Chemistry

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Measurement of ΔU and ΔH Using Calorimetry

Calorimeter is used for measuring the amount of heat change in a chemical or physical change. In calorimetry, the temperature change in the process is measured which is inversely proportional to the heat change. By using the expression C = q/mΔT, we can calculate the amount of heat change in the process. Calorimetric measurements are made under two different conditions.

• At constant volume (q_V)
• At constant pressure (q_P)

(A) ΔU Measurements

For chemical reactions, heat evolved at constant volume, is measured in a bomb calorimeter.

The inner vessel (the bomb) and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws.

A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess oxygen. The bomb is immersed in water, in the inner volume of the calorimeter.

A stirrer is placed in the space between the wall of the calorimeter and the bomb, so that water can be stirred, uniformly. The reaction is started by striking the substance through electrical heating.

A known amount of combustible substance is burnt in oxygen in the bomb. Heat evolved during the reaction is absorbed by the calorimeter as well as the water in which the bomb is immersed. The change in temperature is measured using a Beckman thermometer. Since the bomb is sealed its volume does not change and hence the heat measurements is equal to the heat of combustion at a constant volume (ΔU_c°).

The amount of heat produced in the reaction (ΔU_c°) is equal to the sum of the heat abosrbed by the calorimeter and water.

Heat absorbed by the calorimeter
q₁ = k.∆T

where k is a calorimeter constant equal to m_cC_c (m_c is mass of the calorimeter and C_c is heat capacity of calorimeter)

Heat absorbed by the water
q₂ = m_wC_wΔT
where m_w is molar mass of water

C_w is molar heat capacity of water (75.29 J K^{-1 mol-1)}

Therefore ΔU_c = q₁ + q₂
= k.ΔT + m_w C_w ΔT
= (k + m_wC_w) ΔT

Calorimeter constant can be determined by burning a known mass of standard sample (benzoic acid) for which the heat of combustion is known (- 3227 kJmol^-1)

The enthalpy of combustion at constant pressure of the substance is calculated from the equation (7.17)

ΔH_C°_(Pressure) = ΔU°_(Vol) + Δn_gRT

Applications of Bomb Calorimeter:

1. Bomb calorimeter is used to determine the amount of heat released in combustion reaction.
2. It is used to determine the calorific value of food.
3. Bomb calorimeter is used in many industries such as metabolic study, food processing, explosive testing etc.

(b) ΔH Measurements

Heat change at constant pressure (at atmospheric pressure) can be measured using a coffee cup calorimeter. A schematic representation of a coffee cup calorimeter is given in Figure 7.7. Instead of bomb, a styrofoam cup is used in this calorimeter. It acts as good adiabatic wall and doesn’t allow transfer of heat produced during the reaction to its surrounding.

This entire heat energy is absorbed by the water inside the cup. This method can be used for the reactions where there is no appreciable change in volume. The change in the temperature of water is measured and used to calculate the amount of heat that has been absorbed or evolved in the reaction using the following expression.

q = m_wC_wΔT

where m_w is the molar mass of water and C_w is the molar heat capacity of water (75.29 J K^-1 mol^-1)

Problem 7. 4

Calculate the enthalpy of combustion of ethylene at 300 K at constant pressure, if its heat of combustion at constant volume (ΔU) is – 1406 kJ.

The complete ethylene combustion reaction can be written as,

C₂H₄ (g) + 3O₂ (g) → 2CO₂ (g) + 2H₂O(l)
ΔU = – 1406 kJ
Δn = n_P(g) – n_r(g)
Δn = 2 – 4 = -2
ΔH = ΔU + RTΔn_g
ΔH = – 1406 + (8.314 × 10^-3 × 300 × (-2)
ΔH = – 1410.9 kJ

Applications of the Heat of Combustion:

(1) Calculation of Heat of Formation:

Since the heat of combustion of organic compounds can be determined with considerable ease, they are employed to calculate the heat of formation of other compounds.

For example let us calculate the standard enthalpy of formation ΔH_f° of CH₄ from the values of enthalpy of combustion for H₂, C(graphite) and CH₄ which are – 285.8, – 393.5, and -890.4 kJ mol^-1 respectively.

