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g At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H 2 (g) Br 2 (g) 2HBr (g) A mixture of 0.682 mol of H 2 and 0.440 mol of Br 2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.516 mol of H 2 present. At equilibrium, there are ________ mol of Br 2 present in the reaction vessel.

Sagot :

Answer: At equilibrium , there are 0.274 moles of [tex]Br_2[/tex]

Explanation:

Moles of  [tex]H_2[/tex] = 0.682 mole

Moles of  [tex]Br_2[/tex] = 0.440 mole

Volume of solution = 2.00 L

Initial concentration of [tex]H_2[/tex] = [tex]\frac{0.682}{2.00}=0.341 M[/tex]

Initial concentration of [tex]Br_2[/tex] =  [tex]\frac{0.440}{2.00}=0.220 M[/tex]

Equilibrium concentration of [tex]H_2[/tex] = [tex]\frac{0.516}{2.00}=0.258 M[/tex]

The given balanced equilibrium reaction is,

             [tex]H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)[/tex]

Initial conc.   0.341 M    0.220 M        0 M    

At eqm. conc.   (0.341-x) M   (0.220-x) M   (2x) M

Given : (0.341-x) M = 0.258 M

x= 0.083 M

Thus equilibrium concentartion of [tex]Br_2[/tex] = (0.220-0.083) M = 0.137 M

Thus moles of [tex]Br_2[/tex] at equilibrium = [tex]0.137M\times 2.00L=0.274mol[/tex]

At equilibrium , there are 0.274 moles of [tex]Br_2[/tex]

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