Enthalpy change, (Δr H) of reactions - reaction Enthalpy

Standard states:

A pressure of 1 atm and a constant temperature (25˚C) are taken as standard states.

State	Standard states
Gas	Ideal gas at the given temperature and 1 atm
Liquid	Pure liquid at the given temperature and 1 atm
Solid	Stable crystalline from at given temperature and 1 atm pressure

Heat of reaction:

Heat of reaction is defined as the amount of heat evolved or absorbed when the reacting species, as represented by a balanced chemical equation, have completely reacted. Enthalpy or heat of reaction is given by 

∆Hr = (sum of enthalpies of products) – (sum of enthalpies of reactants)

ΔH = ∑a_iH_{f products} − ∑b_iH_{f reactants}

a_i and b_i = stoichiometric coefficients of products and reactants

In the calculation of heat of reaction, it is a convention to assume that the heat of formation of elements in their standard state is zero.

Factors affecting heat of reaction: 

Physical state of reactants and products: 

H₂(g) +  ½ O₂(g) → H₂O(g);         ΔH = − 57.8 kcal
H₂(g) +  ½ O₂(g) → H₂O(l);          ΔH = − 68.32 kcal 

Allotropic forms of the element: 

C(diamond) + O₂(g) → CO₂(g);              ΔH = − 94.3kcal
C(amorphous) + O₂(g) → CO₂(g);           ΔH = − 97.6kcal

Temperature: 

ΔH_T₂ - ΔH_T₁ / T₂ - T₁ = ΔC_p

ΔU_T₂ - ΔU_T₁ / T₂ - T₁ = ΔC_v

Enthalpies of Phase Transition:

Enthalpy of fusion:

It is the enthalpy change involved in the conversion of one mole of a substance from solid state to liquid state at its melting point.
This equals the latent heat of fusion per gram multiplied by the molar mass.

For example:

H₂O(s) → H₂O(l);         Δ_fus H^θ = 1.44kcal/mol

Enthalpy of vaporization:

It is the enthalpy change involved in converting one mole of the substance from liquid state to gaseous state at its boiling point.

For example:

H₂O(l) → H₂O(g);        Δ_vapH = 10.5kcal/mol

Enthalpy of sublimation:

It is the enthalpy change involved in the conversion of one mole of a solid directly into its vapor at a given temperature below its melting point.

Δ_subH = Δ_fusH + Δ_vapH

For example:

I₂(s) → I₂(g);                Δ_subH = 14.9kcal/mol

Enthalpy or Heat of Formation:

The amount of heat evolved or absorbed when one mole of a substance is formed from its constituent elements is called the heat of formation.
Standard enthalpy of formation.

It is the enthalpy change of a reaction by which a compound is formed from its constituent elements, the reactants and products, at standard state (i.e., at 298K and 1 atm pressure). Its symbol is Δ_f H^⊝.

By convention, standard enthalpy for formation, Δ_f H^⊝ of an element in reference state, i.e., its most stable state of aggregation is taken as zero.

C (s) + O₂ (g) → CO₂ (g); Δ_f H^⊝ = -393.51 kJ/mol

The compounds having positive enthalpies of formation are called endothermic compounds and are less stable than the reactants. 

The compounds having negative enthalpies of formation are known as exothermic compounds and are more stable than the reactants.

Thermochemical equation:

A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH. In variable form, a thermochemical equation would look like this:

A + B → C;     ΔH = (±) x

Where {A, B, C} are the usual agents of a chemical equation with coefficients and “(±) x” is a positive or negative numerical value, usually with units of kJ.

Endothermic reaction: A + B + Heat → C, ΔH > 0

Exothermic reaction: A + B → C + Heat, ΔH < 0

Hess's Law:

If a specified set of reactants is transformed to a specified set of products by more than one sequence of reactions, the total enthalpy change must be the same for every sequence.

The enthalpy change in a chemical or physical process is the same whether the process is carried out in one step or in several steps.

Example:

Calculate the enthalpy of the following chemical reaction:

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

Given:

C(s) + O₂(g) → CO₂(g);          ΔH = -393.5 kJ/mol  --- (i)
S(s) + O₂(g) → SO₂(g);           ΔH = -296.8 kJ/mol  --- (ii)
C(s) + 2S(s) → CS₂(l);            ΔH = +87.9 kJ/mol   ---- (iii)

Solution:

Leave eq 1 untouched (want CO₂ as a product)
multiply second eq by 2 (want to cancel 2S, also want 2SO₂ on product side)
flip 3rd equation (want CS₂ as a reactant)

2) The result:

C(s) + O₂(g) → CO₂(g);                      ΔH = -393.5 kJ/mol
2S(s) + 2O₂(g) → 2SO₂(g);                 ΔH = -593.6 kJ/mol ← ( multiply by 2 on the ΔH )
CS₂(ℓ) → C(s) + 2S(s);                        ΔH = -87.9 kJ/mol ← ( sign change on the ΔH)

3) Add the three revised equations. C and 2S will cancel.

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

4) Add the three enthalpies for the final answer.

ΔH = -1075 kJ/mol

Some things to remember:

If you have to reverse a reaction to get things to cancel, the sign of ΔH must also be reversed.

If you have to multiply an agent to get it to cancel, all other agents and ΔH must also be multiplied by that number.

Generally ΔH values given in tables are under 1atm and 25 °C (298.15 K), so be aware of what conditions your reaction is under.