac{\left|S O_{2} \| O_{2}\right|^{1 / 2}}{\left|S O_{3}\right|}$

$=4.9 \times 10^{-2}$

On taking the square of the above reaction

$\frac{\left[S O_{2}\right]^{2}\left|O_{2}\right|}{\left[S O_{3}\right]^{2}}$

$=24.01 \times 10^{-4}$

now $K_{C}^{\prime}$

for $2 S O_{2}(g)+O_{2}(g) \rightleftharpoons 2 S O_{3}$

$=\frac{\left|S O_{3}\right|^{2}}{\left[S O_{2}\right]^{2}\left[O_{2}\right]}$

$=\frac{1}{24.01 \times 10^{-4}}$

$=416$

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