Use the information provided to determine ΔH°rxn for the following reaction:

Use the information provided to determine ΔH°rxn for the following reaction:





ΔH°f (kJ/mol) CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g) ΔH°rxn = ?


CH4(g) -75


CCl4(g) -96


HCl(g) -92

A) -389 kJ


B) -113 kJ


C) +113 kJ


D) -71 kJ

Hey D!

For this question, use the standard enthalpy of formation values given for each of the given molecules and subtract the sum of these values of the reactants from the products. So basically: ΔH°f = (sum of ΔH°f of products) - ( sum of ΔH°f of reactants)

Now we just add the enthalpies of each side. Products: (-96 + 4(-92)) = -464. Reactants: (-75+4(0))= -75. All elements, diatomic or not, always have standard enthalpies of zero. Now subtract: -464-(-75) = -389. And so ΔH°rxn = -389KJ choice A!

I hope this helped and feel free to ask any more questions :)