Chemistry For F.S.C: Chapter 7th

CHAPTER 7
THERMOCHEMISTRY
MCQs

Q.1 Which of the following statements is contrary to the first law of thermodynamics?

(a) energy can neither be created nor destroyed

(b) one form of energy can be transferred into an equivalent amount of other kinds of energy

(d) continuous production of mechanical work with out equivalent amount of heat is possible

Q.2 The change in heat energy of a chemical reaction at constant temperature and pressure is called

(a) enthalpy change (b) bond energy

Q.3 For the reaction NaOH + HCl ® NaCl + H2O, the change in enthalpy is called as:

(a) heat of reaction (b) heat of formation

Q.4 Calorie is equivalent to

(a) 0.4184 J (b) 41.84 J

Q.5 For a given process, the heat change at pressure (qp) and constant volume (qv) are related to each other as

(a) qp = qv (b) qp < qv

Q.6 The net heat change in a chemical reaction is same whether. It is brought about in two or more different ways in one or several steps. It is known as

(a) Henry’s law (b) Joule’s principle

Q.7 Enthalpy of neutralisation of all the strong acids and strong bases has the same value because

(a) neutralisation leads to the formation of salt and H2O

(b) strong acid and bases are ionic substances

(d) the net chemical change involve the combination of H+ and OH– ions to form water

Q.8 If an endothermic reaction is allowed to take place very rapidly in the air. The temperature of the surrounding air

(a) remains constant (b) increase

Q.9 In endothermic reactions, the heat content of the

(a) products is more than that of reactants

(b) reactants is more than that of products

Q.10 Hess’s law is also called

(a) first law of thermodynamics

(b) second law of thermodynamics

(d) second law of thermochemistry

Q.11 Pressure – volume work is

(a) P D v (b) F x d

Q.12 Kinetic energy of molecules is due to

(a) rotational energy (b) vibrational energy

Q.13 The condition for standard enthalpy change is

(a) 1 atm 30oC (b) 1 atm 0oC

Q.14 The unit of enthalpy change is

(a) calorie (b) joule

Q.15 The sum of all kinds of a system is ions or molecules of a system is

(a) vibrational energy (b) potential energy

Q.16 An endothermic reaction is one is which

(a) enthalpy of reactants and products are same

(b) enthalpy of products is greater than reactant

(d) heat is evolved from system

Q.17 Bomb calorimeter is used to determine

(a) enthalpy of solution

(b) enthalpy of atomization

(d) enthalpy of neutralization

Q.18 Glass calorimeter is used to determine

(a) enthalpy of combustion

(b) enthalpy of reaction

(d) none of above

Q.19 Born–Haber cycle is used to calculate

(a) enthalpy of combustion

(b) lattice energy of ionic camps

(d) none of above

Q.20 Born–Haber cycle is an application of

(a) first law of thermodynamics

(b) second law of thermodynamics

(d) Hess’s law

Q.21 An exothermic reaction is one in which

(a) enthalpy of reactants and products are same

(b) heat is absorbed by system

(d) enthalpy of reactants is lesser than products

Q.22 A substance under observation during an experiment

(a) surrounding (b) system

Q.23 Enthalpy of neutralization is merely

(a) heat of solution (b) heat of atomization

(d) heat of formation of H2O

Q.24 Lattice energy of NaCl is

(a) + 500 kJ (b) – 344 kJ

Q.25 Standard enthalpy of Al2O3 cannot be measured because

(a) it does not catch fire

(b) it reacts with CO2

(d) none of above

Q.26 Ammonium chloride dissolve in water this process is

(a) endothermic process (b) exothermic process

Q.27 First law of thermodynamics is represented as

(a) D E = q + R T (b) D E = q + D P

Q.28 Pumping of water uphill is

(a) spontaneous reaction (b) exothermic reaction

(d) endothermic

Q.29 In exothermic reaction D H is

(a) positive (b) negative

ANSWERS

Questions	1	2	3	4	5
Answers	D	a	c	c	C
Questions	6	7	8	9	10
Answers	c	d	c	c	D
Questions	11	12	13	14	15
Answers	a	d	c	a	d
Questions	16	17	18	19	20
Answers	b	c	b	b	d
Questions	21	22	23	24	25
Answers	c	b	d	c	c
Questions	26	27	28	29
Answers	a	d	b	b

SHORT QUESTION AND ANSWERS

Q.1 Define the following terms and give three examples of each.

