Chemistry Cha 11 Thermochemistry new notes for, kpk, fbise, adamjee, Punjab and Sindh textbook boards jamshoro.
Thermochemistry Class 11th Notes 1st years for kpk 2021 textbook
Q.5 i)Total energy of the system and its surroundings remains constant.
Answer:
According to the law of conservation of energy, “energy can neither be created nor destroyed although it may change from one form to another”. Energy is transferred between the system and surroundings in the form of heat and work. A system may lose energy to the surroundings in the form of heat but at the same time, the same amount of energy is absorbed by the surroundings. Hence, the amount of energy lost by the system is equal to the amount of energy gained by the surrounding and the total energy of the system and its surroundings always remains constant.
Q.5 ii) It is essential to mention the physical states of the reactants and the products in a thermochemical equation.
Answer:
The heat of reaction (heat of formation) depends upon the physical states of the reactants and products. The heat of reaction of the same compound may be different in different physical states. This is because the heat contents of the same reactants and products are different for different physical states. For example; the heat of formation of liquid water and steam are different i.e. –285.8 kJ/mole and –241.8 kJ/mol although the same compound (H2O) is formed. This difference in heat of formation is due to the difference of physical states of H2O. Thus it is necessary to mention the physical states of reactants and products in thermochemical equations.
Q.5 iii) Ionic reactions are very fast.
Answer:
Ionic reactions are the reactions between ionic compounds. As ionic compounds dissociate easily in aqueous solution to form ions, and these ions reunite rapidly to form the product. There is no need for extra energy to break the compound into its ions first. So ionic reactions are very fast.
Q.5 iv) The work is done has positive and negative values.
Answer:
The work done may have a positive or negative value. Work done by the system on the surroundings is negative because it tends to decrease the internal energy of the system. On the other hand, the work done on the system by the surroundings is positive because it increases the internal energy of the system.
Q.5 v) The exothermic reactions are spontaneous.
Answer:
Exothermic reactions are spontaneous because the product created by an exothermic reaction is the most stable. Exothermic reactions release heat or energy. The more energy something has, the more unstable it will be. Therefore kinetics determines that the most stable product will dominate and as a result, exothermic reactions are spontaneous to produce a stable product.
Q.5 vi) Standard heat of combustion is always negative.
Answer:
Standard heat of combustion is always negative because the combustion is an exothermic reaction. Because of the bonding in oxygen and the electronic configuration, oxygen has quite a lot of energy. When it reacts, it releases this excess energy to create compounds with lower total energy, stabilizing itself. The excess energy is released as heat.
Q.5 vii) Enthalpy change is a state function but heat is not.
Answer:
Enthalpy change is a state function because it only depends upon the initial and final state and not on the path taken to accomplish that enthalpy change. On the other hand, heat is not a state function because it depends on how the change is brought about.
Q.5 viii) ΔH for solids and liquids becomes equal to ΔE.
Answer:
At constant pressure, the enthalpy change is written as ΔH = ΔE + PΔV. As the volume change in solids and liquids is very small and the term PΔV is negligible, ΔH for solids and liquids becomes equal to ΔE.
Read more: Chemistry Cha 9 Chemical Kinetics Class 11 Notes KPK
Q.5 ix) Endothermic reactions are always written with a positive sign.
Answer:
Endothermic reactions are those in which the heat is absorbed from the surrounding. As the heat is added to the system and the total heat content of products becomes greater than that of reactants, the enthalpy change for endothermic reactions is always written with a positive sign.
For example, the reaction between hydrogen and iodine is an endothermic process and is written with a positive sign as follows,
H2(g) + l2(s) → 2HI(g) ΔH°= +51.8 kJ mol-1
Q.5 x) Enthalpy of neutralization of strong acids and strong bases has always the same value.
Answer:
Enthalpy of neutralization is the amount of heat evolved when one mole of hydrogen ions H+ from an acid reacts with one mole of hydroxide ions OH- from a base to form one mole of water. Since strong acids and strong bases are completely ionized in water and the same reaction occurs for neutralization of any strong acid with any strong base, hence enthalpy of neutralization is always the same i.e. -57.4 kJ mol-1 approximately.
OH– + H+ → H2O ΔH°n= -57.4 kJ mol-1
Q.5 xi) ΔH° and ΔE have the same value for the reaction taking place in solutions.
Answer:
At constant pressure, the enthalpy change is written as ΔH = ΔE + PΔV. For the reactions taking place in solutions, the volume change ΔV is approximately zero hence ΔH becomes equal to ΔE.
