Question 10 0 / 3.5 points Many high temperature studies have been carried out on the equilibrium of the reaction: 2SO2(g) + O2(g) = 2 SO3(g) In one study the reaction vessel initially contained (5.000x10^-3) M SO2, (2.50x10^-3) MO2, and no SO3. If it was determined that at equilibrium the SO2 concentration was (2.8x10^-3) M, determine Kc at this temperature for the reaction as written. • Answers must be written in scientific notation • Write your answer using ONE decimal place (TWO significant figures), even if this is not the correct number of significant figures (e.g., 3.4E-6 or 3.4 x 10-6). • Do NOT use spaces. • Do NOT include units.

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kobenhavn kobenhavn · Answer 1 · 2019-09-25T12:29:32+02:00

Answer: [tex]4.4\times 10^{2}[/tex]

Explanation:

The chemical reaction follows the equation:

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

t = 0 [tex]5.000\times 10^{-3}[/tex] [tex]2.50\times 10^{-3}[/tex] 0

At eqm [tex](5.000\times 10^{-3}-2x)[/tex] [tex](2.50\times 10^{-3}-x)[/tex] (2x)

The expression for [tex]K_c[/tex] for the given reaction follows:

[tex]K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}[/tex]

[tex]K_c=\frac{[2x]^2}{[5.000\times 10^{-3}-2x]^2[2.50\times 10^{-3}-x]}[/tex]

Given : [tex][SO_2]_{eqm}=2.80\times 10^{-3}[/tex]

[tex]5.000\times 10^{-3}-2x=2.80\times 10^{-3}[/tex]

[tex]x=1.1\times 10^{-3}[/tex]

Putting the values we get:

[tex]K_c=\frac{[2\times 1.1\times 10^{-3}]^2}{[5.000\times 10^{-3}-2\times 1.1\times 10^{-3}]^2[2.50\times 10^{-3}-1.1\times 10^{-3}]}[/tex]

[tex]K_c=4.4\times 10^2[/tex]

Therefore, the equilibrium concentration [tex]4.4\times 10^{2}[/tex]