Q 8.67
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Re: Q 8.67
You need to first make the balanced equation of how the products form from their elemental forms because we are trying to find standard enthalpy of formation which means the products must be elemental forms. Then use the bond enthalpies and subtract the energy released when creating the new product bonds from the energy needed to break each of the bonds in the reactants. For example for part A the equation is H2 + 1/2 O2 -> H20. In this case we need to break one H2 bond and 1/2 of an O2 bond and for two O--H bonds. Thus we do (436 kJ + 1/2(496kJ)) - (2(463kJ)) based on the bond enthalpies in the tables.
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Re: Q 8.67
After you do the above calculations with the bond enthalpies, you should have delta H = -242 kJ/mol.
You then need to take into account that water is in it's gaseous phase here, and the question asks for you to find the enthalpy of formation at a liquid state.
With this in mind, subtract the enthalpy of vaporization, which is 44 kJ/mol, from -242 kJ/mol, to get the answer of -286 kJ/mol.
You then need to take into account that water is in it's gaseous phase here, and the question asks for you to find the enthalpy of formation at a liquid state.
With this in mind, subtract the enthalpy of vaporization, which is 44 kJ/mol, from -242 kJ/mol, to get the answer of -286 kJ/mol.
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