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At a certain temperature, the equilibrium constant, K c , for this reaction is 53.3. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K c = 53.3 At this temperature, 0.800 mol H 2 and 0.800 mol I 2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Respuesta :

Answer: Concentration of HI is present at equilibrium is 1.26 M

Explanation:

Moles of  [tex]H_2[/tex] = 0.800 mole

Moles of  [tex]I_2[/tex] = 0.800 mole

Volume of solution = 1.00 L

Initial concentration of [tex]H_2[/tex] = 0.800 M

Initial concentration of [tex]I_2[/tex] = 0.800 M

The given balanced equilibrium reaction is,

                          [tex]H_2(g)+I_2(g)\rightleftharpoons 2Hl(g)[/tex]

Initial conc.         0.800     0.800             0

At eqm. conc.     (0.800-x) M   (0.800-x) M   (2x) M

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[HI]^2}{[H_2]\times [l_2]}[/tex]

Now put all the given values in this expression, we get :

[tex]53.3=\frac{(2x)^2}{(0.800-x)^2}[/tex]

By solving the term 'x', we get :

x = 0.63

Concentration of [tex]HI[/tex] at equilibrium = 2 x = [tex]2\times 0.63=1.26[/tex]

Concentration of HI is present at equilibrium is 1.26 M

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