CHEM 640: Biological Physical Chemistry

Section 2. Thermochemistry

Krzysztof Kuczera

Contents

• Introduction
• Thermochemistry: Phase changes
  • Phases and phase changes
  • Standard states
  • Work and heat
  • Phase transitions: state of system
  • Manipulating transition enthalpies
  
  
• Thermochemistry: chemical reactions
  • Reaction enthalpy
  • The return of the standard state
  • Standard enthalpies of formation
  • Heat effects of chemical reactions
  • Work effects of chemical reactions
  • Reaction energies
  • Combining reaction enthalpies: Hess's Law
  • Variation of reaction enthalpies with T: Kirchhoff's Law
  • Chemical reactions: state of system
  
  
• Overview of Changes of State

Introduction

In this section we will analyze energy flow in systems for which changes in composition are allowed. We will look at two ways of changing composition: physical change, through phase transitions and chemical change, through chemical reactions. All considerations in this section will deal with closed systems having pressure-volume work only. Our starting point will be the First Law of Thermodynamics:

$$dU =  dq    or     \Delta U =  q       (closed sys; pV work only; V=const)$$

$$dH =  dq    or     \Delta H =  q       (closed sys; pV work only; p=const)$$

These equations tell us that analyzing heat flow is going to be especially simple under conditions of either p=const or V=const, as q becomes equal to the change of a state function, and thus depends only on the initial and final state, but not the path of the process.

In most practical applications in kitchen, lab or industry the condition p=const applies. Thus, the enthalpy becomes a primary quantity of interest in the study of heat effects of phase transitions and chemical reactions. The properties of enthalpy are collected in the Enthalpy section.

Thermochemistry: Phase changes

Phases and phase changes

Substances may exist in different phases, i.e. specific states that are uniform throughout in physical state and chemical composition. Examples of phases are the solid, liquid and gaseous phases of water. Some substances have several different phases which are all solids - e.g. water has at least eight solid forms, most of which exist at very high pressures, while carbon has at least three solid phases - diamond, graphite and fullerene. Different phases of a given substance differ in the physical organization of the constituent atoms or molecules, but have the same chemical structure. Thus, transitions between phases are often called physical changes of state.

The process of conversion of a substance from one phase to another is called a phase transition. To simplify matters we will usually consider phase transitions that occur at constant values of both temperature and pressure. This may be schematically represented as

$$Phase a  \rightarrow  Phase b        (p,T=const)$$

For this process we can define an enthalpy change

$$\Delta H =  H_f -  H_i =  H(b) -  H(a)$$

The enthalpy of transition $\Delta_{trs} H$ is the enthalpy change per mole for this process, i.e.

$$\Delta_{trs} H =  \frac{\Delta H}{n}      or       \Delta H =  n\Delta_{trs} H$$

Another way of writing this is

$$\Delta_{trs} H =  H_m(b) -  H_m(a)$$

where $H_m(a), H_m(b)$ are the molar enthalpies of a and b.

Example. To vaporize a sample of 250 g of liquid water at 100 $^oC$ and 1 bar required input of 565 kJ of heat. What is the enthalpy of vaporization of water under these conditions?

This is a closed system, with pV work only, and the process occurs at a constant pressure (equal to 1 bar). Thus

$$\Delta H =  q$$

$\Delta_{vap} H =  \frac{\Delta H}{n} =  \frac{q}{n} =  \frac{565 kJ}{13.88 mol} =  40.7 \frac{kJ}{mol}$

where we used n = (250 g)/(18.015 g/mol) = 13.88 mol

Most often discussed phase changes are

• $solid \rightarrow  liquid$   freezing
• $liquid \rightarrow  solid$   melting or fusion
• $liquid \rightarrow  gas$   vaporization
• $gas \rightarrow  liquid$   condensation
• $solid \rightarrow  gas$   sublimation

Standard states

In chemical thermodynamics we start by considering substances in the simplest kind of state, called a standard state. For a substance existing as a solid, liquid or gas, the standard state is the pure substance at the standard pressure of 1 bar.

