Enthalpy

It is the thermodynamic function that measures the total heat of the system at constant pressure (q_p). In this case, volume is variable, so work is done. Therefore, enthalpy measures the work done also. Hence enthalpy is qualitatively equal to the sum of internal energy change and work done. It is denoted by H.

H = E + PV---(i)

Here, E, P and V all are state functions. Therefore the absolute value of it cannot be measured but a change in value between two sides is measurable.

Let H₁ = Enthalpy of the thermodynamic system at an initial state
H₂ = Enthalpy of the thermodynamic system at a final state
Then from (i),

\begin{align*} H_{1} &= E_{1} + P_{1}V_{1}---(ii)\\ H_{2} &= E_{2} + P_{2}V_{2}---(iii)\\ So,\ Change\ &in\ enthalpy,\\ H_{2} - H_{1} &= (E_{2} - E_{1}) + (P_{2}V_{2} - P_{1}V_{1})\\ Since,\ P_{2} - P_{1} &= P\\ so,\ H_{2} - H_{1} &= E_{2} - E_{1} + P(V_{2} - V_{1})\\ Thus,\ \Delta H &= \Delta E + P\Delta V---(iv) \end{align*}

where ΔE is a change in internal energy and ΔV is a change in volume. Equation (iv) is the general expression for enthalpy change.
Now, we have from the first law of thermodynamics,

\begin{align*} E &= q - W--- (v)\\ &Substituting\ (v)\ in\ (iv),\\ \Delta H &= q - W + P \Delta V\\ Since,\ w = &P\Delta V,\ \Delta H = q\ so,\ \Delta H = q_{p}\\ But,\ q_{v} = &\Delta E,\ Then\ (iv)\ becomes\\ q_{p} &= q_{v} + P\Delta V \end{align*}

*Chemical reactions are usually carried out at constant pressure. Therefore, it is better to express energy change in terms of enthalpy rather than internal energy.

Energy profile diagram for exothermic and endothermic reaction

In the case of an exothermic reaction, the enthalpy of the product (H_p) is less than the enthalpy of the reactant (H_R). So change in enthalpy is negative.

\begin{align*} H_{p} &< H_{R}\\ \Delta H &= H_{P} - H_{R}\\ i.e.\ \Delta H &= -ve \end{align*}

In the case of an endothermic reaction, the enthalpy of the product (HP) is greater than the enthalpy of the reactant (H_R). So, the change in enthalpy is positive.

\begin{align*} H_{P} &> H_{R}\\ \Delta H &= H_{P} - H_{R}\\ i.e.\ \Delta H &= +ve \end{align*}

Enthalpy change or heat of reaction

It is defined as the amount of heat evolved or absorbed when the number of moles of reactant is converted into the product as represented in the balanced chemical reaction.

\begin{align*} CH_{4} + 2O_{2} &\rightarrow CO_{2} + 2H_{2}O,\ \Delta H = -890.35 KJ\\ C + O_{2} &\rightarrow CO_{2},\ \Delta H = -393.5 KJ\\ C + H_{2}O &\rightarrow CO + H_{2},\ \Delta H = 131.6 KJ \end{align*}

As we know that enthalpy is a function of temperature and pressure. As the pressure and temperature change, the heat of the reaction also varies. In order to compare the heat of reaction at various temperatures and pressure, it is necessary to fix a certain temperature and pressure as a standard temperature and pressure. So, according to IUPAC recommendation, the temperature of 25°C (298K) and pressure of 1 atm (760 mm of Hg) are taken as standard reference temperature and pressure. Hence standard heat of reaction or standard enthalpy change in a chemical reaction is defined as the amount of heat evolved or absorbed when the number of moles of reactant is converted into the product as represented in the balanced chemical reaction at 25°C temperature and 1 atm pressure. It is denoted by ΔH°.

Different forms of enthalpy change

A) Chemical change

1. Heat of formation

Heat of formation of a compound is defined as the change in enthalpy that takes place when one mole of a compound is formed from its constituent elements.
Eg: when 1 mole of hydrogen is reacted with 1 mole of chlorine, it gives 2 moles of HCl gas by the evolution of 44 KCal heat.

H_{2} + Cl_{2} \rightarrow 2HCl,\ \Delta H = -44KCal

This reaction is exothermic in nature.
The heat of formation of HCl = -44/2 = -22 KCal

Similarly, 1 mole of Hydrogen gas reacts with 1 mole of iodine vapour to give 2 moles of hydrogen iodide gas by the absorption of 12.4 KCal heat.

H_{2} + I_{2} \rightarrow 2Hl,\ \Delta H = 12.4KCal

This reaction is endothermic in nature.
The heat of formation of HI = 12.4/2 = 6.2 KCal.

The data of heat of formation is very useful to relate the thermodynamic stability of a compound. It is found that a substance having a lower value of heat of formation is relatively more stable than a substance having a high value of heat of formation. In the above equation, HCl is more stable than HI.

