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

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

$WO_3(s) + 3H_2(g) -> W(s) + 3H_2O(g)$
from the following equations:
$2W(s) + 3O_2(g) -> 2WO_3(s)$
$Delta H = -1685.4 kJ$
$2H_2(g) + O_2(g) -> 2H_2O(g)3$
$Delta H = -477.84 kJ$

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Answer 1 · 2016-04-16T10:40:14

$2WO_3(s)->2W(s)+3O_2(g),DeltaH =+1685.4kJ$
Dividing the eq by 2 we have
$WO_3(s)->W(s)+3/2O_2(g),DeltaH =+842.7kJ....(.1)$
Again
$2H_2(g)+O_2(g)->2H_2O(g),DeltaH=-477.8kJ$
Multiplying this eq by $3/2$
$3H_2(g)+3/2O_2(g)->3H_2O(g),DeltaH=-716.7kJ.....(2)$

Adding Eq(1) with eq (2)

$WO_3(s)+3H_2(g)->W(s)+H_2O(g), Delta H =842.7-716.7=126kJ$

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

$WO_3(s) + 3H_2(g) -> W(s) + 3H_2O(g)$
from the following equations:
$2W(s) + 3O_2(g) -> 2WO_3(s)$
$Delta H = -1685.4 kJ$
$2H_2(g) + O_2(g) -> 2H_2O(g)3$
$Delta H = -477.84 kJ$

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

WO_3(s) + 3H_2(g) -> W(s) + 3H_2O(g)WO_3(s) + 3H_2(g) -> W(s) + 3H_2O(g) from the following equations: 2W(s) + 3O_2(g) -> 2WO_3(s)2W(s) + 3O_2(g) -> 2WO_3(s) Delta H = -1685.4 kJDelta H = -1685.4 kJ 2H_2(g) + O_2(g) -> 2H_2O(g)32H_2(g) + O_2(g) -> 2H_2O(g)3 Delta H = -477.84 kJDelta H = -477.84 kJ

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$WO_3(s) + 3H_2(g) -> W(s) + 3H_2O(g)$
from the following equations:
$2W(s) + 3O_2(g) -> 2WO_3(s)$
$Delta H = -1685.4 kJ$
$2H_2(g) + O_2(g) -> 2H_2O(g)3$
$Delta H = -477.84 kJ$