The situation with phase transitions of pure substances is especially simple, as each phase forms a separate, pure subsystem. For example when solid ice melts to form liquid water, we can see with the naked eye where the solid ends and the liquid begins. Even if the solid phase consists of many small ice crystals (like crushed ice), we can identify macroscopic regions that belong to either the solid or liquid phase. There is no mixing at the microscopic level.

Thus, for any phase transition occurring at 1 bar we can define the standard enthalpy of transition $\Delta_{trs} H^o$ as the enthalpy change per mole for the process:

$$Phase a  \rightarrow  Phase b        (p=p^o=1 bar; T=const)$$

$$\Delta_{trs} H^o =  H^o_m(b) -  H^o_m(a)$$

where $H^o_m(a), H^o_m(b)$ are the standard molar enthalpies of the two phases and the special symbol $p^o$ was introduced for the standard pressure of 1 bar.

The take-home message from this section is: phases exist as pure substances, so standard enthalpies are appropriate to describe real phase transitions. Remember that the standard enthalpy change refers to a transition between standard states, i.e. starting in one kind of pure substance at 1 bar and ending in another pure substance at 1 bar.

Note. The definition of the standard state does not set the temperature.

Work and heat

We consider phase transitions at p,T=const. Remember that a practical realization of the constant pressure condition is $p =  p_{ex} =  const$ (see p=const section).
In a closed system with pV work only at p=const we can use the following two thermodynamic properties:

1. The heat transfer is equal to the enthalpy change

$$q =  \Delta H =  n \Delta_{trs} H       (closed sys; pV work only; p=const)$$

We can determine enthalpy changes by measuring heat transfer (calorimetry).

2. The work is performed reversibly, since p=const implies mechanical equilibrium $(p =  p_{ex} =  const)$

$$w =  -p_{ex}\Delta V =  -p \Delta V$$

We can calculate the work done by measuring the volume change.

Additionally, using the First Law we get at p=const:

$$\Delta U =  q +  w =  \Delta H -  p \Delta V$$

which allows us to calculate the internal energy change in a phase transition.

Phase transitions: state of system

In a system containing a pure substance undergoing a phase transition, we have to use four variables to describe the state, e.g.

$$(n_a,n_b,p,T)$$

i.e. not only can we change the temperature T and pressure p, but also the system composition, defined by the amounts of the two phases $n_a and  n_b$. Since our system is closed, the total amount of substance must remain constant:

$$n_a +  n_b =  const$$

The amounts of substance cannot change in an arbitrary way, if the amount of one phase decreases, the amount of the other phase must increase by the same value:

$$\Delta n_a =  -\Delta n_b$$

A phase transition is reversible if the two phases are in equilibrium. The solid and liquid phases of water are at equilibrium at 0 $^oC$ and 1 atm, while the liquid and vapor phases of water are at equilibrium at 100 $^oC$ and 1 atm. What does this mean practically? Take a cube of ice floating in liquid water at 1 atm and 0 $^oC$. If we add a small amount heat to the system, some ice will melt. If we take away a small amount of heat, some of the liquid water will freeze. Thus, by applying small changes in an external variable (heat transfer) we can change the direction of the process. The situation is different if the phases are not at equilibrium. E.g. if we place an ice cube in a glass of water at 25 $^oC$ and 1 atm, the transition will proceed only in one direction - the ice will melt. We will learn more about reversible processes when we cover the Second Law and entropy.

An interesting question is how we can keep a system at constant temperature while it is undergoing various state changes, such as compression, phase transitions or chemical reactions. It turns out that the practical way to do this is to put the system in contact with surroundings having a very large heat capacity and the desired temperature $T_{ex} =  T$. Thus T = const practically means

$$T =  T_{ex} =  const$$

See the Constant temperature section for details.