2. Heat of combustion or oxidation

The heat of combustion of a compound is the change in enthalpy when one mole of a compound is completely oxidized.

\begin{align*} CH_{4} + 2O_{2} &\rightarrow 2H_{2}O + CO_{2},\ \Delta H = -212 KCal\\ C + O_{2} &\rightarrow CO_{2},\ \Delta H = -93.4 KCal\end{align*}

During combustion, heat is always evolved out. So it is only an exothermic process.

The data of heat of formation is very useful for rating the quality of fuel. It is found that the fuel having a higher value of heat of combustion is better than the fuel having a low value of heat of combustion. It is also used to calculate the calorific value of fuel ( the amount of heat produced by burning 1 gram of fuel).

3. Heat of neutralization

It is the change in enthalpy that takes place when 1 gm equivalent of acid is neutralized by 1 gm equivalent of base or vice versa in dilute solution.

\begin{align*} HCl + NaOH &\rightarrow NaCl + H_{2}O, \Delta H = -13.7 KCal\\ HNO_{3} + NaOH &\rightarrow NaNO_{3} + H_{2}O, \Delta H = -13.69 KCal\\ HCl + NH_{4}OH &\rightarrow NH_{4}Cl + H_{2}O, \Delta H = -12.4 KCal\\ HCOOH + NH_{4}Cl &\rightarrow HCOONH_{4} + H_{2}O, \Delta H = -11.9 KCal \end{align*}

Physical change

1. Heat of vaporization

It is the enthalpy change when one mole of a substance changes from a liquid state into a vapour state.

H_{2}O(l) \rightarrow H_{2}O(g), \Delta H = 9.7 KCal

2. Heat of solution

It is the enthalpy change when one mole of a substance completely dissolve in a solvent which on further addition does not give any change in enthalpy.

NH_{4}NO_{3}(s) \rightarrow NH_{4}NO_{3}(aq), \Delta H = 6.2KCal

Bond energy

It is defined as the average amount of energy required to break all bonds in 1 mole of a particular substance. The formation of a bond is an exothermic process and breaking of a bond is an endothermic process.
Bond energy (ΔH) = Bond energy of reactant – Bond energy of the product

Factors affecting bond energy

• Size of atom: Smaller the size of existing atoms in a compound, strong would be the bond length which results the higher value of bond energy.
• Electronegativity: Higher the value of electronegativity of existing atoms of a compound, greater will be the value of bond energy.
• Bond length: Shorter the bond length of existing atoms in a compound, greater would be the value of bond energy.

Hess law of constant heat summation (Law of thermochemistry)

It states that “The enthalpy change in a chemical reaction always remains constant whether the reaction is carried out directly through a single step or indirectly through different multiple intermediate steps”.

Illustration of Hess Law

Here, the reactant(A) is converted in the product(Z) either by path I or by path II.
In Path I, A is converted to Z by the evolution of Q₁ KCal of heat.

A \rightarrow Z,\ \Delta H = -Q_{1}\ KCal

In Path II, A is converted into Z through intermediate steps B and C by the evolution of q₁, q₂ and q₃ KCal of heat respectively.

\begin{align*} A &\rightarrow B, H_{1} = -q_{1}\ KCal\\ B &\rightarrow C, H_{2} = -q_{2}\ Kcal\\ C &\rightarrow Z, H_{3} = -q_{3}\ KCal \end{align*}

According to the statement,

\begin{align*} \Delta H = \Delta &H_{1} + \Delta H_{2} + \Delta H_{3}\\ or,\ -Q_{1} &= -q_{1} -q_{2} -q_{3}\\ But,\ Q_{2} &= q_{1} + q_{2} + q_{3}\\ So,\ Q_{1} &= Q_{2} \end{align*}

Illustration

Let us take an example of the burning of CO₂. Carbon can be burnt to give CO₂ directly or it may be changed first to CO and then to CO₂ as shown below:

\begin{align*} Path\ i:\ C + O_{2} &\rightarrow CO_{2},\Delta H = -94 KCal\\ Path\ ii:\ C + 1/2O_{2} &\rightarrow CO, H_{1} = -26 KCal\\ CO + 1/2O_{2} &\rightarrow CO_{2},\Delta H_{2} = -68 KCal\\ Adding\ H_{1}\ and\ H_{2},\\ C + O_{2} \rightarrow CO_{2},\ &\Delta H_{1} +\Delta H_{2} = -94 KCal\\ So, H &= H_{1} + H_{2} \end{align*}

Hence, enthalpy change in a single step is the same as that of a multiple-step.

