Question

What is the enthalpy for this reaction? Report your answer in kJ/mol to 1 decimal place.

Answer 1

Answer:∆H reactants = 1 mol x 0 kJ/mol + (1mol x -393.5 kJ/mol) = -393.5 kJ ∆H rxn = -221 kJ - (-393.5 kJ) = 172.5 kJ The molar enthalpy (in kJ/mol) for this reaction is 172.5 kJ/mol

Step-by-step explanation: