Question

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"`

Answer 1

`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`