Using data from the back of the textbook, calculate the ΔH°rxn​ for the formation of 6.00 moles of product?

Question

2SO2 + O2 --> 2SO3

in the appendix in the back of the book it states that the ∆Hf in kJ/mol of SO2 is -296.8 and the SO3 is -395.7. the O2 is 0 kJ/mol. and then i get a hint that states that since I have these values, that I then need an equation that we know which relates them to the ΔH°rxn?

Answer 1

Hi Daisy,

Reaction enthalpy can be calculated by subtracting the heats of formation for the reactants from the heats of formation for the products (all weighted by the coefficients for each substance). The equation I believe you are referring to is as follows:

ΔH°rxn = ∑ΔHf (products) - ∑ΔHf (reactants)

In your example, that would be ΔH°rxn = (2 mol * -395.7 kJ/mol) - ((2 mol * -296.8 kJ/mol) + (1mol * 0 kJ/mol))

You can take it from there. I hope this helps.

Jack B.