Ch. 15: Energy and Chemical Change

Ch. 15: Energy and Chemical Change Sec. 15.4: Calculating Enthalpy Change

Objectives • Use Hess’s Law of summation to calculate the enthalpy change for a reaction. • Explain the basis for the table of standard enthalpies of formation • Calculate ΔHrxn using thermochemical equations. • Determine the enthalpy change for a reaction using enthalpies of formation data.

Calculating Enthalpy Change • C(s, diamond) C(s, graphite) • How can ΔH be determined when a reaction is very slow or can’t be duplicated in a lab?

Hess’s Law • Chemists use Hess’s law of heat summation to determine a theoretical value for ΔH. • Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction.

Hess’s Law A + 2B --> D A + B --> C _________________________________________ 3) A + 2B --> D ΔHo = ??? C + B --> D

Hess’s Law • If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for the individual steps. • The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.

Hess’s Law • Thus we can use the information tabulated for a relatively small number of reactions to calculate DH for a large number of different reactions.

Hess’s Law • Consider the combustion reaction of methane to form CO2 and liquid H2O CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) DH = ?

Hess’s Law • This reaction can be thought of as occurring in two steps: • In the first step methane is combusted to produce water vapor: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) DH = -802 kJ

Hess’s Law • In the second step water vapor condenses from the gas phase to the liquid phase: 2H2O(g) -> 2H2O(l) DH = -88 kJ

Hess’s Law • Hess’s Law says we can combine these equations and the enthalpy changes: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) DH = -802 kJ 2H2O(g) -> 2H2O(l) DH = -88 kJ CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) DH = - 890 kJ

Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • First, look over the equations that contain the substances found in the desired equation and have known enthalpy changes. • S(s) + O2(g) --> SO2(g) DH = -297 kJ • 2SO3(g)--> 2SO2(g) + O2(g) DH = 198 kJ

Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • Next, rewrite the found equations so that they are in agreement with the original equation in terms of coefficients and position of the reactants/products. • Equation #1 must be doubled because the original equation has 2 moles of S. S(s) + O2(g) --> SO2(g) DH = -297 kJ becomes 2S(s) + 2O2(g) --> 2SO2(g) DH = -594 kJ

Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • Equation #2 must be reversed because you want SO3 to be a product. 2SO3(g)--> 2SO2(g) + O2(g) DH = 198 kJ becomes 2SO2(g) + O2(g --> 2SO3(g) DH = -198 kJ

Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • Now add the rewritten equations to obtain the equation for the desired reaction. • 2S(s) + 2O2(g) --> 2SO2(g) DH = -594 kJ • 2SO2(g) + O2(g --> 2SO3(g) DH = -198 kJ • 2S(s) + 3O2(g) --> 2SO3(g) DH = -792 kJ

Hess’s Law • Please note that sometimes thermochemical equations are written with fractional coefficients to determine DH for one mole of product. • 2S(s) + 3O2(g) --> 2SO3(g) DH = -792 kJ can be written as • S(s) + 3/2 O2(g) --> SO3(g) DH = -396 kJ

Practice Problems • Use the equations below to determineDH for the decomposition of H2O2: 2H2O2 --> 2H2O + O2 a) 2H2 + O2 --> 2H2O DH = -572 kJ b) H2 + O2 --> H2O2DH = -188 kJ • Complete practice problems #28 & 29 on pg. 508.

Standard Enthalpy (Heat ) of Formation • Hess’s Law is useful in calculating unknown ΔH values. However, recording ΔH values for all known reactions would be extremely hard. • Scientists have chosen to record ΔH values for one type of reaction - the reaction in which a compound is formed from its elements under standard conditions.

Standard Enthalpy (Heat ) of Formation • The ΔH value for such a reaction is called ΔHfo. • It is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements under standard conditions. • Every free element in its standard state has a ΔHfo = 0 kJ.

Standard Enthalpy (Heat ) of Formation • For example: 2S + 3O2 --> 2SO3 ΔH = -792 kJ becomes S + 3/2O2 --> SO3 ΔHfo = -396 kJ • In this reaction, ΔHfo of S = 0 kJ & ΔHfo of O2 = 0 kJ. • The energy content of the product SO3 is 396 kJ less than the energy content of the elements from which it is formed.

Using ΔHfo • Find ΔHrxno for H2S + 4F2 --> 2HF + SF6 • To apply Hess’s Law using ΔHfo, you must have one equation for the formation of each compound in the given equation. • Locate these equations in given reference tables: Table 5, pg. 538 or Table R-11, pg. 975.

Using ΔHfo H2S + 4F2 --> 2HF + SF6 • H2 + S --> H2S ΔHfo = -20.6 kJ • 1/2H2 + 1/2F2 --> HF ΔHfo = -273.3 kJ • S + 3F2 --> SF6 ΔHfo = -1220.5 kJ These equations must now be manipulated so they can be “added” according to Hess’s law. • Equation 1 must be reversed. • Equation 2 must be doubled.

Using ΔHfo H2S + 4F2 --> 2HF + SF6 H2S --> H2 + S ΔHfo = 20.6 kJ H2 + F2 --> 2HF ΔHfo = -546.6 kJ S + 3F2 --> SF6 ΔHfo = -1220.5 kJ H2S + 4F2 --> 2HF + SF6 ΔHrxno = -1746.5 kJ Guess what? There’s an easier way!

Using ΔHfo • Use the following formula: ΔHrxno = ∑ ΔHfo(products) - ∑ ΔHfo(reactants) • The formula says to subtract the sum of heats of formation of the reactants from the sum of the heats of formation of the products.

Using ΔHfo • Find ΔHrxno for H2S + 4F2 --> 2HF + SF6 ΔHrxno = ∑ ΔHfo(products) - ∑ ΔHfo(reactants) = [(2)ΔHfo(HF) + ΔHfo(SF6)] - [ΔHfo(H2S) + (4)ΔHfo(F2)] = [(2)(-273.3 kJ) + (-1220.5 kJ)] - [-20.6 kJ + (4)(0 kJ)] = -1746.5 kJ

Practice Problems • Use standard enthalpies of formation to calculate ΔHrxno for the following: • CH4 + 2O2 --> CO2 + 2H2O • CaCO3 --> CaO + O2 • 2H2O2 --> 2H2O + O2

Ch. 15: Energy and Chemical Change