Answer:
The reaction enthalpy is 1080 kJ/mol
Explanation:
The given reaction is:
[tex]QX_{3} + 3H_{2}0 \rightarrow Q(OH)_{3} + 3HX[/tex]
The reaction enthalpy is given by the difference between the number of bonds broken and the number of bonds formed.
[tex]\Delta H_{rxn}=\sum (bonds\ broken)-\sum (bonds\ formed)[/tex]
For the given reaction:
[tex]\Delta H_{rxn}=[3(Q-X)+6(O-H)]-[3(Q-O)+3(H-X)][/tex]
from the given bond energies:
[tex]\Delta H_{rxn}=[3(240)+6(464)]-[3(359)+3(449)]=1080kJ/mol[/tex]
