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An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc equals 1.4 × 102 for the reaction: 2 CO(g) + O2(g) ⇌ 2 CO2(g). What is the equilibrium concentration of CO?

Respuesta :

ajeigbeibraheem

Answer:

[tex]5.35 *10^{-4}[/tex]M

Explanation:

Equation for the reaction is as follows:

[tex]2CO_{(g)}[/tex]      +      [tex]O_{2(g)}[/tex]        ⇄       [tex]2CO_{2(g)}[/tex]

By Applying the ICE Table; we have

                             [tex]2CO_{(g)}[/tex]      +      [tex]O_{2(g)}[/tex]        ⇄       [tex]2CO_{2(g)}[/tex]

Initial                      x                  0.0025 M              0.0010 M

Change                  0                       0                            0

Equilibrium             x                  0.0025 M              0.0010 M

[tex]K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}[/tex]

Given that [tex]K_c = 1.4*10^2[/tex] ; Then:

[tex]1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}[/tex]

[tex]1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}[/tex]

[tex]3.5 =( \frac{(0.001)}{(x)})^2[/tex]

[tex]\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}[/tex]

[tex]1.87=\frac{(0.001)}{(x)}[/tex]

[tex](x)= \frac{(0.001)}{1.87 }[/tex]

[tex]x = 5.35 *10^{-4}[/tex]M

∴ The equilibrium concentration of CO = [tex]x = 5.35 *10^{-4}[/tex]M

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