ΔH = ΔE + PΔV
For reactions taking place in solution ΔV=0, so the equation becomes
ΔH = ΔE + P(0)
ΔH = ΔE + 0
ΔH = ΔE
Long Question Chemistry 1st-Years Notes for KPK 2021
Q.2) Applying Hess’s law, calculate ΔH° for the sublimation of one mole of iodine from the following equations.
i) H2(g) + l2(s) → 2HI(g) ΔH°= 51.8 kJ mol-1
ii) H2(g) + I2(g) → 2HI(g) ΔH°= -10.5 kJ mol-1
Answer:
Calculations:
(i) H2(g) + I2(s) → 2HI(g) ΔH°= 51.8 kJ mol-1
(ii) H2(g) + I2(g) → 2HI(g) ΔH°= -10.5 kJ mol-1
Subtracting equation (ii) from equation (i) and similarly subtracting the corresponding enthalpies,
I2(s) – I2(g) → 0 ΔH°= 51.8 – (-10.5) kJ mol-1
I2(s) → I2(g) ΔH°= 62.3 kJ mol-1
As sublimation is a process in which the solid directly converts from solid to vapours, hence the enthalpy of sublimation of iodine (I2) is 62.3 kJ mol–1.
Q.3) Calculate the heat of formation of an aqueous solution of NH4CI from the following data.
i) NH3(g) + aq → NH3(aq) ΔH° = -35.16 kJ mol-1
ii) HCl(g) + aq → HCl(aq) ΔH° = -72.41 kJ mol-1
iii) NH3(aq) + HCl(aq) → NH4Cl(aq) ΔH° = -51.48 kJ mol-1
Answer:
Calculations:
(i) NH3(g) + aq → NH3(aq) ΔH° = -35.16 kJ mol-1
(ii) HCl(g) + aq → HCl(aq) ΔH° = -72.41 kJ mol-1
(iii) NH3(aq) + HCl(aq) → NH4Cl(aq) ΔH° = -51.48 kJ mol-1
The heat of formation of an aqueous solution of NH4Cl will be the heat evolved when the components NH3 and HCl are dissolved in water to from NH4Cl. From the given data, the heat of formation of NH4Cl can be calculated as follows,
Adding equations (i), (ii) and (iii) and also their corresponding enthalpies,
NH3(g) + aq + HCl(g) + aq → NH4Cl(aq) ΔH° = -35.16 – 72,41 – 51.48 kJ mol-1
NH3(g) + HCl(g) + aq → NH4Cl(aq) ΔH° = -159.05 kJ mol-1
Hence the heat of formation of NH4Cl is -159.05 kJ mol-1.
Q.4) Liquid ethanol when burnt in oxygen at 25°C shows ΔH°= -1402.14 kJ mol-1. The heats of formation of CO2 and H2O are -393.50 and -285.81 kJ mol-1 respectively at the same temperature. Calculate the heat of formation of ethanol at 25°C.
Answer:
Calculations:
Liquid ethanol when burnt in oxygen at 25oC shows ΔH°= -1402.14 kJ mol–1. The chemical equation for this reaction is as follows,
(i) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g) ΔH° = -1402.14 kJ mol-1
Reverse of this reaction can be written with positive value of ΔH°. That is,
(ii) 2CO2(g) + 3H2O(g) → C2H5OH(l) + 3O2(g) ΔH° = +1402.14 kJ mol-1
The heats of formation of CO2 and H2O are -393.50 and -285.81 kJ mol-1 respectively, so we can write,
(iii) C(s) + O2(g) → CO2(g) ΔH° = -393.50 kJ mol-1
(iv) H2(g) + 1/2O2(g) → H2O(l) ΔH° = -285.81 kJ mol-1
Multiplying equation (iii) with ‘2’ and equation (iv) with ‘3’,
(v) 2C(s) + 2O2(g) → 2CO2(g)
ΔH° = 2(-393.50) = -787 kJ mol-1
(vi) 3H2(g) + 3/2O2(g) → 3H2O(l)
ΔH° = 3(-285.81) = -857.43 kJ mol-1
Adding equations (ii), (v) and (vi) and their corresponding enthalpies,
2C(s) + 2O2(g) + 3H2(g) + 3/2O2(g) → C2H5OH(l) + 3O2(g)
ΔH° = 1402.14-787-857.43 kJmol-1
2C(s) + 2O2(g) + 3H2(g) + 3/2O2(g) → C2H5OH(l) + 3O2(g)
ΔH° = -242.29 kJmol-1
Hence the enthalpy of formation of liquid ethanol is -242.29 kJ mol-1.