Points to be noted while studying the enthalpy of reaction

1. To express the enthalpy of a reaction, the chemical reaction must be balanced.

2. While balancing the chemical reaction, the coefficients of chemical species need not always be a whole number.

C(s) + 1/2O_{2}(g) \rightarrow CO(g),\ \Delta H = -26 KCal

3. In a chemical reaction, the physical state of all chemical constituents must be written because heat change will be different for different states of matter.

\begin{align*} H_{2}(g) + 1/2O_{2}(g) &\rightarrow H_{2}O(l),\Delta H = -68.4 KCal\\ H_{2}(g) + 1/2O_{2}(g) &\rightarrow H_{2}O(g),\Delta H = -57.8 KCal \end{align*}

4. For an exothermic reaction, the enthalpy change is negative and for an endothermic reaction, the enthalpy change is positive.

5. The value of enthalpy change should correspond to the balanced chemical equation. It is usually expressed in per mole.

H_{2}(g) + Cl_{2}(g) \rightarrow 2HCl(g), \Delta H = -44 KCal

Enthalpy of formation of 2 moles of HCl = -44 KCal
Enthalpy of formation of 1 mole of HCl = -22 KCal
So, enthalpy of formation of HCl = -22 KCal per mole

6. The same chemical equation can be expressed in different ways.

\begin{align*} \frac{1}{2}H_{2}(g) + \frac{1}{2}Cl_{2}(g)&\rightarrow HCl(g), \Delta H = -22 KCal\\ \frac{1}{2}H_{2}(g) + \frac{1}{2}Cl_{2}(g)&\rightarrow HCl(g) + 22 KCal\\ \frac{1}{2}H_{2}(g) + \frac{1}{2}Cl_{2}(g) &-22 KCal \rightarrow HCl(g) \end{align*}

7. Enthalpy of reaction will be different at different temperatures and pressure. To compare, the enthalpy of reactions at 25°C and 1 atm are taken as standard reference and the chemical substances are should be in pure and standard state. Enthalpy of one mole of a pure substance at 25°C temperature (298K) and 1 atm pressure (760 mm of Hg) is called standard enthalpy. It is denoted by H°.

The quantity of heat change involved when the number of moles of reactant is converted into the product as represented in a balanced chemical reaction at 25°C temperature (298K) and 1 atm pressure (760 mm of Hg) is called standard heat of reaction or standard enthalpy of reaction. It is denoted by H°.

Solved Examples

1. The standard heat of formation of SO₂(g) and SO₃(g) are -296 KJ and -396 KJ resp. Calculate ΔH for the reaction.
Solution:

\begin{align*} S(s) + O_{2}(g)&\rightarrow SO_{2}(g),\Delta H=-296 KJ---(i)\\ S(s) + \frac{3}{2} O_{2}(g)&\rightarrow SO_{3}(g),\Delta H=-396 KJ---(ii)\\ SO_{2}(g) + &\frac{1}{2}O_{2}(g)\rightarrow SO_{3}(g), \Delta H=?\\ Subtracting\ &(i)\ from\ (ii),\ we\ get,\\ \frac{1}{2}O_{2}(g) &\rightarrow SO_{3}(g) - SO_{2}(g),\Delta H = -99.4 KJ\\ So,\ SO_{2}(g) + &\frac{1}{2}O_{2}(g)\rightarrow SO_{3}(g),\Delta H = -99.4 KJ \end{align*}

2. Enthalpy of combustion of carbon to CO₂ is -393 KJ. Calculate the heat released upon formation of 33 gm of CO₂ from carbon and oxygen.
solution:

C + O_{2} \rightarrow \underset{\substack{1\ mole\\ 44\ gm}}{CO_{2}},\ \Delta H = -393 KJ

44 gm of CO₂ requires -393 KJ heat energy
33 gm of CO₂ requires (-393/44) x 33 = 294.75KJ heat energy

3. Heat of combustion of methane, carbon and hydogen are -210 KCal, -94 KCal and -68 KCal resp. Calculate the heat of formation of methane.
solution:

\begin{align*} CH_{4} + 2O_{2}&\rightarrow CO_{2} + 2H_{2}O,\Delta H = -210 KCal.....(i)\\ C + O_{2}&\rightarrow CO_{2},\Delta H = -94 KCal......(ii)\\ H_{2} + 1/2O_{2}&\rightarrow H_{2}O,\Delta H = -68 KCal......(iii)\\ C + 2H_{2}&\rightarrow CH_{4},\Delta H = ?\\ (iii) \times 2 &+ (ii) - (i)\\ C + 2H_{2}&\rightarrow CH_{4},\Delta H = -20 KCal \end{align*}

4. Calculate the heat of combustion of glucose from given data:
C + O₂ → CO₂, ΔH = -395 KJ
H₂ + 1/2O₂ → H₂O, ΔH = -269 KJ
6C + 6H₂ + 3O₂ → C₆H₁₂O₆, ΔH = -1169 KJ
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, ΔH = ?
solution:

\begin{align*} \Delta H &=\Delta H_{product} - \Delta H_{reactant}\\ &= [6 (-395) + 6 (-269)] - [(-1169) + 6(0)]\\ &= -2815\ KJ \end{align*}