(D) The chemical equation is: $2SO_3 \rightleftharpoons 2SO_2 + O_2$Initial moles: $3 \quad 0 \quad 0$Degree of dissociation $(\alpha)$ $= 0.4$Moles r

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(D) The chemical equation is: $2SO_3 \rightleftharpoons 2SO_2 + O_2$Initial moles: $3 \quad 0 \quad 0$Degree of dissociation $(\alpha)$ $= 0.4$Moles reacted $= 3 \times 0.4 = 1.2$At equilibrium:$SO_3 = 3 - 1.2 = 1.8$$SO_2 = 1.2$$O_2 = \frac{1.2}{2} = 0.6$Total moles at equilibrium $= 1.8 + 1.2 + 0.6 = 3.6$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Degree of dissociation and Vapour density

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$2SO_3 \rightleftharpoons 2SO_2 + O_2$. Initially,$3$ moles of $SO_3$ were heated. If the degree of dissociation of $SO_3$ is $40\%$,find the total number of moles at equilibrium.

A
$3.3$
B
$6.6$
C
$3.15$
D
$3.6$

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6-1.Equilibrium (Chemical Equilibrium)

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Degree of dissociation and Vapour density

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$2SO_3 \rightleftharpoons 2SO_2 + O_2$. Initially,$3$ moles of $SO_3$ were heated. If the degree of dissociation of $SO_3$ is $40\%$,find the total number of moles at equilibrium.

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