What do you mean by state of a system? What are state functions?
State:
“The condition of the system in terms of temperature, pressure, volume and composition is called the state of the system”.
State Function:
“A state function is that property of the system which depends on the initial and final states and is independent of the path adopted to bring about the change”.
For example, if we heat a sample of water from 0oC to 25oC, the change in temperature is equal to the difference between the final and initial temperatures. The way the temperature change is brought about has no effect on the result.
ΔT = Tfinal – Tinitial = 25oC
Q.6 b) A chemical reaction takes place in a container of crosssectional area 100 cm2 fitted with a weightless and frictionless piston.The piston is moved up through 10 cm against an external pressure of 1 atmosphere as a result of the reaction. Calculate the work done by the system.
Answer:
Data:
Area of container = 100 cm2
Distance moved = Δl = 10 cm
Pressure = P = 1 atm
Required:
Work done = W =?
Calculations:
ΔV = A x Δl = 100 x 10 = 1000 cm3
Using the formula to calculate pressure-volume work,
W = P ΔV = (1 atm)(1000cm3) = 1000 atm-cm3
Q.7 a) State and explain Hess’s law of constant heat summation. Show that it is a direct consequence of the first law of thermodynamics.
Answer:
Hess’s Law:
“The amount of energy evolved or absorbed in a chemical reaction is the same whether the reaction takes place in a single or several steps”.
Explanation:
According to Hess’s law, the net heat of reaction depends only on the initial and final states and not on intermediate steps. Let a substance A changes to B in two ways.
i) A changes directly to B and Q is the amount of heat absorbed in this process.
A → B ΔH° = +Q
ii) A changes to B indirectly in three steps as represented in the diagram given below. According to Hess’s law, the total absorption of heat is the sum of heats involved in different steps. That is,
Q = q1 + q2 + q3
Hess’s law is the direct consequence of the first law of thermodynamics because the amount of energy remains conserved either the reaction takes place in single step or several different steps.
Q.7 b) Determine ΔH° for the following reaction.
2C(S) + 2H2(g) → C2H4 (g) ΔH0 = ?
i) C(S) + O2(g) → CO2 (g) ΔH°1= -393.29 kJ mol-1
ii) H2(g) + ½O2(g) → H2O (g) ΔH°2= -285.7 kJ mol-1
iii) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) ΔH°3= -1430.92 kJ mol-1
Answer:
2C(s) + 2H2(g) → C2H4(g) ΔH0 = ?
i) C(s) + O2(g) → CO2 (g) ΔH°1= -393.29 kJ mol-1
ii) H2(g) + ½O2(g) → H2O (g) ΔH°2= -285.7 kJ mol-1
iii) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) ΔH°3= -1430.92 kJ mol-1
ΔH0 of this reaction can be calculated by using Hess’s law
Multiplying reaction (i) and (ii) with ‘2’,
iv) 2C(s) + 2O2(g) → 2CO2 (g) ΔH°4= 2(-393.29) = -786.58 kJ mol-1
v) 2H2(g) + O2(g) → 2H2O (g) ΔH°5= 2(-285.7) = -571.4 kJ mol-1
Reversing the reaction (iii),
vi) 2CO2(g) + 2H2O(g) → C2H4(g) + 3O2(g) ΔH°6= +1430.92 kJ mol-1
Now adding the reactions (iv), (v) and (vi) and their corresponding enthalpies,
2C(S) + 2H2(g) → C2H4(g) ΔH0 = -786.58-571.4+1430.92 = 72.94 kJ
Q.8 a) What is lattice energy? How does Born Haber cycle help to calculate it?
Answer:
Lattice Energy:
The amount of heat given out when one mole of ionic compound is formed from gaseous ions is called lattice energy.
Calculation of Lattice Energy from Born Haber Cycle:
The born Haber cycle is based on the principle, that the sum of energy change which occurs in a closed cycle, from the same initial and final states, is zero. The principle involves the law of conservation of energy in accordance with the first law of thermodynamics.
