8.63
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8.63
For part c does the h20 (1) have a different formation enthalpy because when i calculate evreything my value is off
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Re: 8.63
The enthalpy of formation for H2O(l) is -285.83, but you would have to multiply that number by 2 because there are 2 moles in the reaction.
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Re: 8.63
The standard enthalpy of formation for H2O(l) should be -285.83 kJ/mol.
standard reaction enthalpy = [(-471.5 kJ/mol) * 1 mol + 2 mol * (-285.83 kJ/mol)] - [(-39.7 kJ/mol) * 1 mol + 2 mol * (-482.37 kJ/mol)]
=-38.72 kJ
standard reaction enthalpy = [(-471.5 kJ/mol) * 1 mol + 2 mol * (-285.83 kJ/mol)] - [(-39.7 kJ/mol) * 1 mol + 2 mol * (-482.37 kJ/mol)]
=-38.72 kJ
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Re: 8.63
There is a mistake in the solution manual as Dr. Lavelle pointed out on his website: the enthalpy of formation of K2S is actually -471.5 kJ/mol but it says -417.5 kJ/mol in the manual. The correct answer is then -38.72 kJ.
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