Given the standard enthalpy changes for the following two reactions:
(1) 4C(s) + 5H2(g) -> C4H10(g)...... ΔH° = -125.6 kJ
(2) C2H4(g) -> 2C(s) + 2H2(g)......ΔH° = -52.3 kJ
what is the standard enthalpy change for the reaction:
(3) 2C2H4(g) + H2(g) -> C4H10(g)......ΔH° = ?
__________kJ
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To determine the standard enthalpy change for reaction (3), we need to use Hess's law.
We can manipulate reactions (1) and (2) so that the reactants and products of reaction (3) are formed.
First, we multiply reaction (2) by 2 so that the number of carbon atoms and hydrogen atoms match reaction (3):
2[C2H4(g) -> 2C(s) + 2H2(g)]
Now, we need to flip reaction (1) to match the formation of reactants in reaction (3):
[C4H10(g) -> 4C(s) + 5H2(g)]
Now, we can add the manipulated reactions together to get reaction (3):
2[C2H4(g) -> 2C(s) + 2H2(g)] + [C4H10(g) -> 4C(s) + 5H2(g)]
Cancelling out the intermediate species, we have:
2C2H4(g) + H2(g) -> C4H10(g)
Now, we can add the enthalpy changes of each reaction to get the enthalpy change of reaction (3):
ΔH°(3) = 2ΔH°(2) + ΔH°(1)
ΔH°(3) = 2(-52.3 kJ) + (-125.6 kJ)
ΔH°(3) = -104.6 kJ - 125.6 kJ
ΔH°(3) = -230.2 kJ
Therefore, the standard enthalpy change for reaction (3) is -230.2 kJ.
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