Manipulating transition enthalpies

Consider the reverse transition

$$Phase b  \rightarrow  Phase a        (p,T=const)$$

The enthalpy change

$$\Delta_{trs} H(b\rightarrow a) =  H_m(a) -  H_m(b)  =  -(H_m(b) -  H_m(a)) =  -\Delta_{trs} H(a\rightarrow b)$$

Now let's think about combining two phase transitions:

$$Phase a  \rightarrow  Phase b        (p,T=const)$$

$$Phase b  \rightarrow  Phase c        (p,T=const)$$

$$\Delta_{trs} H(a\rightarrow b) =  H_m(b) -  H_m(a)    ;     \Delta_{trs} H(b\rightarrow c) =  H_m(c) -  H_m(b)$$

The overall transition is

$$Phase a  \rightarrow  Phase c        (p,T=const)$$

To obtain the enthalpy change we can use the definition:

$$\Delta_{trs} H(a\rightarrow c) =  H_m(c) -  H_m(a)$$

or we can combine the enthalpies of the two stages

$$\Delta_{trs} H(a\rightarrow c) = \Delta_{trs} H(a\rightarrow b) +  \Delta_{trs} H(b\rightarrow c)$$

$$= H_m(b) -  H_m(a) +  H_m(c) -  H_m(b) =  H_m(c) -  H_m(a)$$

Of course we get the same result, as H is state function, and it's change depends only on the initial and final state (here a and c) and not on the path (i.e. whether we go from a to c directly or through an intermediate b).

Our bonus is the combination equation:

$$\Delta_{trs} H(a\rightarrow c) =  \Delta_{trs} H(a\rightarrow b) +  \Delta_{trs} H(b\rightarrow c)$$

Example.
At 1 atm pressure and 0 $^oC$ the enthalpy of fusion of water is 6.0 kJ/mol and the enthalpy of vaporization is 45.1 kJ/mol. Calculate
a) The enthalpy of freezing of water
b) The enthalpy of sublimation of ice

Problem deals with enthalpies of phase transitions.

Solution a)
Freezing is the opposite of melting, or fusion, so

$$\Delta_{freeze} H =  -6.0 kJ/mol    (at 0  ^oC and  1 atm)$$

enthalpy change is negative = heat leaves system.

Solution b)
We can break up the sublimation process (solid to gas) into two steps: fusion (solid to liquid) and vaporization (liquid to gas). We have data for both these steps, so we calculate

$$\Delta_{sub} H =  \Delta_{fus} H +  \Delta_{vap} H =  6.0 kJ/mol +  45.1 kJ/mol =  51.1 kJ/mol    (at 0  ^oC and  1 atm)$$

Note. The equations in this section are completely analogous to Hess's Law for chemical reactions described below.

Question to ponder. Why are the enthalpies of melting and vaporization positive for all substances?

Thermochemistry: chemical reactions

A chemical reaction is a process in which the chemical composition of a substance changes. The system has the same elemental composition before and after the reaction, but the chemical structure changes. We will consider chemical reactions at constant temperature and pressure, in order to simplify the analysis. We can use several schemes to represent chemical reactions

• The general scheme
  
  
  $$(Reactants)  \rightarrow   (Products)      (p,T=const)$$
  
  

  The chemical species present before the reaction (Reactants) are converted to different species after the reaction (Products)

• The semi-general scheme
  
  
  $$a A +  b B  \rightarrow   c C +  d D$$
  
  

  In this case there are two reactant species (A, B) and two product species (C,D). The numbers a, b, c, d are called stoichiometric coefficients.

• The abstract scheme
  
  
  $$0 =  \sum_J \nu_J J$$
  
  
  In this scheme the arrow is replaced by an "=" sign, so the reaction scheme becomes an abstract equation; reactants are moved from left-hand-side to the right-hand-side with change of sign of stoichiometric coefficients. The sum is over all species present, $\nu_J$ is the stoichiometric coefficient of species J (positive for products, negative for reactants).

Example.
Express this chemical reaction in the three schemes

$$H_2(g) +  \frac{1}{2}O_2(g) \rightarrow  H_2O(l)$$

(the g and l refer to the phase of the species - gas and liquid).

