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yuradex [85]
3 years ago
11

At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall

y placed into a 4.9-L vessel, calculate the equilibrium concentrations of all species.
Chemistry
1 answer:
mestny [16]3 years ago
3 0

Answer:

concentration of [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 0.0248 = 2.48 ×10⁻² M

concentration of [CO_2] = 0.4442 M

Explanation:

Equation for the reaction:

2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Concentration of   CO_2_{(g) = \frac{2.3}{4.9}  = 0.469

For our ICE Table; we have:

                       2CO_2_{(g)                ⇄          2CO_{(g)       +       O_2_{(g)

Initial                 0.469                              0                           0

Change              - 2x                                +2x                      +x

Equilibrium       (0.469-2x)                       2x                         x

K = \frac{[CO]^2[O]}{[CO_2]^2}

K = \frac{[2x]^2[x]}{[0.469-2x]^2}

4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}

Since the value pf K is very small, only little small of  reactant goes into product; so (0.469-2x)² = (0.469)²

4.1*10^{-6} = \frac{2x^3}{(0.938)}

2x^3 =3.8458*10^{-6

x^3 =\frac{3.8458*10^{-6}}{2}

x^3=1.9229*10^{-6

x=\sqrt[3]{1.9929*10^{-6}}

x = 0.0124

∴ at equilibrium; concentration of  [O_2] = 0.0124 = 12.4 ×10⁻³ M

concentration of [CO] = 2x  = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of [CO_2] = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

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