## Thermochemistry

### Introduction

Every chemical change and every physical change involves an energy change. These energy changes that accompany the transformation of matter help us understand better the nature of chemical and physical changes.

First law of thermodynamics : Energy can be neither created nor destroyed.

#### Thermodynamic Quantities

Chemical thermodynamics is the study of the energy effects accompanying chemical and physical changes. Three of these thermodynamic quantities are: enthalpy, H, entropy, S, and Gibbs free energy, G.

#### Enthalpy

The enthalpy of formation of a compound is a change in energy that occurs when it is formed from its element.
CH4 + 2O2 --> CO2 + 2H2O , H = -890.2 kJ/mol
4Fe(s) + 3O2(g) --> 2Fe2O3(s) , H = -1647 kJ/mol
N2O4(g) --> 2NO2(g) , H = 57.24 kJ/mol
The enthalpies of formation may be either positive or negative. A positive enthalpy of formation indicates that energy has to be provided in order for the reaction to proceed. A negative enthalpy of formation indicates that energy is evolved.
What the enthalpy of formation tells us is the change of enthalpy that would occur if a reaction pathway could be found that takes one of the elements to the compound of interest.

Hess's law: The overall reaction enthalpy is the sum of reaction enthalpies of each step into which the reaction may formally be divided.

#### Entropy

The entropy, S, of a system is a measure of its randomness or, its disorder.

Second law of thermodynamics: The entropy of the universe increases in the course of every natural change.

Consider the oxidation of iron (25oC).:
4Fe(s) + 3O2(g) --> 2Fe2O3(s) S = 2(87.37J/(K mol)) - 4(27.29J/(K mol)) - 3(205.2J/(K mol)) = -550.12J/(K mol) ( H = -1647kJ/mol)
The negative enthalpy in this reaction (exothermic reaction), causes the S(surroundings) to be positive: S(surroundings) = - H/T S(surroundings) = -(-1647 kJ/mol)/(298.15 K) = +5524 J/(K mol)
This increase in entropy arises because the reaction releases energy into the surroundings.

In contrast to oxidation of iron, the reaction: N2O4(g) --> 2NO2(g)
is an endothermic reaction ( H = +57.2 kJ/mol): S = 2(240.1J/(K mol)) - 304.38J/(K mol) = 175.82J/(K mol) S(surroundings) = -(57.24 kJ/mol)/(298.15 K) = -192 J/(K mol)
All endothermic reactions decrease the entropy of the surroundings.

The sum of the total entropy changes S(total) = - H/T + S(surroundings) is a measure of the total change of energy in the universe.
For every 4 mol of Fe converted to 2 mol Fe2O3 there is an overall increase in the entropy of the universe: S(total) = +5524J/(K mol) + (-550.12J/(K mol)) = +4973.88 J/(K mol)
This large positive quantity indicates that the reaction is spontaneous and a thermodynamic reason why steel corrodes.

The dissociation of N2O4 indicates that the total entropy change: S(total) = -192J/(K mol) + (175.82J/(K mol)) = -16.18 J/(K mol)
Because this is a small, negative value, we can conclude that this reaction does not go to completion.

#### Gibbs Free Energy.

When a reversible chemical change occurs, the difference between H and T S represents the amount of available energy released or absorbed as a result of the chemical change.It is expressed by:
-T S(total) = H - T S
The new quantity -T S(total) is now given a new symbol and a new name. It is called the Gibbs function of reaction or Gibbs free energy and denoted G. The last equation therefore becomes: G = H - T S
This energy represents the driving force of the reaction. When G is negative the reaction will be a spontaneous one; when G is positive the reverse of the reaction as written will be the spontaneous one; when G is zero equilibrium exists. The magnitude of G is a measure of the extent to which the reaction will go to completion. Thus, a knowledge of G values enables us to predict the course of a reaction. For example, we can predict that the reaction: Cl2 (g) + 2I-(aq) --> 2Cl-(aq) + I2(g) G = -38 kcal will be a spontaneous one since G is negative. Had G been positive, the spontaneous reaction would have been the reverse of the reaction as written. -

### State symbols

Tab. Thermochemistry (state symbols)
State symbols Example Interpretation
(g) H2(g) Gas
(l) H2O(l) Liquid
(s) Cu(s) Solid
(aq) Cu+2(aq) Aqua

CHEMIX School - Thermochemistry Related topics:

Signs and symbols for Molecular Weight Calculator
Signs and symbols for Balancing Chemical Equations

### Example

Ammonium nitrate ,NH4NO3, decomposes explosively:
NH4NO3(s) > N2O(g) + H2O(g)
a) Calculate H for the balanced reaction.
b) If 36 grams of H2O are formed from the reaction, how much heat was released?

Solution:
a) Insert and calculate: NH4NO3(s) > N2O(g) + H2O(g)
b) Arg. H2O by 36g, insert and calculate: NH4NO3(s) > N2O(g) + 36gH2O(g)

Here's an example of an equation using all the state symbols (g),(l),(s) and (aq)
CaCO3(s) + H2SO4(aq) > Ca+2(aq) + CO2(g) + SO4-2(aq) + H2O(l)

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