Question

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 initially placed into a 4.9-L vessel, calculate the equilibrium concentrations of all species.

Answer 1

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