Born Haber cycle helps to calculate lattice energy of ionic compound. For example, lattice energy of NaCl is calculated as follows. The standard heat of formation (△Hof) of NaCl is the amount of heat given out when one mole of NaCl is formed from its elements, solid Na and gaseous chlorine. Lattice energy cannot be measured directly but it can be determined indirectly from the other steps that lead to the formation of Na+ (g) from Na (s) and Cl– (g) from ½ Cl2 (g). We measure △Hof for NaCl as follows:
Direct Method
Na (s) + ½ Cl2 (g) ⟶ NaCl (s) △Hof = -411 kJ/mol
Stepwise
Solid Na is ionized in two steps.
i) Na (s) ⟶ Na (g) △Hosub = 109 kJ/mol
ii) Na (g) ⟶ Na+ (g) + e– △HoIE = 496 kJ/mol
Similarly,
iii) Cl2 (g) ⟶ 2Cl (g) △HoDISS = 242 kJ/mol
iv) Cl (g) + e– ⟶ Cl- (g) △HoEA = -348 kJ/mol
v) Na+ (g) + Cl– (g) ⟶ NaCl (s) △HoLat = U =?
Various energies involved in Born Haber cycle for NaCl can be summarized as follows:
According to Hess’s law the sum of all five steps lead to the heat of formation of NaCl (s).
△Hof = △Hosub + △HoIE + ½ △HoDISS + △HoEA + △HoLat
△HoLat = △Hof – △Hosub – △HoIE – ½ △HoDISS – △HoEA
△HoLat = -411 kJ/mol – 109 kJ/mol – 496 kJ/mol – 121 kJ/mol –(-348 kJ/mol) = -789 kJ/mol
Q.8 b) Calculate the electron affinity of LiF from the following data: The heat of formation of LiF is -618 kJ mole-1, the heat of sublimation of Li is 161 kJ mole-1. The ionization energy of Li is 520 kJ mole-1, the dissociation energy of fluorine is 158 kJ mole-1, while the lattice energy is -1056 kJ mole-1.
Answer:
According to Hess’s law, sum of all the steps involved leads to the heat of formation.
△Hof = △Hosub + △HoIE + ½ △HoDISS + △HoEA + △HoLat
Rearranging equation for △HoEA and putting the values,
△HoEA = △Hof – △Hosub – △HoIE – ½ △HoDISS – △HoLat
△HoEA = -618 kJ/mol – 161 kJ/mol – 520 kJ/mol – ½ (158 kJ/mol) – (-1056 kJ/mol)
△HoEA = -322 kJ/mol
Hence the electron affinity for LiF is -322 kJ/mol.
Q.9) What is meant by heat of a reaction? Explain how the heat of a reaction at constant volume differs from the heat of a reaction at constant pressure?
Answer:
The heat of Reaction:
“The enthalpy change during a reaction is called the heat of reaction”. It is the difference in enthalpies of products and reactants. The heat of reaction is positive for endothermic reactions and negative for exothermic reactions.
The change in enthalpy can be written as,
ΔH = ΔE + Δ(PV)
ΔH = ΔE + PΔV + VΔP
Constant pressure:
For constant pressure, ΔP = 0 and hence VΔP = 0 and the equation becomes
ΔH = ΔE + PΔV (Heat of reaction at constant pressure)
Constant Volume:
For constant volume, the change in volume ΔV = 0 and hence PΔV = 0 and equation becomes
ΔH = ΔE (Heat of reaction at constant volume)
For a constant pressure process, all the heat supplied to the system is used to increase the internal energy of the system. While in constant pressure system a part of this energy is also used to do pressure-volume work.
Answer:
Pressure-Volume Work:
“The work done by a system when it expands is called pressure volume work”. The only work done in thermodynamics is pressure volume work.
Explanation:
Consider a gas contained in a cylinder of cross-sectional area A, fitted with a weightless and frictionless piston. The pressure on the piston is P. Since, pressure is force per unit area so P can be written as,
P = F/A or F = P x A
We know that,
Work = Force x Distance
If piston moves through a distance Δl during expansion, the work done will be,
Work = P x A x Δl
Since, A x Δl = ΔV so we can write
Work = PΔV
Pressure-volume work is related with the first law of thermodynamics which states that the total amount of energy remains constant although it changes from one form to other. So the heat supplied to the system may be used to do pressure-volume work by the system or if the system works its energy decreases because a part of the energy is used in performing the pressure-volume work.
Work is positive (PΔV) when it is done on the system. So we can write,
q = ΔE + PΔV
Work is negative (-PΔV) when it is done by the system. So we can write,
q = ΔE – PΔV