Using the provided table, determine the molar enthalpy (in kJ/mol) for the reaction C(s) + CO₂ (g)→ 2 CO (g)

Question

Substances given in provided table=

C (s) = 0 KJ/mol

CO₂ (g) = -393.5 KJ/mol

CO = -110..5 KJ/mol

Answer 1

C(s) + CO₂ (g) ==> 2 CO (g)

∆H_rxn = ∑∆H_products - ∑∆H_reactants

∆H_products = (2 mol x -110.5 kJ/mol) = -221 kJ

∆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