Using Hess's law, determine the enthalpy change for the overall reaction?

#(i)# #"H"_3"BO"_3(aq) -> "HBO"_2(aq) + "H"_2"O"(l)#, #DeltaH_1 = -"0.02 kJ/mol"#

#(ii)# #"H"_2"B"_4"O"_7(aq) + "H"_2"O"(l) -> 4"HBO"_2(aq)#, #DeltaH_2 = -"11.3 kJ/mol"#

#(iii)# #"H"_2"B"_4"O"_7(aq) -> 2"B"_2"O"_3(s) + "H"_2"O"(l)#, #DeltaH_3 = "17.5 kJ/mol"#

1 Answer

Answer:

#DeltaH=+14.36# #kJ*mol^-1#

Explanation:

The given equation can be obtained by

#2xx(i)-((ii))/2+((iii))/2#,

since we have:

#2("H"_3"BO"_3(aq) -> cancel("HBO"_2(aq)) + "H"_2"O"(l))#

#-1/2 (cancel("H"_2"B"_4"O"_7(aq)) + "H"_2"O"(l) -> cancel(4"HBO"_2(aq)))#

#1/2 (cancel("H"_2"B"_4"O"_7(aq)) -> 2"B"_2"O"_3(s) + "H"_2"O"(l))#
So,

#DeltaH=2*(-0.02)-((-11.3))/2+((17.5))/2#

#=>DeltaH=+14.36# #kJ*mol^-1#