Let us interpret the information about enthalpy of formation by writing out the equations. It is important to note that the standard enthalpy of formation of pure elemental gases and elements is assumed to be zero under standard conditions. Thermochemical equation for the formation of methane from its constituent elements is,

C_(graphite) + 2H₂(g) → CH₄(g)
ΔH_f° = X kJ mol^-1 ………….. (1)

Thermo chemical equations for the combustion of given substances are,
H₂(g) + \(\frac{1}{2}\)O₂ → H₂O(l)
ΔH° = – 258.8 kJ mol^-1 ……………… (2)

C_{(graphite) + O2 → CO2
ΔH° = – 393.5 kJ mol^-1 ……………… (3)}

CH₄(g) + 2O₂ → CO₂(g) + 2H₂O(l)
ΔH° = – 890.4 kJ mol^-1 ……………… (4)

Since methane is in the product side of the required equation (i), we have to reverse the equation (iv)

CO₂(g) + 2 H₂O (l) → CH₄(g) + 2 O₂
ΔH° = + 890.4 kJ mol^-1 ……………… (5)

In order to get equation (i) from the remaining,

(i) = [(ii) × 2] + (iii) + (v)
X = [(- 285.8) × 2] + [- 393.5] + [+ 890.4]
= – 74.7 kJ

Hence, the amount of energy required for the formation of 1 mole of methane is – 74.7 kJ

The heat of formation methane = – 74.7 kJ mol^-1

Calculation of Calorific Value of Food and Fuels:

The calorific value is defined as the amount of heat produced in calories (or joules) when one gram of the substance is completely burnt. The SI unit of calorific value is J kg^-1. However, it is usually expressed in cal g^-1.

Heat of Solution:

Heat changes are usually observed when a substance is dissolved in a solvent. The heat of solution is defined as the change in enthalpy when one mole of a substance is dissolved in a specified quantity of solvent at a given temperature.

Heat of Neutralisation:

The heat of neutralisation is defined as “The change in enthalpy when one gram equivalent of an acid is completely neutralised by one gram equivalent of a base or vice versa in dilute solution”.

HCl(aq) + NaOH(aq) → NaCl (aq) + H₂O(l); ΔH = – 57.32 kJ
H⁺(aq) + OH^–(aq) → H₂O(l); ΔH = – 57.32 kJ

The heat of neutralisation of a strong acid and strong base is around – 57.32 kJ, irrespective of nature of acid or base used which is evident from the below mentioned examples.

HCl (aq) + KOH(aq) → KCl (aq) + H₂O(l)
ΔH = – 57.32 kJ

HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l)
ΔH = – 57.32 kJ
H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2 H₂O(l)
ΔH = – 57.32 kJ × 2

The reason for this can be explained on the basis of Arrhenius theory of acids and bases which states that strong acids and strong bases completely ionise in aqueous solution to produce H⁺ and OH^– ions respectively. Therefore in all the above mentioned reactions the neutralisation can be expressed as follows.

H⁺(aq) + OH^–(aq) → H₂O(l); ΔH = – 57.32 kJ

Molar Heat of Fusion

The molar heat of fusion is defined as “the change in enthalpy when one mole of a solid substance is converted into the liquid state at its melting point”. For example, the heat of fusion of ice can be represented as

Molar Heat of Vapourisation

The molar heat of vaporisation is defined as “the change in enthalpy when one mole of liquid is converted into vapour state at its boiling point”. For example, heat of vapourisation of water can be represented as

Molar Heat of Sublimation

Sublimation is a process when a solid changes directly into its vapour state without changing into liquid state. Molar heat of sublimation is defined as “the change in enthalpy when one mole of a solid is directly converted into the vapour state at its sublimation temperature”. For example, the heat of sublimation of iodine is represented as

Another example of sublimation process is solid CO₂ to gas at atmospheric pressure at very low temperatures.

Heat of Transition

The heat of transition is defined as “The change in enthalpy when one mole of an element changes from one of its allotropic form to another. For example, the transition of diamond into graphite may be represented as

Similarly the allotropic transitions in sulphur and phosphorous can be represented as follows,

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Thermochemical Equations:

A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (ΔH). The following conventions are adopted in thermochemical equations:

1. The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.

2. The enthalpy change of the reaction ΔH_r has to be specified with appropriate sign and unit.

3. When the chemical reaction is reversed, the value of ΔH is reversed in sign with the same magnitude.

4. The physical states (gas, liquid, aqueous, solid in brackets) of all species are important and must be specified in a thermochemical reaction, since ΔH depends on the physical state of reactants and products.

