THERMOCHEMISTRY

ENTHALPY:

The Heat of Reaction and Hess’s Law

Enthalpy

• The term used for the total energy of a system when it is at constant pressure is called the enthalpy of a system. It is symbolized as H.

• The word enthalpy comes from Greek endon - in, within and thalpein - to warm

• Enthalpy is a state function so when a system reacts at constant pressure and absorbs or evolves energy we say that it experiences an enthalpy change.

Enthalpy: a state function• As a state function, its value depends solely on the difference in enthalpy between the initial

and final states and not on the mechanism by which the system undergoes this change.

H = Hfinal - Hinitial• For a chemical reaction, the initial state refers to the reactants and the

final state to the products, so for a chemical reaction we can write:

H = Hproducts - Hreactants

The Importance of H

• Our formal definition of H, in an absolute sense, serves only one purpose. This purpose is to define the meaning of positive and negative values of H.

• It also gives us an idea of the magnitude of the energy absorbed or released.

Sign Conventions for Enthalpy

• a) If H is negative

this reaction is

(-) Exothermic

• b) If H is positive

this reaction is (+)

Endothermicsystem

surroundings

system

surroundings

Energy is leaving the system

Energy is being absorbed by the system

b) if Hproducts < Hreactants then H is__ (+/-)

and the reaction is ________________.

Endothermic or Exothermic?

• Determine whether a reaction would be endothermic or exothermic under the following conditions:

a) if Hproducts > Hreactants then H is__ (+/-)

and the reaction is ________________.

Thermochemical Equations• Thermochemical reactions are generally

written at standard thermochemical temperature and pressure. This is adapted to the normal laboratory conditions of 25C and 1 atmosphere.

• To show that H is at this standard condition it is designated as: H

• The units of H are usually kilojoules (kJ)

Example: A Thermochemical Equation

N2(g) + 3 H2(g) 2 NH3(g) H= -92.38 kJ

How can we calculate the enthalpy of this reaction given only 1.5 mol of H2?Divide the entire equation by 2 including H

Notice that all physical statesof the reactants and products are cited in this equation.

This sign indicates an Exothermic Reaction

H= - 92.38 kJ/2 = - 46.19 kJ

RULES FOR MANIPULATING THERMOCHEMICAL EQUATIONS:

1. When an equation is reversed (written in the opposite direction) the sign of H must also be reversed.

2. Formulas canceled from both sides of an equation must be for the substance in identical physical states.

3. If all the coefficients of an equation are multiplied or divided by a common factor, the value of H must be likewise changed.

Example of a Hess’s Law Calculation

• Given two chemical reactions (#1 and #2) and the H for each, find the H for a third (target) reaction.

#1: Fe2O3 + 3CO(g) 2 Fe(s) + 3CO2(g) H=-26.7 kJ

#2: CO(g) + 1/2 O2(g) CO2(g) H=-283.0 kJ

Calculate the H for the following reaction:2 Fe(s) + 3/2 O2(g) Fe2O3(s) H= ?

STEP 1:

TARGET

EQUATION: 2 Fe(s) + 3/2 O2(g) Fe2O3(s) H= ?

EQ #1: Fe2O3 + 3CO(g) 2 Fe(s) + 3CO2(g) H=-26.7 kJ

To match the target equation correctly, equation #1 must have 2 Fe on the left. However, it has 2 Fe on the product side of the reaction arrow.SOLUTION: Reverse the entire equation

as well as the +/- sign on H

Begin by reviewing the position of the reactants and the products in the desired reaction and compare them with the given equations.

Solution:

2 Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g)

Fe2O3(s) + 3CO(g) 2 Fe(s) +

3CO2(g) H = - 26.7 kJ

H = + 26.7 kJ

Given equation:

Reversing the equation:

Reverses the sign of enthalpy:

STEP 2:• There must be 3/2 O2 on the left, and we must

be able to cancel 3 CO and 3 CO2 when the equations are added. If we multiply the second equation by 3 we will obtain the necessary coefficients.

H = 3 (-283.0 kJ) = - 849.0 kJ

Remember to multiply the H by 3 also.

3CO(g) + 3/2 O2(g) 3 CO2(g)

#2: 3 [CO(g) + 1/2 O2(g) CO2(g) ] H=

-283.0 kJ

STEP 3:• Now put the equations together and cancel out

anything that appears on both sides of the equations (very similar to adding redox equations together).

2 Fe(s) + 3CO2(g) Fe2O3(s) + 3CO(g) H = + 26.7 kJ 3CO(g) + 3/2 O2(g) 3 CO2(g) H = - 849.0 kJ

2Fe(s) +3CO2(g) +3CO(g) + 3/2O2(g) Fe2O3(s) + 3CO(g)+ 3

CO2(g) NET: 2 Fe(s) + 3/2 O2(g) Fe2O3(s) H = - 822.3 kJ

Hess’s Law

• Stated in terms of standard enthalpies of formation:

H = (sum of Hf of all the products) -

(sum of Hf of all the reactants)

HEATS OF FORMATION AND HESS’S LAW:

• The standard enthalpy of formation, Hf , is the amount of heat absorbed or evolved when one mole of the substance is formed at 25C and 1 atm from its elements (Standard Thermochemical Conditions) and in its most stable form (solid, liquid, gas). Tables are provided.

Using Hf Tables to Calculate

H • Sample Hf Table – Note that Hf is in

kJ/molSubstance Hf Substance Hf

Al(s)

BaCO3(s)

CO(g)

CO2(g)

CS2(g)

0-1219-110-394+117

Cl2(g)

H2O(l)

H2O(g)

NaHCO3(s)

Na2CO3(s)

0-286-242 -947.7-1131

Note also that Hf for elements in their standard states = 0

Example of Hess’ Law• Some chefs keep baking soda, NaHCO3

handy to put out grease fires. When thrown on the fire, baking soda partly smothers the fire and the heat decomposes it to give CO2, which further smothers the flame. The equation for the decomposition of NaHCO3 is:

2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)

Calculation Process

• Recalling that:H = (sum of Hf of all the products) -

(sum of Hf of all the reactants)• And utilizing the Table Values of all

products and reactants:-1131 kJ/mol Na2CO3

-286 kJ/mol H2O -394 kJ/mol CO2

-947.7 kJ/mol NaHCO3

Note the minus sign

Complete the Final Calculation 2 NaHCO3(s) Na2CO3(s) + H2O(l) +

CO2(g)

-947.7 kJ/mol -1131 kJ/mol -286 kJ/mol -394 kJ/mol

H =(- 1131 kJ/mol + - 286 kJ/mol + - 394 kJ/mol)

- (2 x - 947.7 kJ/mol) = -1131 + -286 + -394 + 1895.4 = + 84 kJ

Note that you must multiply the enthalpy by the coefficient

Note also that you must correct for a sign change: -(-) = +

Summary• Enthalpy, H, is a state function and is expressed in kiloJoules.• There are two types of calculations based on Hess’s Law that

deal with enthalpy illustrated in this tutorial:

1. Those in which one manipulates several different chemical reactions to combine and form the desired reaction. H’s for each reaction has been provided to calculate the desired H.

2. Enthalpy calculations utilizing a Table of Standard Enthalpies of Formation which can be used with the following equation: H = (sum of Hf of all the products) -

(sum of Hf of all the reactants)