Answered

Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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]

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.