In the general scheme we can say :
one mole of reactants = $H_2(g) ~+~ \frac{1}{2}O_2(g)$
one mole of products = $H_2O(l)$

In the semi-general scheme we would write:
$A =  H_2(g)     B =  O_2(g)      C =  H_2O(l)$
$a =  1     b =  \frac{1}{2}     c =  1      d =  0$

In the abstract scheme:

$$0 =  H_2O(l) -  H_2(g) -  \frac{1}{2}O_2(g)$$

where the reactants and products may be identified by the signs of the stoichiometric coefficients.

Reaction enthalpy

The reaction enthalpy $\Delta_r H$ is the enthalpy change per mole of the reaction process as written. At the start we will think of the simplest possible situation, when each reactant and product species exists as a separate, pure substance. We can then break up the reaction enthalpy into independent contributions from the individual species.

In the general scheme

$$\Delta H =  H_f -  H_i =  H(Products) -  H(reactants)$$

$$\Delta_r H =  \frac{\Delta H}{n}$$

where n is the number of moles of reaction that took place.

In the semi-general scheme

$$\frac{H(Products)}{n} =  cH_m(C) +  dH_m(D)$$

$$\frac{H(Reactants)}{n} =  aH_m(A) +  bH_m(B)$$

$$\Delta_r H =  c H_m(C) +  d H_m(D) -  a H_m(A) -  b H_m(B)$$

where $H_m(A),...$ are molar enthalpies of the species.

In the abstract scheme the equation is especially simple

$$\Delta_r H =  \sum_J  \nu_J H_m(J)$$

where $H_m(J)$ is the molar enthalpy of species J.

The return of the standard state

Most chemical data refers to standard states of substances, i.e. the pure substance at 1 bar pressure. In this part of the class we will consider exclusively standard reaction enthalpies, which are evaluated for the case when all reactant and product species are in their respective standard states. The only difference compared to the previous section is that we have added the condition

$$p =  p^o =  1 bar$$

We denote this by adding a subscript "o" to the quantities. The corresponding equations are:
in the general scheme

$$\Delta_r H^o =  \frac{H^o(Products) -  H^o(reactants)}{n}$$

In the semi-general scheme

$$\Delta_r H^o =  c H_m^o(C) +  d H_m^o(D) -  a H_m^o(A) -  b H_m^o(B)$$

In the abstract scheme

$$\Delta_r H^o =  \sum_J  \nu_J H_m^o(J)$$

Reference states.
There is a special kind of standard state used only for elements. The reference state of an element is its most stable state at 1 bar and the given temperature. The term "most stable" refers to the form having the lowest molar enthalpy. For example, graphite at 1 bar is the reference state for solid carbon.

VERY IMPORTANT NOTE. Most actual chemical reactions occur in mixtures, in which reactant and product species do not exist as separate, pure substances. In mixtures it is not possible to define the reaction enthalpy solely in terms of independent contributions from the standard states. We will work on this in the Chemical equilibrium section. Our treatment of chemical reactions here describes a special kind of chemical reaction, in which separated pure reactants are converted into separated pure products. This captures most important effects contributing to enthalpy change and is quite useful, though not complete.

Standard enthalpies of formation

The standard enthalpy of formation $\Delta_f H^o$ of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states.

$$\left( \begin{array}{c} Elements in reference states \end{... ...n{array}{c} compound in standard state \end{array} \right)$$

The First Law only allows us to calculate changes in U and H in processes, not absolute values for individual states. Because of this, only differences of H (or U) have physical meaning and we can arbitrarily choose zero of enthalpy. In chemical thermodynamics we choose the enthalpies of elements in their reference states to be zero.

$$\Delta_f H^o =  0         (all  elements in reference states)$$

The reactants and products in any reaction have the same elemental composition. Changing the enthalpies of elements from zero to different values (even a different value for each element) would shift the enthalpy of the reactant and product state by the same amount, and would not change enthalpy differences.

