# How do you use Hess's Law to calculate the enthalpy change for the reaction?

## $W {O}_{3} \left(s\right) + 3 {H}_{2} \left(g\right) \to W \left(s\right) + 3 {H}_{2} O \left(g\right)$ from the following equations: $2 W \left(s\right) + 3 {O}_{2} \left(g\right) \to 2 W {O}_{3} \left(s\right)$ $\Delta H = - 1685.4 k J$ $2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(g\right) 3$ $\Delta H = - 477.84 k J$

Apr 16, 2016

$2 W {O}_{3} \left(s\right) \to 2 W \left(s\right) + 3 {O}_{2} \left(g\right) , \Delta H = + 1685.4 k J$
Dividing the eq by 2 we have
$W {O}_{3} \left(s\right) \to W \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) , \Delta H = + 842.7 k J \ldots . \left(.1\right)$
Again
$2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(g\right) , \Delta H = - 477.8 k J$
Multiplying this eq by $\frac{3}{2}$
$3 {H}_{2} \left(g\right) + \frac{3}{2} {O}_{2} \left(g\right) \to 3 {H}_{2} O \left(g\right) , \Delta H = - 716.7 k J \ldots . . \left(2\right)$

$W {O}_{3} \left(s\right) + 3 {H}_{2} \left(g\right) \to W \left(s\right) + {H}_{2} O \left(g\right) , \Delta H = 842.7 - 716.7 = 126 k J$