5.4 Enthalpy
When the heat is absorbed or gained under the condition of constant pressure, then the change of the heat is called Enthalpy. The enthalpy (H) is under the constant pressure condition will equal to the sum of the internal energy (U) and the product of pressure and volume (PV).
H = U + PV
The change in enthalpy ∆ H will equal the change in the internal energy and the volume.
∆ H = ∆ U + P ∆ V [Pressure P is constant].
The internal energy (U) of a system is recognized as the random, unorganized motion of molecules. The total internal energy in a system includes potential and kinetic energy. It is the sum of all the microscopic energies such as: translational kinetic energy, vibrational and rotational kinetic energy etc.
.During a system change, its internal energy will change and the internal energy can be transferred from the system to the surrounding.
Energy is transferred into the system when the system is gaining energy (heat) q from the surrounding or when the surrounding is doing work (w) on the system.
Energy is transferred out of the system (given off) when the system is losing energy (heat) q to the surrounding or when the system is doing work (w) on the surrounding.
P ∆ V represents the work (w) with the ∆ V is positive. Then the P ∆ V term will equal:
P ∆ V = – w
By substitution, the enthalpy change ∆ H then equals:
∆ H = ∆ U + P ∆ V = ∆ U – w
The molar enthalpy ∆ Hm is defined as the enthalpy change per a mole of substance.
.Enthalpy is a state function which does not depend on the path of the change. It is an extensive property that depends on mass of the system. If the system moves from one state to another state, the enthalpy change is independent of the path between the two states of the system.
The combustion reaction of methane will yield 802.3 kJ/mol
CH4 + O2 ….> CO2 + 2 H2O ∆ H = – 802.3 kJ/mol [for 1 mole]
If 2 moles of CH4 are used then:
2 CH4 + 2 O2 …..> 2 CO2 + 4 H2O ∆ H = – 2 x 802.3 kJ/mol [for 2 mole]
The thermochemical equation for the 2 moles of CH4 combustion:
2 CH4 + 2 O2 ⟶ 2 CO2 + 4 H2O ∆ H = – 1604.6 kJ/mol
Example:
When 1.50 g Mg(s) reacts with 50.0 mL of 0.850 M HCl(aq), 5.15 kJ of heat are produced. Determine the enthalpy change per mole of magnesium reacting for the reaction and write the complete thermochemical equation:
Mg(s) + 2 HCl(aq) ⟶ MgCl2(aq) + H2(g)
Number of moles of Mg = (1.50 g) / (24.3 g Mg /mole) = 0.0617 moles
∆ H = 1 mole Mg x [5.15 kJ / 0.0617 moles] = – 83.5 kJ / mole [Heat is produced]
The thermochemical equation:
Mg(s) + 2 HCl(aq) ⟶ MgCl2(aq) + H2(g) ∆ H = – 83.5 kJ / mole
The heat absorbed or gained will equal ∆ H = q if there is no expansion work on the system and the pressure P is constant.
With the help of formula ∆ H = q, the process of exothermic or endothermic processes will be determined.
In other words:
By the endothermic process: the system is gaining heat from the surrounding and q will be greater than zero and ∆ H will be greater than zero as well.
By the exothermic process: the system is giving off heat to the surrounding and q will be less than zero and ∆ H will be less than zero as well.
Effect of the temperature on the enthalpy:
When temperature increases, the molecular interactions increase the internal energy of the system as well as the enthalpy of the system.
c = q / ∆ T
At the constant pressure:
cp = [∆ H / ∆ T]p The subscript indicates the constant pressure condition.
Enthalpy of phase changes
A plot of the enthalpy of a system as a function of its temperature is called an enthalpy diagram. The slope of the line is given by Cp. The enthalpy diagram of a pure substance such as water shows that this plot is not uniform, but is interrupted by sharp breaks at which the value of Cp is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change; you already know that the temperature the water boiling in a kettle can never exceed 100 until all the liquid has evaporated, at which point the temperature of the steam will rise as more heat flows into the system.
