Using the table of standard enthalpies of formation at 298 K provided in the pdf
ID: 478045 • Letter: U
Question
Using the table of standard enthalpies of formation at 298 K provided in the pdf, calculate the H° for the following reaction: NaOH(aq) + HCl(g) NaCl(aq) + H2O(l) Please give your answer in kJ. Do NOT input the units. Give your answer to ONE decimal place. For example, if your answer is 12.62581, please input 12.6. If you need to input a negative value please use the hyphen "-" followed directly by the number, NO SPACE. Do not use the underscore ("_") For example, if your answer is -15.1611, please input -15.2.
Explanation / Answer
Enthalpy of reaction = Enthalpy of products - Enthalpy of reactants.
Standard Enthalpy of formation of reactants:
NaOH(aq)= -470.1 kJ/mol
HCl(g)= -93.3 kJ/mol
Standard Enthalpy of formation of products:
NaCl(aq)= -407 kJ/mol
H2O(l)= -285.8 kJ/mol
Enthalpy of reaction at 298K=
(-407-285.8)kJ/mol - (-470.1-93.3) kJ/mol
= - 692.8 kJ/mol - (-563.4) kJ/mol = -692.8 + 563.4 kJ/mol =-129.4kJ/mol
Standard heat of reaction at 298K = -129.4 kJ/mol
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