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

## $\left(i\right)$ $\text{H"_3"BO"_3(aq) -> "HBO"_2(aq) + "H"_2"O} \left(l\right)$, $\Delta {H}_{1} = - \text{0.02 kJ/mol}$ $\left(i i\right)$ ${\text{H"_2"B"_4"O"_7(aq) + "H"_2"O"(l) -> 4"HBO}}_{2} \left(a q\right)$, $\Delta {H}_{2} = - \text{11.3 kJ/mol}$ $\left(i i i\right)$ $\text{H"_2"B"_4"O"_7(aq) -> 2"B"_2"O"_3(s) + "H"_2"O} \left(l\right)$, $\Delta {H}_{3} = \text{17.5 kJ/mol}$

Aug 11, 2017

$\Delta H = + 14.36$ $k J \cdot m o {l}^{-} 1$

#### Explanation:

The given equation can be obtained by

$2 \times \left(i\right) - \frac{\left(i i\right)}{2} + \frac{\left(i i i\right)}{2}$,

since we have:

$2 \left(\text{H"_3"BO"_3(aq) -> cancel("HBO"_2(aq)) + "H"_2"O} \left(l\right)\right)$

$- \frac{1}{2} \left(\cancel{{\text{H"_2"B"_4"O"_7(aq)) + "H"_2"O"(l) -> cancel(4"HBO}}_{2} \left(a q\right)}\right)$

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

$\Delta H = 2 \cdot \left(- 0.02\right) - \frac{\left(- 11.3\right)}{2} + \frac{\left(17.5\right)}{2}$

$\implies \Delta H = + 14.36$ $k J \cdot m o {l}^{-} 1$