Respuesta :
Explanation:
Volume of container = 2.50 L
Initial number of moles of [tex]SO_{3}[/tex] = 0.760 mol
Initial concentration of [tex]SO_{3}[/tex] = [tex]\frac{0.760 mol}{2.50 L}[/tex]
= 0.304 M
No. of moles of [tex]O_{2}[/tex] present at equilibrium = 0.110 mol
Volume of container, V = 2.50 L
Equilibrium concentration of [tex]O_{2}[/tex] = [tex]\frac{0.110 mol}{2.50 L}[/tex]
= 0.044 M
[tex]2SO_{3}(g) \rightleftharpoons 2SO_{2}(g) + O_{2}(g)[/tex]
Initial: 0.304 0 0
Change: -2x 2x x
Equilibrium: 0.304 - 2x 2x x
Since, equilibrium concentration of [tex]O_{2}[/tex] = x
= 0.044 M
Equilibrium concentration of [tex]SO_{2}[/tex] = 2x
= [tex]2 \times 0.044[/tex]
= 0.088 M
Equilibrium concentration of [tex]SO_{3}[/tex] = (0.304 - 2x)
= (0.304 - 2 \times 0.044)
= 0.216 M
Hence, calculate equilibrium constant ([tex]K_{c}[/tex]) as follows.
[tex]K_{c} = \frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}[/tex]
= [tex]\frac{(0.088)^{2} \times 0.044}{(0.216)^{2}}[/tex]
= [tex]7.40 \times 10^{-3}[/tex]
Thus, we can conclude that equilibrium constant of the reaction is [tex]7.40 \times 10^{-3}[/tex].
