Thermodynamics | Chemistry
Answer the following questions
26. State the first law of thermodynamics.
"Energy can neither be created nor destroyed, but may be converted from one form to another"
27. Define Hess's law of constant heat summation.
The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps.
28. Explain intensive properties with two examples
Intensive properties are the properties that is independent of the mass or size of the system.
Example : Temperature and boiling point.
29. Define the following terms:
a. isothermal process
b. adiabatic process
c. isobaric process
d. isochoric process
a. Isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final states. ie., dT = 0
b. The process in which no heat can flow into or out of the system is called adiabatic process ie., q = 0
c. Isobaric process is defined as one in which the Pressure of the system remains constant during its change from the initial to final state ie., dp = 0
d. Isochoric process is defined as one in which the volume of system remains constant during its change from initial to final state of the process, i.e., dv = 0
30. What is the usual definition of entropy? What is the unit of entropy?
Entropy is a measure of randomness or disorderliness of the system. Entropy (s) is equal to heat energy (q) divided by constant temperature.
S = q/T
UNITS OF ENTROPY : SI Unit: JK−1
CGS Unit: cal K−1
31. Predict the feasibility of a reaction when
i) both ΔH and ΔS positive
ii) both ΔH and ΔS negative
iii) ΔH decreases but ΔS increases
i) when both ΔH and ΔS are positive, then ΔG to be −ve only reaction is carried out at high temperature it is feasible.
In otherwords, TΔS > ΔH, then the reaction is feasible.
ii) When both ΔH and ΔS are negative, then ΔG to be −ve only if reaction is carried out at low temperature is, it is feasible. In other words,
TΔS < ΔH, then the reaction is feasible.
iii) When ΔH is −ve but ΔS is +ve, the ΔG becomes −ve is feasible at constant temperature.
32. Define is Gibb’s free energy.
"Gibbs free energy is defined as the part of the total energy of a system that can be converted (or) available for conversion into work".
G is mathematically defined as G =H − TS When H − enthalpy, T- temperature, S − entropy.
33. Define enthalpy of combustion.
Enthalpy 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 ΔHC.
34. Define molar heat capacity. Give its unit.
The molar heat capacity may be defined as " The amount of heat absorbed by one mole of the substance to raise in temperature by 1 kelvin"
Unit : JK−1 mol−1 (SI)
35. Define the calorific value of food. What is the unit of calorific value?
The Calorific value of food is defined as :
The amount of heat produced in calories (or joules) when one gram of a food is completely burnt. It is expressed in cal g −1. In SI unit it is expressed in J Kg −1
36. Define enthalpy of neutralization.
The enthalpy of neutralizations is defined as "The change in enthalpy of the system when one gram equivalent of an acid is neutralised by one gram equivalent of a base or vice versa in dilute solution".
37. What is lattice energy?
Lattice energy is defined as the amount of energy required to completely separate one mole of a solid ionic compound into gaseous constituent. It is also referred as lattice enthalpy.
38. What are state and path functions? Give two examples.
The variables like P,V,T and 'n' that are used to describe the state of a system are called as state functions. A state function is a thermodynamic property of a system, which has a specific value for each state of the system and does not depend on the path (or manner ) in which a particular state is reached.
Example: Pressure(P), Volume (V), Temperature (T), Internal energy(U), Enthalpy (H), free energy (G) etc.
A path function is a thermodynamic property of the system whose value depends on the path or manner by which the system goes from its initial to final states. Example: Work (w), Heat (q).
39. Give Kelvin statement of second law of thermodynamics.
It is impossible to take heat from a hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to a cooler reservoir.
40. The equilibrium constant of a reaction is 10, what will be the sign of ∆G? Will this reaction be spontaneous?
We know that, ΔG = −2.303RT log K
= −2.303 RT log l0
ΔG = −2.303RT [log 10 =1]
since, ΔG is −ve, the reaction is spontaneous.
41. Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
According to Arrhenius theory of electroytic dissociation, strong acid and strong base ionizes completely in dilute solution to produce H+ ions and OH− ions respectively results in neutralisation to give net reaction water. Hence, enthalpy of neutralisation of strong acid and strong base is always a constant ie., 57.32 kJ
H+(aq) + OH− (aq) → H2O(l) ΔH = -57.32 KJ
42. State the third law of thermodynamics.
The third law of the thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero (or) otherwise it can be states as " it is impossible to lower the temperature of an object to absolute zero in a finite number of steps. Mathematically
lim T → 0 S = 0
43. Write down the Born-Haber cycle for the formation of CaCl2
Born - Haber cycle for the formation of CaCl2 as follows:
According to Hess's law of constant heat of summation;
ΔHf = ΔH4 + ΔH2 + ΔH3 + 2ΔH4 + U
Where ΔHf = Enthalpy change for formation of CaCl2
ΔH1 = Enthalpy change for sublimation
ΔH2 = Enthalpy change for dissociation
ΔH3 = Ionisation Energy
ΔH4 = Electron Affinity
U = Lattice enthalpy for formation of CaCl2.
44. Identify the state and path functions out of the following: a)Enthalpy b)Entropy c) Heat d) Tem-perature e) Work f)Free energy.
45. State the various statements of second law of thermodynamics.
1. Entropy statements:
"whenever a spontaneous process takes place, it is accompanied by an increase in the total entropy of the Universe".
2. Kelvin - planck statement:
It is impossible to take heat from hotter reservoir and convert it completely into work by a cyclic process without transferring a part of heat to a cooler reservoir.
3. Clausius statement:
Heat flows spontaneously from hot objects to cold objects and to get it flow in the opposite direction, we have to expend some work.
46. What are spontaneous reactions? What are the conditions for the spontaneity of a process?
A reaction which takes place on their own accord without proper initiation is called spontaneous reaction. All natural process are spontaneous.
The necessary condition for a reaction to be spontaneous is Δ H –TΔS < 0, then only ΔG will be negative.
47. List the characteristics of internal energy.
● Characteristics of internal energy (U):
Internal energy of a system is an extensive property. It depends on the amount of the substances present in the system. If the amount is doubled, internal energy is also doubled.
● Internal energy of a system is a state function. It depends only upon the state variables ( T, P, V, n) of the system. The change in internal energy does not depend on the path by which the final state is reached.
● The change in internal energy of a system is expressed as ΔU = U2 – U1
● In a cyclic process, there is no energy change. ΔU(cyclic) = 0.
● If the internal energy of the system at final state ( Uf) is less than the internal energy of the system at its initial state (Ui), then ΔU = Uf − Ui = −veUf < Ui
● If the internal energy of the system at final state (Uf) is greater than the internal energy of the system at its initial state (Ui), then Δ U would be negative.
ΔU = Uf − Ui = − ve Uf < Ui
48. Explain how heat absorbed at constant volume is measured using bomb calorimeter with a neat diagram.
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 weighted 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 lowered in water, which is placed inside 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 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 the water.
Heat absorbed by the calorimeter q1 = kΔT where k is a calorimeter constant equal to mc Cc (mc is mass of the calorimeter and Cc is heat capacity of calorimeter)
Heat absorbed by the water q2 = mwCwΔT where mw is molar mass of water Cw is molar heat capacity of water (4,184 kJ K−1 mol −1)
Therefore ΔUC = q1 + q2
= kΔT + mwCwΔT
= (k + mwCw)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 kJ mol−1)
The enthalpy of combustion at constant pressure of the substance is calculated from the equation
ΔHCo (pressure) = ΔUCo (vol) + ΔngRT
49. Calculate the work involved in expansion and compression process.
● In most thermodynamic calculations we are dealing with the evaluation of work involved in the expansion or compression of gases.
● The essential condition for expansion or compression of a system is that there should be difference between external pressure (Pext) and internal pressure (Pint).
● For understanding pressure volume work, let us consider a cylinder which contains 'n' moles of an ideal gas fitted with a frictionless piston of cross sectional area A.
● The total volume of the gas inside is Vi and pressure of the gas inside is Pint.
● If the external pressure Pext is greater than Pint, the piston moves inward till the pressure inside becomes equal to Pext. Let this change be achieved in a single step and the final volume be Vf.
In this case, the work is done on the system (+ w). It can be calculated as follows
w = −F. Δx ………………… (1)
where dx is the distance moved by the piston during the compression and F is the force action on the gas.
F = PextA ………………. (2)
Substituting (2) in (1)
w = − PextA. Δx
A.Δx = change in volume = Vf − Vi
w = −Pext. (Vf − Vi) …………….. (3)
w = − Pext. (-ΔV) ……………. (4)
= Pext. ΔV
Since work is done on the system, it is a positive quantity.