From now on we will use the standard enthalpies of formation of compounds as their standard molar enthalpies.

$$\Delta_f H^o(A) =  H^o_m(A)$$

Our equations for the reaction enthalpy will thus from now on involve standard enthalpies of formation of compounds: in the general scheme

$$\Delta_r H^o =  \Delta_f H^o(Products) -  \Delta_f H^o(reactants)$$

in the semi-general scheme

$$\Delta_r H^o =  c \Delta_f H^o(C) +  d \Delta_f H^o(D)  -  a\Delta_f H^o(A) -  b\Delta_f H^o(B)$$

in the abstract scheme

$$\Delta_r H^o =  \sum_J  \nu_J \Delta_f H^o(J)$$

Example.
When 250 g of liquid water is produced from gaseous oxygen and hydrogen at 298 K and 1 bar, 2606 kJ of heat was released. Calculate the standard enthalpy of formation of liquid water.

Solution.
The reaction for the formation of liquid water is

$$H_2(g) +  \frac{1}{2}O_2(g) \rightarrow  H_2O(l)$$

The amount of reaction to produce 250 g of liquid water is

$$n =  \frac{m}{M} =  \frac{250 g}{18.015 g/mol} =  13.877 mol$$

For a closed system with pV work only at constant pressure the enthalpy change is equal to the heat transfer and the reaction enthalpy is

$$\Delta_r H^o =  \frac{\Delta H^o}{n} =  \frac{q}{n} =  \frac{ -2606 kJ}{13.877 mol} =  -187.8  kJ/mol$$

where we use the superscript "o" to denote that the process involves conversion of products in their standard states to the reactant in it's standard state at 1 bar, and the "-" sign appears because heat is released, i.e. process is exothermic. Since our reaction is the formation reaction of the compound from elements in reference states, the standard enthalpy of formation is

$$\Delta_f H^o(H_2O,l) =  \Delta_r H^o =  -187.8  kJ/mol$$

Note.
Suppose we used this reaction as the basis of our calculation:

$$2H_2(g) +  O_2(g) \rightarrow  2H_2O(l)$$

Since each mole of this reaction produces 2 moles of liquid water (M = 36.030 g/mol), the amount of reaction would now be

$$n =  \frac{m}{M} =  \frac{250 g}{36.030 g/mol} =  6.939 mol$$

and the reaction enthalpy refers to the reaction as written:

$$\Delta_r H^o =  \frac{\Delta H^o}{n} =  \frac{q}{n} =  \frac{ -2606 kJ}{6.939 mol} =  -375.6  kJ/mol$$

The formation reaction creates 1 mol of the compound from elements, so we would have

$$\Delta_f H^o(H_2O,l) =  \frac{1}{2}\Delta_r H^o =  -187.8  kJ/mol$$

Heat effects of chemical reactions

For a closed system at with pV work only at constant pressure, the enthalpy change for any process is equal to the heat transfer.

$$\Delta H =  q$$

A process with negative enthalpy change, $\Delta H <  0$, is called exothermic. An exothermic process leads to heat flow from system to surroundings (heat released). The final state (products) has lower enthalpy than initial state (reactants) - we say the final state is more stable.

A process with positive enthalpy change, $\Delta H >  0$, is called endothermic. An endothermic process leads to heat flow from surroundings to system (heat absorbed). The final state (products) has higher enthalpy than initial state (reactants) - we say the final state is less stable.

To understand the concept of stability, think of a rock on a hillside. We start with the rock at the bottom of the hill. To move it up the hill requires work input $(w >  0)$, as we increase the energy of the system by moving the rock up hill, we make the system less stable.

For closed system with pV work only at V = const the work done is zero and

$$\Delta U =  q$$

An example of a system in which the condition V = const applies is the bomb calorimeter (see textbook).

Heat obtained from reactions has great practical importance. Burning fuel such as wood, natural gas, oil or coal is the basis of energy production. The endo- or exothermic nature of a reaction is crucial for designing industrial processes. Biochemical reactions releasing heat generate the energy needed to sustain living organisms, and their energy balance has to be considered in designing diets.

