Answer:
concentration of [tex][O_2][/tex] = 0.0124 = 12.4 ×10⁻³ M
concentration of [tex][CO][/tex] = 0.0248 = 2.48 ×10⁻² M
concentration of [tex][CO_2][/tex] = 0.4442 M
Explanation:
Equation for the reaction:
[tex]2CO_2_{(g)[/tex] ⇄ [tex]2CO_{(g)[/tex] + [tex]O_2_{(g)[/tex]
Concentration of [tex]CO_2_{(g)[/tex] = [tex]\frac{2.3}{4.9}[/tex] = 0.469
For our ICE Table; we have:
[tex]2CO_2_{(g)[/tex] ⇄ [tex]2CO_{(g)[/tex] + [tex]O_2_{(g)[/tex]
Initial 0.469 0 0
Change - 2x +2x +x
Equilibrium (0.469-2x) 2x x
K = [tex]\frac{[CO]^2[O]}{[CO_2]^2}[/tex]
K = [tex]\frac{[2x]^2[x]}{[0.469-2x]^2}[/tex]
[tex]4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}[/tex]
Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²
[tex]4.1*10^{-6} = \frac{2x^3}{(0.938)}[/tex]
[tex]2x^3 =3.8458*10^{-6[/tex]
[tex]x^3 =\frac{3.8458*10^{-6}}{2}[/tex]
[tex]x^3=1.9229*10^{-6[/tex]
[tex]x=\sqrt[3]{1.9929*10^{-6}}[/tex]
x = 0.0124
∴ at equilibrium; concentration of [tex][O_2][/tex] = 0.0124 = 12.4 ×10⁻³ M
concentration of [tex][CO][/tex] = 2x = 2 ( 0.0124)
= 0.0248
= 2.48 ×10⁻² M
concentration of [tex][CO_2][/tex] = 0.469-2x
= 0.469-2(0.0124)
= 0.469 - 0.0248
= 0.4442 M
