Question

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.

Answer 1

Answer: At equilibrium , there are 0.274 moles of $Br_2$

Explanation:

Moles of  $H_2$ = 0.682 mole

Moles of  $Br_2$ = 0.440 mole

Volume of solution = 2.00 L

Initial concentration of $H_2$ = $\frac{0.682}{2.00}=0.341 M$

Initial concentration of $Br_2$ =  $\frac{0.440}{2.00}=0.220 M$

Equilibrium concentration of $H_2$ = $\frac{0.516}{2.00}=0.258 M$

The given balanced equilibrium reaction is,

             $H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)$

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 $Br_2$ = (0.220-0.083) M = 0.137 M

Thus moles of $Br_2$ at equilibrium = $0.137M\times 2.00L=0.274mol$

At equilibrium , there are 0.274 moles of $Br_2$