5 moles of $SO_{2}$and 5 moles of $O_{2}$ are allowed to react to forms $SO_{3}$ in a closed vessel. At the equilibrium stage 60% of $SO_{2}$ is used up. The total number of moles of $SO_{2},O_{2}$ and $SO_{3}$ in the vessel now is

Question

Dr. Abubacker · Accepted Answer

(a)
$SO_{2} + \ \ \ \ \ \frac{1}{2}O_{2}\ \ \ \  ightleftharpoons {SO}_{3}$
$5\ mol\ \ \ \ \ \ 5\ mol\ \ \ \ \ \ \ \ \ \ \ 0\$ initially
$(5 - x)\left( 5 - \frac{1}{2}x ight)\ x$ at equilibrium
$x = \frac{60}{100} 	imes 5 = 3$
Total number of moles $= (5 - x) + \left( 5 - \frac{1}{2}x ight) + x$
$= (5 - 3) + \left( 5 - \frac{1}{2} 	imes 3 ight) + 3$
$= 8.5$

Q. 5 moles of $SO_{2}$and 5 moles of $O_{2}$ are allowed to react to forms $SO_{3}$ in a closed vessel. At the equilibrium stage 60% of $SO_{2}$ is used up. The total number of moles of $SO_{2},O_{2}$ and $SO_{3}$ in the vessel now is

Explanation

Reviewed by Dr. Abubacker

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