Answer:
Explanation:
From the information given:
The Chemical equation is:
[tex]SO_2Cl_{2(g)} \iff SO_{2(g)} + Cl_{2(g)}\\[/tex]
since 43.6% of the initial concentration remains at equilibrium
Then; the amount of [tex]SO_2Cl_2[/tex] that is being reacted is:
= 0.543 × (100 -43.6)%
= 0.306 M
The ICE table can be computed as follows:
[tex]SO_2Cl_2[/tex] ⇔ [tex]SO_{2(g)[/tex] + [tex]Cl_{2(g)[/tex]
I 0.543 0 0
C 0.306 +0.306 0.306
E 0.237 0.306 0.306
[tex]K_c = \dfrac{[SO_2] [Cl_{2}]}{[SO_2Cl_2]}[/tex]
[tex]K_c = \dfrac{0.306 \times 0.306}{0.237}[/tex]
[tex]K_c = 0.995[/tex]
Thus; the concentration at equilibrium for the species are:
[tex]SO_2Cl_2[/tex] = 0.237 M
[tex]SO_{(2g)[/tex] = 0.306 M
[tex]Cl_{2(g)[/tex] = 0.306 M