5. If the thermochemical equation is multiplied throughout by a number, the enthalpy change is also multiplied by the same number.

6. The negative sign of ΔH_r indicates that the reaction is exothermic and the positive sign of ΔH_r indicates an endothermic reaction.

For example, consider the following reaction,

2H₂O(g) → 2 H₂O(g) ΔH_r° = – 967.4 KJ
2H₂O(g) → 2H₂(g) + O₂(g) ΔH_r° = + 967.4 kJ

Standard Enthalpy of Reaction (ΔH_r°) from standard enthalpy of formation (ΔH_f°)

The standard enthalpy of a reaction is the enthalpy change for a reaction when all the reactants and products are present in their standard states. Standard conditions are denoted by adding the superscript 0 to the symbol (ΔH°)

We can calculate the enthalpy of a reaction under standard conditions from the values of standard enthalpies of formation of various reactants and products. The standard enthalpy of reaction is equal to the difference between standard enthalpy of formation of products and the standard enthalpies of formation of reactants.

Problem 7.2

The standard enthalpies of formation of C₂H₅OH(l), CO₂(g) and H₂O(l) are – 277, – 393.5 and – 285.5 kJ mol^-1 respectively. Calculate the standard enthalpy change for the reaction

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3 H₂O(l)

The enthalpy of formation of O₂(g) in the standard state is Zero, by definition

Solution:

For example, the standard enthalpy change for the combustion of ethanol can be calculated from the standard enthalpies of formation of C₂H₅OH(l), CO₂(g) and H₂O(l). The standard enthalpies of formation of are – 277, – 393.5 and – 285.5 kJ mol^-1 respectively.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

ΔH_r° = [2 mol (-393.5)kJ mol^-1]
[+ 3 mol (0) kJ mol^-1
= [- 787 – 856.5] – [- 277]
ΔH_r° = – 1366.5 kJ

Heat of Combustion

The heat of combustion of a substance is defined as “The change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen”. It is denoted by ΔH_C. For example, the heat of combustion of methane is – 87.78 kJ mol^-1

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
ΔH_C = – 87.78 kJ mol^-1

For the combustion of carbon,
C (s) + O₂(g) → CO₂(g)
ΔH_C = – 394.55 kJ mol^-1

Combustion reactions are always exothermic. Hence the enthalpy change is always negative.

Molar Heat Capacities

When heat (q) is supplied to a system, the molecules in the system absorb the heat and hence their kinetic energy increases, which in turn raises the temperature of the system from T₁ to T₂. This increase (T₂ – T₁) in temperature is directly proportional to the amount of heat absorbed and inversely proportional to mass of
the substance. In other words,

q α mΔT
q = c m ΔT
c = q/m ΔT

The constant c is called heat capacity.

…………….. (7.8)

when m = 1 kg and (T₂ – T₁) = 1 K then the heat capacity is referred as specific heat capacity. The equation 7.18 becomes

c = q

Thus specific heat capacity of a system is defined as “The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a specified temperature”.

The heat capacity for 1 mole of substance, is called molar heat capacity (c_m). It is defined as “The amount of heat absorbed by one mole of the substance to raise its temperature by 1 kelvin”.

Units of Heat Capacity:

The SI unit of molar heat capacity is JK^-1mol^-1

The molar heat capacities can be expressed either at constant volume (C_v) or at constant pressure (C_p).
According to the first law of thermodynamics
U = q + w or U = q – PdV
q = U + PdV …………….. (7.19)

Differentiate (7.19) with respect to temperature at constant volume i.e dV = 0,

……………… (7.20)

Thus the heat capacity at constant volume (C_V) is defined as the rate of change of internal energy with respect to temperature at constant volume. Similarly the molar heat capacity at constant pressure (C_p) can be defined as the rate of change of enthalpy with respect to temperature at constant pressure.

C_P = (\(\frac{∂H}{∂T}\))_p …………. (7.21)

Relation between Cp and Cv for an ideal gas.

From the definition of enthalpy

H = U + PV …………… (7.8)
for 1 mole of an ideal gas
PV = nRT …………… (7.22)
By substituting (7.22) in (7.8)
H = U + nRT …………. (7.23)
Differentiating the above equation with respect to T,

At constant pressure processes, a system has to do work against the surroundings. Hence, the system would require more heat to effect a given temperature rise than at constant volume, so C_P p is always greater than C_v.

