Given the standard enthalpy changes for the following two reactions:
(1) 2Pb+O2 -> 2PbO ΔH° = -434.6 kj
(2) Pb+Cl2 -> PbCl2 ΔH° =-359.4kj
what is the standard enthalpy change for the reaction:
(3) 2PbCl2+O2 -> 2PbO+2Cl2 ΔH° =?
ΔH° = _______ kJ
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To determine the standard enthalpy change for reaction (3), we need to combine the enthalpy changes of reactions (1) and (2).
First, let's flip reaction (1) and multiply it by 2 to match the stoichiometric coefficients in reaction (3):
2PbO -> 2Pb + O2 ΔH° = +434.6 kJ (Note that we flipped the sign of the enthalpy change to make it reactants - products)
Next, let's multiply reaction (2) by 2 to match the stoichiometric coefficients in reaction (3):
2Pb + 2Cl2 -> 2PbCl2 ΔH° = -718.8 kJ
Now, let's add these two reactions together to get reaction (3):
2PbO + 2Pb + 2Cl2 -> 2PbO + 2Cl2
Cancelling out the common species on each side of the reaction, we're left with:
2PbO + 2Pb -> 2PbO + 2Cl2 ΔH° = 434.6 kJ + (-718.8 kJ)
ΔH° = -284.2 kJ
Therefore, the standard enthalpy change for reaction (3) is -284.2 kJ.
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