Example.
Is heat released or absorbed when graphite is converted to carbon at 298 K and 1 bar?

Solution.
We can use the data in Appendix 1 of text:

$\Delta_f H^o(C,diamond) =  +1.895 kJ/mol$

We can also calculate the enthalpy of formation using the reaction of formation of diamond from solid carbon in the reference state (i.e. graphite):

$$C(graphite) \rightarrow  C(diamond)    (298 K, 1 bar)$$

and data for graphite and diamond

$$\Delta_f H^o(C,diamond) =  \Delta_f H^o(C,diamond) -  \Delta_f H^o(C,graphite) =  (1.895 -0)kJ/mol =  1.895 kJ/mol$$

Answer: enthalpy change is positive, heat is absorbed by system.

Note: this is what is meant by saying that graphite is the most stable form of solid carbon at 298 K and 1 bar: it has the lowest enthalpy of formation of all known forms of solid carbon.

Work effects of chemical reactions

For a closed system with pV work only the work will be

$$w =  0      (V=const)$$

$$w =  -p\Delta V      (p=const)$$

The constant pressure work term can be easily calculated in two cases.

1. Reaction involves solids and liquids only

.
The volume changes are negligible under standard pressure

$$\Delta V \approx  0     and     w \approx 0     (p=const, solids/liquids)$$

Note. This might not work for very high pressures, like inside planets or stars.

2. Reaction involves gaseous components.

We ignore the volumes of solids and liquids compared to gases and assume that gases behave ideally

$$p\Delta V = p(V(products) -  V(reactants)) \approx  V(gas products) -  V(gas reactants) =$$

$$= n\Delta\nu_{gas} RT     (p=const; gases)$$

Where $\Delta\nu_{gas}$ is the difference between the sums of the stoichiometric coefficients of the gas-phase species between the products and reactants:

$$\Delta\nu_{gas} = (sum of gas \nu_J:products) -  (sum of gas \nu_J:reactants)$$

or, in terms of the abstract scheme of chemical reactions

$$\Delta\nu_{gas} =  \sum_{J, gases} \nu_J$$

n is the amount of moles of reaction and we have used the ideal gas equation pV = nRT, with T - the temperature at which the reaction takes place.

Example.
Calculate the work done in the formation of one mole of liquid water at 298 K and 1 bar

$$H_2(g) +  \frac{1}{2}O_2(g) \rightarrow  H_2O(l)$$

Solution.
We will use our formula for the case of presence of gaseous species:

$$w =  -p\Delta V =  -n\Delta\nu_{gas} RT = -(1 mol)(-\frac{3}{2}) (8.314 \frac{J}{mol K})(298 K) =  3.716 kJ$$

The product state occupies a smaller volume than the reactants, system is compressed by surroundings.
See Problem Orgy #2 for more.

Reaction energies

If either $\Delta H$ or $\Delta U$ are known, the other quantity can be calculated:

$$\Delta H =  \Delta U +  \Delta (pV)$$

$$\Delta (pV) =  pV(products) -  pV(reactants)$$

At constant pressure:

$$\Delta(pV) =  p\Delta V$$

$$\Delta H =  \Delta U +  p\Delta V$$

This is the same formula we obtained before We can identify the same two cases when the results are easy to obtain.

1. Reaction involves solids and liquids only

. The volume changes are negligible under standard pressure

$$\Delta V \approx  0     and     \Delta H \approx  \Delta U     (p=const, solids/liquids)$$

2. Reaction involves gaseous components.

We ignore the volumes of solids and liquids compared to gases and assume that gases behave ideally

$$p\Delta V =  p(V(products) -  V(reactants)) \approx   =  n\Delta\nu_{gas} RT    (p=const; gases)$$

$$\Delta H =  \Delta U +  n\Delta\nu_{gas} RT    (p=const; gases)$$

where all quantities are defined in the previous section. If the reaction occurs under standard conditions we get the following formula linking the reaction energy and reaction enthalpy

$$\Delta_r H^o =  \frac{\Delta H^o}{n} =  \Delta_r U^o +  \Delta\nu_{gas} RT (p=const; gases)$$

Where $\Delta_r U^o$ is the reaction energy.

