Thermo-chemistry (enthalpy) / Hess Law

Determine the change in enthalpy (∆H) for the combustion of 100.0 g of ethanol . C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H = -1370 kJ

100 g x (1mol / 46.08 g) x ( -1370 kJ / mol) = - 2973.1 kJ ∆H = - 2970 kJ

Determine the ∆H of the reaction: C → E + 3D Use the following reactions to determine ∆H 2A → 1/2B + C ∆H1 = 5 kJ/mol (3/2)B + 4C → 2A + C + 3D ∆H2 = -15 kJ/mol E + 4A → C ∆H3 = 10 kJ/mol

3A → 3/2B + 3C ∆H1 = 15 kJ/mol (3/2)B + 4C → 2A + C + 3D ∆H2 = -15 kJ/mol C → E + 4A ∆H3 = -10 kJ/mol ∆H = -10 kJ/mol

Determine the change in enthalpy (∆H) for the production of 5oo.o g of CO2 from the combustion of ethanol. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H = -1370 kJ

500 x ( 1mol / 44.01 g) x ( -1370 kJ / 2 mol) = -7782.3 kJ ∆H = - 7780 kJ

Change in enthalpy calculated from sum of several reactions. This is another application of Hess's Law

A + 2B → 3C ∆H = -500 kJ /mol 2C + A → 2D + B ∆H = -350 kJ /mol 2A + B → C + 2D ∆H = ??? ∆H = -850 kJ /mol

Determine the heat of formation of ethanol (∆H°f) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H = -1370 kJ ∆H°f (kJ/mol) CO2(g) -393.5 H2O(l) -285.8

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H = -1370 kJ (-393.5 (2) + (-285.5(3)) - (x-0) = -1370 kJ x = -273.5 kJ/mol ∆H°f ethanol = -273.5 kJ/mol

Determine the ∆H of the reaction using the heats of formation: CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g) ∆H°f (kJ/mol) CH4(g) -75 CCl4(g) -96 HCl(g) -92

CH4(g) + 4 Cl2(g) → CCl4(g) + 4 HCl(g) Products - reactants = ∆H (-96 + -92(4)) - (-75 + 0) = -389 kJ ∆H = -389 kJ

Break down of a chemical reaction with respect to enthalpy.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O First the bonds must be broken in the reactants (C₃H₈ + 5O₂) this is an endothermic process as energy is required. (activation energy) secondly bonds must be formed to make the products ( 3CO₂ + 4H₂O) This is an exothermic process, energy is released from the formation of bonds. All reactions are both endothermic and exothermic, the overall character of the reaction is based on which one is greater in magnitude for that given reaction.

Hess's Law

Regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function, and independent of pathway.

Enthalpy

The heat absorbed or released during a chemical reaction where the only work done is the expansion of a gas at constant pressure. expresses most energy changes resulting from chemical reactions. Because it is impractical to measure total enthalpy, we measure changes in enthalpy.

Change in enthalpy calculated from heats of formation. This is one application of Hess's Law

ΔH°rxn = (sum ofΔH°f products - sum ofΔH°f reactants)

Heat of formation ∆H⁰f

∆H for a chemical reaction in which one mole of a compound is formed from its constituent elements. Example: C + 2H₂ → CH₄ ∆H = -74.7 kJ /mol