Calculation of ΔU and ΔH

For one mole of an ideal gas, we have

C_V = \(\frac{dU}{dT}\)
dU = C_VdT
For a finite change, we have
ΔU = C_VΔT
ΔU = C_V(T₂ – T₁)
and for n moles of an ideal gas we get
ΔU = n C_V(T₂ – T₁) …………… (7.25)
Similarly for n moles of an ideal gas we get
ΔH = n C_P(T₂ – T₁) ………….. (7.26)

Problem 7.3

Calculate the value of ΔU and ΔH on heating 128.0 g of oxygen from 0°C to 1000°C. C_V and C_P on an average are 21 and 29 J mol^-1K^-1. (The difference is 8Jmol^-1 K^-1 which is approximately equal to R)

Solution:

We know
ΔU = n C_V(T₂ – T₁)
ΔH = n C_P(T₂ – T₁)
Here n = \(\frac{128}{32}\) = 4 moles;
T₂ = 100°C = 373K; T₁ = 0° C = 273K
ΔU = n C_v(T₂ – T₁)
ΔU = 4 × 21 × (373 – 273)
ΔU = 8400 J
ΔU = 8.4 kJ
ΔH = n Cp (T₂ – T₁)
ΔH = 4 × 29 × (373 – 273)
ΔH = 11600 J
ΔH = 11.6 kJ

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Enthalpy (H)

The enthalpy (H), is a thermodynamic property of a system, is defined as the sum of the internal energy (U) of a system and the product of pressure and volume of the system. That is,

H = U + PV …………………. (7.8)

It reflects the capacity to do mechanical work and the capacity to release heat by the system. When a process occurs at constant pressure, the heat involved (either released or absorbed) is equal to the change in enthalpy.

Enthalpy is a state function which depends entirely on the state functions T, P and U. Enthalpy is usually expressed as the change in enthalpy (ΔH) for a process between initial and final states at constant pressure.

ΔH = ΔU + PΔV …………………. (7.9)

The change in enthalpy (ΔH) is equal to the heat supplied at the constant pressure to a system (as long as the system does no additional work).

ΔH = q_P

In an endothermic reaction heat is absorbed by the system from the surroundings that is q>0 (positive). Therefore, ΔH is also positive. In an exothermic reaction heat is evolved by the system to the surroundings that is, q<0 (negative). If q is negative, then ΔH will also be negative.

Relation Between Enthalpy ‘H’ and Internal Energy ‘U’

When the system at constant pressure undergoes changes from an initial state with H₁, U₁ and V₁ to a final state with H₂, U₂ and V₂ the change in enthalpy ΔH, can be calculated as follows:

H = U + PV

In the initial state

H₁ = U₁ + PV₁ ………….. (7.10)

In the final state

H₂ = U₂ + PV₂ …………… (7.11)

change in enthalpy is (7.11) – (7.10)

(H₂ – H₁) = (U₂ – U₁) + P(V₂ – V₁)
ΔH = ΔU + PΔV ……………… (7.12)
As per first law of thermodynamics,
ΔU = q + w
Equation 7.12 becomes
ΔH = q + w + PΔV
w= – PΔV
ΔH = q_p – PΔV + PΔV
ΔH = q_p ……………….. (7.13)

q_p – is the heat absorbed at constant pressure and is considered as heat content. Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperature and pressure with V_i and V_f as the total volumes of the reactant and product gases respectively, and n_i and n_f as the number of moles of gaseous reactants and products, then,

For Reactants (Initial State):

PV_i = n_iRT ……………….. (7.14)

For Products (Final State):

PV_f = n_fRT ………………. (7.15)

(7.15) – (7.14)

P (V_f – V_i) = (n_f – n_i) RT
PΔV= Δn_(g)RT ………………… (7.16)

Substituting in 7.16 in 7.12
ΔH = ΔU + Δn_(g)RT ……………… (7.17)

Enthalpy Changes for Different Types of Reactions and Phase Transitions:

The heat or enthalpy changes accompanying chemical reactions is expressed in different ways depending on the nature of the reaction. These are discussed below.

Standard Heat of Formation

The standard 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 elements, present in their standard states (298 K and 1 bar pressure)”. By convention the standard heat of formation of all elements is assigned a value of zero.