A plot of the enthalpy of carbon tetrachloride as a function of its temperature provides a concise view of its thermal behavior. The slope of the line is given by the heat capacity Cp. All H-vs.-C plots show sharp breaks at which the value of Cp is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change; you already know that the temperature of the water boiling in a kettle can never exceed 100°C until all the liquid has evaporated, at which point the temperature (of the steam) will rise as more heat flows into the system.
The lowest-temperature discontinuity on the CCl4 diagram corresponds to a solid-solid phase transition associated with a rearrangement of molecules in the crystalline solid.
Figure 5.25 Enthalpy Diagram of CCl4
Standard Enthalpy:
A standard stateis defined by IUPAC as standard state by which the system has the pressure of 1 bar (0.987 atm) and in general 1 atm is adopted and 1 M (molar) concentrations of the reactants.
The standard ∆ H is given the superscript “o”.
The symbol of the standard enthalpy is ∆ Ho
There is no standard temperature but usually the temperature used and considered is 298.15 K.
∆ H with the superscript “o” indicates the enthalpy under no standard conditions.
The videos below illustrate the concept of the standard enthalpy and the standard enthalpy of combustion:
5.1 Standard enthalpy change of combustion (SL)
Practice Problem: Enthalpy of Combustion
The table below shows the standard enthalpy of the combustions:
Example of standard heat of combustion calculation:
Consider the combustion reaction below:
CH4 + O2 ⟶ CO2 + 2 H2O ∆ H = – 802.3 kJ/mol
If 1.50 L of CH4 is used for the combustion, calculate the standard enthalpy of combustion.
The density of methane (gas) is: 0.72 g/L
Molar mass of methane: 16.0 g/mole
∆ Ho = [1.50 CH4 L] x [0.72 CH4 g / L] x [1 mole CH4 / 16.0 g CH4] x [802.3 kJ / mole CH4] = – 54.2 kJ
How are enthalpy changes determined experimentally?
First, you must understand that the only thermal quantity that can be observed directly is the heat q that flows into or out of a reaction vessel, and that q is numerically equal to ΔH° only under the special condition of constant pressure. Moreover, q is equal to the standard enthalpy change only when the reactants and products are both at the same temperature, normally 25°C.
The measurement of q is generally known as calorimetry.
The most common types of calorimeters contain a known quantity of water which absorbs the heat released by the reaction. Because the specific heat capacity of water (4.184 J g–1 K–1) is known to high precision, a measurement of its temperature rise due to the reaction enables one to calculate the quantity of heat released.
The calorimeter constant
In all but the very simplest calorimeters, some of the heat released by the reaction is absorbed by the components of the calorimeter itself. It is therefore necessary to “calibrate” the calorimeter by measuring the temperature change that results from the introduction of a known quantity of heat. The resulting calorimeter constant, expressed in J K–1, can be regarded as the “heat capacity of the calorimeter”.
For reactions that can be initiated by combining two solutions, the temperature rise of the solution itself can provide an approximate value of the reaction enthalpy if we assume that the heat capacity of the solution is close to that of the pure water — which will be nearly true if the solutions are dilute.
For example, a very simple calorimetric determination of the standard enthalpy of the reaction H+(aq) + OH–(aq) → H2O(l) could be carried out by combining equal volumes of 0.1M solutions of HCl and of NaOH initially at 25°C. Since this reaction is exothermic, a quantity of heat q will be released into the solution. From the tmperature rise and the specific heat of water, we obtain the number of joules of heat released into each gram of the solution, and q can then be calculated from the mass of the solution. Since the entire process is carried out at constant pressure, we have ΔH° = q. If reactions are carried out in nonstandard conditions then
ΔH= q/mols of limiting reactant
Calculating Enthalpy Change of Reaction by Calorimetry
For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonly placed within a larger insulated container of water. During the reaction, heat passes between the inner and outer containers until their temperatures become identical. Again, the temperature change of the water is observed, but in this case we need to know the value of the calorimeter constant described above.