● If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV.
● In such a case we can calculate the work done on the gas by the reaction
a compression process, Pext the external pressure is always greater
than the pressure of the system.
i.e Pext = (Pint + dP).
In an expansion process, the external pressure is always less than the pressure of the system
i.e Pext = (Pint − dP).
● When pressure is not constant and changes in infinitesimally small steps (reversible conditions) during compression from Vi to Vf, the P-V plot looks like in fig work done on the gas is represented by the shaded area. In general case we can write,
For a given system with an ideal gas
If Vf >Vi (expansion), the sign of work done by the process is negative.
If Vf <Vi (compression) the sign of work done on the process is positive.
50. Derive the relation between ∆H and ∆U for an ideal gas. Explain each term involved in the equa-tion.
● Consider a closed system of gaseous reactants which are chemically reacting to produce gaseous products at constant temperature and pressure with Vi and Vf as the total volumes of gaseous reactants and products respectively, and n’2 and nf as the number of moles of gaseous reactants and products, then
Reactants → products
For reactants, PVi = ni RT …………….. (1)
For products PVf = nf RT ……………… (2)
(2) − (1) gives
P(Vf -Vi) = (nf − ni ) RT
PΔV = Δng RT …………… (3)
We know, ΔH = ΔU + PΔV ………….. (4)
Substitute (3) in (4) we get,
ΔH = ΔU + Δng RT
Where ΔH = Enthalpy change at constant pressure
ΔU = Enthalpy change at constant volume
Δng = Change in number of moles ( ∑npg − ∑nrg)
R = Gas constant
T = Temperature
51. Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal.
Born Haber cycle applies Hess's law which is used to calculate lattice enthalpy of sodium chloride
● Let us use the Born - Haber cycle for determining the lattice enthalpy of NaCl as follows:
● Since the reaction is carried out with reactants in elemental forms and products in their standards states, at 1 bar the overall enthalpy of formation of NaCl.
● Also, the formation of NaCl can be considered in 5 steps.
● The sum of the enthalpy changes of these steps is equal to the enthalpy changes for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride using Born - Haber cycle
ΔHf = heat of formation of sodium chloride = −411.3 kJ mol−1
ΔH1 = heat of sublimation of Na(S) = 108.7 kJ mole−1
ΔH2 = ionisation energy of Cl2(s) = 495kJ mol−1
ΔH3 = dissociation energy of Cl2 (s)= 244 kJ mol−1
ΔH4 = Electron affinity of Cl(s) = −349.0 kJ mol−1
U = Lattice energy of NaCl
Δ Hf = ΔH1 + ΔH2 + ½ ΔH3 + ΔH4 + U
∴ U = (Δ Hf) − (ΔH1 + ΔH2 + ½ ΔH3 + ΔH4)
U = (−411.3) − (108.7 + 495.0 + 122 − 349)
U = (−411.3) − (376.7)
∴ U = − 788 kJ mol−1
This negative sign in lattice energy indicates that the energy is released when sodium is formed from its constituent gaseous ions Na+ and Cl−.
52. List the characteristics of Gibbs free energy.
1. "Gibbs free energy is defined as the part of total energy of a system is defined as the part of (or) available for conversion into Work"
G is mathematically defined as G = H-Ts
where, H - Energy, T- Temperature, S - Entropy.
2. G is a state Function and is a single valued function
3. G is an extensive property, whereas ΔG becomes intensive property for a closed system. Both G and ΔG Values correspond to the system only.
4. ΔG = ΔH − TΔS expression is used to predict the spontaneity of a process.
5. ΔG gives a criteria for spontaneity at constant pressure and temperature.
(i) If ΔG is negative (ΔG < 0), the process is spontaneous.
(ii) If ΔG is positive (ΔG > 0), the process is spontaneous.
(iii) If ΔG = 0 (zero) - process is in equilibrium.
6. For any system at constant pressure and temperature
− ΔG = −w − P ΔV = net work
But −P ΔV represents the work done due to expansion against a constant external pressure. Therefore, it is clear that the decrease in free energy (−ΔG) accompanying a process taking place at constant temperature and pressure is equal to the maximum work obtainable from the system other than the work of expansion.
7. Standard free energy of formation of elements are taken as zero.