Example.
Calculate the work and reaction energy for conversion of one mole of graphite to diamond at 298 K and 1 bar.

$$C(graph) \rightarrow  C(diam)    (298 K, 1 bar)$$

Solution
The quick solution is that this process involves solids only, so we can assume that the volume change and work is zero, and the enthalpy change is the same as energy change:

$$w =  -p \Delta V =  0$$

$$\Delta H =  \Delta U     and     \Delta_r U =  \Delta_r H$$

See Problem Orgy #2 for more.

Combining reaction enthalpies: Hess's Law

This is easiest to understand using the general scheme. Start with this reaction:

$$(Reactants_1)  \rightarrow   (Products_1)      (p,T=const)$$

$$\Delta H_1 =  H_f -  H_i =  H(Products_1) -  H(Reactants_1)$$

$$\Delta_r H_1 =  \frac{\Delta H_1}{n}$$

where n is the amount of reaction that took place.

Next consider the reverse reaction

$$(Products_1)  \rightarrow   (Reactants_1)      (p,T=const)$$

$$\Delta H_{rev} =  H(Reactants_1) -  H(Products_1) =  -\Delta H_1$$

$$\Delta_r H_{rev} =  \frac{\Delta H_{rev}}{n} =  -\Delta_r H_1$$

i.e the reaction enthalpy of the reverse reaction is obtained by reversing the sign of the reaction enthalpy of the forward reaction.

Now look at this reaction, where we have scaled up the amount of reactants and products by a factor x

$$x\cdot (Reactants_1)  \rightarrow   x\cdot (Products_1)      (p,T=const)$$

$$\Delta H_x =  x\cdot H(Products_1) -  x\cdot H(Reactants_1) =  x\cdot \Delta H_1$$

$$\Delta_r H_x =  \frac{\Delta H_x}{n} =  x\cdot\Delta_r H_1$$

The factor x may be positive or negative. A negative factor x corresponds to scaling of the reverse reaction by $\vert x\vert$.

Finally let's add a second reaction

$$(Reactants_2)  \rightarrow   (Products_2)      (p,T=const)$$

$$\Delta H_2 =  H(Products_2) -  H(Reactants_2)$$

$$\Delta_r H_2 =  \frac{\Delta H_2}{n}$$

and combine reactions 1 and 2 in the most general way:

$$x\cdot (Reactants_1) +  y\cdot (Reactants_2) \rightarrow  x\cdot (Products_1) +  y\cdot (Products_2)      (p,T=const)$$

$$\Delta H =  x\cdot \Delta H_1 +  y\cdot \Delta H_2$$

$$\Delta_r H =  \frac{\Delta H}{n} =  x\cdot \Delta_r H_1  +  y\cdot \Delta_r H_2$$

These observations may be summarized in the form of Hess's Law: The overall reaction enthalpy is the combination of the reaction enthalpies of the component reactions.

Note. We will use Hess's Law to combine standard reaction enthalpies, since these kind of data are most widely available. However, the law works also under non-standard conditions.

Note. All these properties are a consequence of the fact that H is an extensive function of state, i.e. $\Delta H$ is proportional to the amount of reaction and is independent of the path, but depends only on the initial and final state. This allows us to break a reaction into intermediate steps in order to calculate reaction enthalpies.

Note. Since these properties do not depend on the formula for H but only on the fact that it is an extensive state function, the same rules will work for any extensive function of state, such as energy U, volume V, mass m, entropy S or Gibbs free energy G!

For examples see Problem Orgy #2 .

Variation of reaction enthalpies with T: Kirchhoff's Law

Typical thermochemical data give standard enthalpies of formation of substances at 298 K. This allows easy calculation of reaction enthalpies at 298 K. The question arises: do we have to make separate measurements at different temperatures, or is there some simple way of calculating the temperature dependence of reaction enthalpies? The answer is yes, and it involves heat capacities.

