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CO(g ) + H2O(g ) <---> CO2(g ) + H2(g ), Kc = 0.0611 at 2000 K . A reaction mixture initially contains a CO partial pressure of 2.50 atm, an H2O partial pressure of 2.50 atm, a CO2 partial pressure of 1.00 atm, and an H2 partial pressure of 1.00 atm at 2000 K. Calculate the equilibrium partial pressure of CO

Respuesta :

Answer:  Thus equilibrium partial pressure of CO is 1.306 atm

Explanation:

[tex]K_p[/tex] is the constant of a certain reaction at equilibrium.

For the given chemical reaction:

        [tex]CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)[/tex]

at t=0     2.50      2.50                  1.00         1.00

at em     (2.50-x)    (2.50-x)         (1.00+x)       (1.00+x)

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{(p_{CO_2})\times p_{H_2}}{p_{CO}\times p_{H_2O}}[/tex]

Putting values in above equation, we get:

[tex]0.0611=\frac{(1.00+x)\times (1.00+x)}{(2.50-x)\times (2.50-x)}[/tex]

[tex]x=0.306[/tex]

Thus equilibrium partial pressure of CO = (1.00+x) = (1.00+ 0.306) = 1.306 atm

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