Fe(s) + S(s) → FeS(s)
ΔH_f° = – 100.42 kJ mol^-1
2C(s) + H₂(g) → C₂H₂(g)
ΔH_f° = + 222.33 kJ mol^-1
\(\frac{1}{2}\)Cl₂(g) + \(\frac{1}{2}\)H₂(g) → HCl (g)
ΔH_f° = – 92.4 kJ mol^-1

The standard heats of formation of some compounds are given in Table 7.4.

Standard Heat of Formation of Some Compounds

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First Law of Thermodynamics

The first law of thermodynamics, known as the law of conservation of energy, states that the total energy of an isolated system remains constant though it may change from one form to another.

When a system moves from state 1 to state 2, its internal energy changes from U₁ to U₂. Then change in internal energy

∆U = U₂ – U₁.

This internal energy change is brought about by the either absorption or evolution of heat and/or by work being done by/on the system.

Because the total energy of the system must remain constant, we can write the mathematical statement of the

First Law as:

ΔU = q + w ………………… (7.7)

Where q – the amount of heat supplied to the system; w – work done on the system

Other statements of first law of thermodynamics

1. Whenever an energy of a particular type disappears, an equivalent amount of another type must be produced.

2. The total energy of a system and surrounding remains constant (or conserved)

3. “Energy can neither be created nor destroyed, but may be converted from one form to another”.

4. “The change in the internal energy of a closed system is equal to the energy that passes through its boundary as heat or work”.

5. “Heat and work are two ways of changing a system’s internal energy”.

Mathematical Statement of the First Law

The mathematical statement of the first law of thermodynamics is

ΔU = q + w ……………… (7.7)

Case 1:

For a cyclic process involving isothermal expansion of an ideal gas,
ΔU = 0.
eqn (7.7) ⇒ ∴ q = – w

In other words, during a cyclic process, the amount of heat absorbed by the system is equal to work done by the system.

Case 2:

For an isochoric process (no change in volume) there is no work of expansion. i.e. ΔV = 0
ΔU = q + w
= q – PΔV
ΔV = 0
ΔU = q_V

In other words, during an isochoric process, the amount of heat supplied to the system is converted to its internal energy.

Case 3:

For an adiabatic process there is no change in heat. i.e. q = 0. Hence q = 0

eqn (7.7) ⇒ ΔU = w

In other words, in an adiabatic process, the decrease in internal energy is exactly equal to the work done by the system on its surroundings.

Case 4:

For an isobaric process. There is no change in the pressure. P remains constant. Hence

ΔU = q + w
ΔU = q – P ΔV

In other words, in an isobaric process a part of heat absorbed by the system is used for P-V expansion work and the remaining is added to the internal energy of the system.

Problem: 7.1

A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.
Solution:
Given data q = 400 J V₁ = 5L V₂ = 10L
Δu = q – w (heat is given to the system (+q); work is done by the system(-w)
Δu = q – PdV
= 400 J – 1 atm (10-5)L
= 400 J – 5 atm L
[∴1L atm = 101.33 J]
= 400 J – 5 × 101.33 J
= 400 J – 506.65 J
= – 106.65 J

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Zeroth Law of Thermodynamics

The law states that ‘If two systems are separately in thermal equilibrium with a third one, then they tend to be in thermal equilibrium with themselves’. According to this law, if systems B and C separately are in thermal equilibrium with another system A, then systems B and C will also be in thermal equilibrium with each other. This is also the principle by which thermometers are used.

The zeroth law of thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. This says in essence that the three bodies are all the same temperature.

Why is it called the zeroth law of thermodynamics? Answer: There were three law of thermodynamics originally established and named. Since the law is the fundamental one, the scientist Raplh H Fowler came with an alternative and numbered the new law as a lower number zero and the law is called the “Zeroth law.”

Similarly, another example of the zeroth law of thermodynamics is when you have two glasses of water. One glass will have hot water and the other will contain cold water. Now if we leave them in the table for a few hours they will attain thermal equilibrium with the temperature of the room.

Zeroth law of thermodynamics states that when two systems are in thermal equilibrium through a third system separately then they are in thermal equilibrium with each other also. Heat flow happens between systems A and C, and between B and C, due to which all 3 systems attain thermal equilibrium.

The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. The thermometer is therefore also in thermal equilibrium with the second cup of water.

Zeroth law of Thermodynamics:

If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. First law of thermodynamics – Energy can neither be created nor destroyed. In any process, the total energy of the universe remains the same.