Kaplan MCAT General Chemistry Review

Enthalpy is a state function and is a property of the equilibrium state, so the pathway taken for a process is irrelevant to the change in enthalpy from one equilibrium state to another. As a consequence of this, Hess’s law states that enthalpy changes of reactions are additive. When thermochemical equations (chemical equations for which energy changes are known) are added to give the net equation for a reaction, the corresponding heats of reaction are also added to give the net heat of reaction. You can think of Hess’s law as being embodied in the enthalpy equations we’ve already introduced. For example, we can describe any reaction as the result of breaking down the reactants into their component elements, then forming the products from these elements. The enthalpy change for the reverse of any reaction has the same magnitude, but the opposite sign, as the enthalpy change for the forward reaction. Therefore,

H (reactants elements) = - H (elements reactants)

The H_rxn written as

H_rxn = H (reactants elements) + H (elements products)

becomes

H_rxn = - H (elements reactants) + H (elements products) = H (elements products) - H (elements reactants)

Thus,

H°_rxn = Σ(H°_f of products)- Σ(H°_f of reactants)

Consider the following phase change:

Br₂ (l) Br₂ (g) H°_rxn = 31 kJ/mol

The enthalpy change for the phase change is called the heat of vaporization (H°_vap). As long as the initial and final states exist at standard conditions, the H°_rxn will always equal the H°_vap, irrespective of the particular pathway that the process takes in vaporization. For example, it’s possible that Br₂ (l) could first decompose to Br atoms, which then recombine to form Br₂ (g), giving the following reaction mechanism. However, because the net reaction is the same as the one shown previously, the change in enthalpy will be the same.

Br₂ (l) Br₂ (g) H = (31 kJ/mol)(1 mol) = 31 kJ

Example: Given the following thermochemical equations:

a)C₃H₈ (g) + 5 O₂ (g) 3 CO₂ (g) + 4 H₂O (l)H_a = -2,220.1 kJb)C (graphite) + O₂ (g) CO₂ (g)H_b = -393.5 kJc)H₂ (g) + 1/2 O₂ (g) H₂O (l)H_c = -285.8 kJ

Calculate H for this reaction:

d) 3 C (graphite) + 4 H₂ (g) C₃H₈ (g)

Solution: Equations a, b, and c must be combined to obtain equation d. Since equation d contains only C, H₂, and C₃H₈, we must eliminate O₂, CO₂, and H₂O from the first three equations. Equation a is reversed to get C₃H₈ on the product side (this gives equation e). Next, equation b is multiplied by 3 (this gives equation f) and c by 4 (this gives equation g). The following addition is done to obtain the required equation d: 3b + 4c + e.

e) 3 CO₂ (g) + 4 H₂O (l) C₃H₈ (g) + 5 O₂ (g) H_e = 2,220.1 kJ

f) 3 × [C ( graphite) + O₂ ( g) CO₂ ( g)] H_f = 3 × -393.5 kJ

g) 4 × [H₂ ( g) + _1/2 O₂ (g) H₂O (l )] H_g = 4 × -285.8 kJ

3 C ( graphite) + 4 H₂ (g) C₃ H₈ ( g) H_d = -103.6 kJ where H_d = H_e + H_f + H_g.

MCAT Expertise

When doing a problem like this on the MCAT, make sure to switch signs when you reverse the equation. Also, make sure to multiply by the correct stoichiometric coefficients when performing your calculations.

It is important to realize that Hess’s law applies to any state function, including entropy and Gibbs free energy.

Bond Dissociation Energy

Hess’s law can also be expressed in terms of bond enthalpies, also called bond dissociation energies, when these are given. Bond dissociation energy is the average energy that is required to break a particular type of bond between atoms in the gas phase (remember, bond dissociation is an endothermic process). Bond dissociation energy is given as kJ/mol of bonds broken. For example, the bond enthalpy of the double bond in the diatomic oxygen molecule O₂ is 498 kJ/mol (of double bonds broken). The tabulated bond enthalpies for bonds found in compounds other than diatomic molecules are the average of the bond energies for the bonds in many different compounds. For example, the C–H bond enthalpy (415 kJ/mol) that you will find listed in a table of bond enthalpies is averaged from measurements of the individual C–H bond enthalpies of thousands of different organic compounds. Please note that bond formation, the opposite of bond breaking, has the same magnitude of energy but is negative rather than positive; that is, energy is released when bonds are formed. The enthalpy change associated with a reaction is given by

H_rxn = ΣH_{bonds broken} + ΣH_{bonds formed} = total energy absorbed - total energy released

For example,

H₂ (g) 2H (g) H = 436 kJ/mol

Key Concept

Because it takes energy to pull two atoms apart, bond breakage is always endothermic. The reverse process, bond formation, must always be exothermic.