Let us use the general scheme to describe a chemical reaction occurring at a fixed pressure p and temperature T; in the general scheme

$$(Reactants)  \rightarrow   (Products)      (p,T=const)$$

$$\Delta_r H^o_T =  \frac{H^o(Products(T)) -  H^o(reactants(T))}{n}$$

and the same reaction at p and T'

$$(Reactants)  \rightarrow   (Products)      (p,T'=const)$$

$$\Delta_r H^o_{T'} =  \frac{H^o(Products(T')) -  H^o(reactants(T'))}{n}$$

The derivation. leads to the following equation:

$$\Delta_r H^{o}_{T'} =  \Delta_r H^{o}_T +  \Delta_r C_P\cdot\Delta T$$

where

$$\Delta_r C_P =  C_P(Products) -  C_P(Reactants)$$

represents the heat capacity difference per mole of products and reactants in the reaction as written.

In the semi-general scheme

$$a A +  b B  \rightarrow   c C +  d D$$

$$\Delta_r C_P =  c{C}_{P,m}(C) +  d{C}_{P,m}(D) -  a{C}_{P,m}(A) -  b{C}_{P,m}(B)$$

and in the abstract scheme

$$0 =  \sum_J  \nu_J J$$

$$\Delta_r C_P =  \sum_J  \nu_J C_{P,m}(J)$$

Chemical reactions: state of system

How many variables do we need to describe a system in which a chemical reaction is occurring? Start with two : pressure p and temperature T. To that we have to add the amount of each species present, e.g. for the semi-general reaction scheme

$$a A +  b B  \rightarrow   c C +  d D$$

we would have 6 variables

$$(n_A,n_B, n_C,n_D,p,T)$$

The system composition cannot change in an arbitrary way, but in a manner consistent with the stoichiometry. We will introduce a new variable to help describe the composition changes - the reaction progress $\xi$, which has units of mol and varies between 0 (system in reactant state) and 1 mol (system in product state). The coupling between the changes in amounts of A, B, C and D due to the stoichiometry may then be described as:

$$\Delta n_A =  -a \Delta \xi$$

$$\Delta n_B =  -b \Delta \xi$$

$$\Delta n_C =  c \Delta \xi$$

$$\Delta n_D =  d \Delta \xi$$

or, more generally:

$$\Delta n_J =  \nu_J \delta \xi$$

where $\nu_J$ is the stoichiometric coefficient of species J.

Note.
As we will see later, chemical reactions are quintessentially irreversible processes. As long as a reaction is going on, the system is not in equilibrium, and when the system reaches equilibrium, all reactions have stopped. This makes chemical reactions very different from heating/cooling, expansion/compression processes or phase transitions, for which we can find reversible paths.

Overview of Changes of State

Thus far, we have applied the First Law to three kinds of processes in closed systems:

pVT changes at constant composition.

Simple processes, involving change of macroscopic variables p,V,T etc., while internal structure of matter remains unchanged. Example:

• heating a sample of liquid water at p=const

  
  

  $$H_2O (l),  T_1, p  \rightarrow   H_2O (l),  T_2, p$$
  
  

System does not change physical state or chemical structure.

Also : compression-expansion work, pVT changes for ideal gas.

Phase changes.

Processes at p,T = const involving change in physical structure of matter, while chemical structure remains unchanged. Example

• evaporation of liquid water

  
  

  $$H_2O (l),  T, p  \rightarrow   H_2O (g),  T, p$$
  
  

Chemical reactions

Processes at p,T = const involving change in both physical and chemical structure of matter.

• decomposition of water at T,p =const

  
  

  $$H_2O (l)  \rightarrow  \frac{1}{2} O_2(g) +  H_2(g)$$
  
  

We will not introduce any new kinds of processes as we go on with the course. Rather, we will take a look at another aspect of these same phenomena, through the lens of the Second Law, entropy and free energy.

Go back to the main Lecture notes section.

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