Ans.

(i) System:

The substance which is under experiment or under observation is called as system.

Examples:

(i) Pb(NO3)2 in decomposition of Pb(NO3)2.

(ii) Zn and CuSO4 solution, the reaction mixture in the vessel.

(iii) CaCO3 in thermal decomposition of CaCO3

(ii) Surroundings:

Everything around the system which is not a part of system is called surroundings.

For example

During the reaction between Zn and CuSO4 solution vessel and air etc are surroundings.

(iii) State function:

A macroscopic property of a system which has some definite value for initial and final state and independent of the path followed e.g.

(i) Pressure (ii) Temperature (iii) Internal energy.

Note: Heat is not a state function.

Q.2 Describe the units of energy.

Ans.

Mostly Joule and calorie are used for the measurement of energy.

Calorie:

The amount of energy required to raise the temperature of one gram of water from 14.5oC to 15.5oC is called one calorie.

Joule:

It is SI unit of energy and defined as energy expanded when a force of one Newton moves a body through one meter in the direction in which force is applied.

Joule = Force x distance

1J = 1 N x 1 m

Q.3 What are rhermochemical reactions?

Ans:

Exothermic reaction:

Those thermochemical reactions in which heat is evolved as a result of reaction are called as exothermic reactions.

C(s) + O2(g) ® CO2(g) DH = – 393.7 kJ/mole

H2(g) + O2(g) ® H2O(l) DH = – 285.5 kJ/mole

N2 + 3H2(g) ® 2NH3(g) DH = – 41.6 kJ/mole

Endothermic reactions:

Those thermochemical reactions in which heat is absorbed as a result of reaction are called as endothermic reaction.

N2(g) + O2(g) ® 2NO(g) DH = + 180.51 kJ/mole

H2O(l) ® H2(g) + O2(g) DH = + 285.58 kJ/mole

H2(g) + I2(g) ® 2H I DH = + 52.96 kJ/mole

Q.4 Differentiate between internal energy and enthalpy of a system?

Ans:

Internal energy:

The total of all kinds of K.E and P.E of all the particles of a system is called as internal energy. It is denoted by “E’ e.g., kinetic energy may be in the form of translation, vibrational and rotational motion and potential energy is intermolecular and intramolecular forces of attraction.It is a sate function of system.

E= K.E + P.E

Enthalpy of the system:

The total heat contents of a system and denoted by H. The increase in the internal energy of a system plus work done is called as enthalpy i.e.

H = E + Pv

Q.5 Define the followings:

(i) Enthalpy of reaction

(ii) Enthalpy of neutralization

(iii) Enthalpy of combustion

Ans:

Standard Enthalpy of reaction:

The enthalpy change when no. of moles of reactants as indicated by the balanced chemical equation react completely together to give the products under the standard conditions.

H2(g) + O2(g) ® H2O(l) DH = – 285.5 kJ/m

N2(g) + O2(g) ® 2NO(g) DH = + 180.5 kJ/m

Standard Enthalpy of Combustion:

The enthalpy change when one mole of a substance is completely burnt in excess of oxygen under standard conditions.

` C2H5OH(l) + 3O2(g) ® 2CO2(g) + 3H2O(l) DH

= – 1368 kJ/m

C(s) + O2(g) ® CO2(g) DH = – 393.7 kJ/m

2Al(s) + O2(g) ® Al2O3(s) DH = – 1675.7 kJ/m

Standard enthalpy of atomization:

The enthalpy change when one mole of gaseous atoms are formed from the elements under the standard conditions is called enthalpy of atomization.

® H(g) DH = 218 kJ/mole

Cl2 ® Cl(g) DH = + 121 kJ/mole

Q.6 Why it is essential to mention the physical states of reactants and products in a thermochemical equation?

Ans.

The heat of reaction depends upon the physical states of the reactants and products, heat of reaction is different in different physical states therefore, while writing a thermochemical equation it is essential to mention the physical states of the reactants and products.e.g

H₂₍_g) + 1/2O_2(g) → H₂O_(g) ∆H = -241.5 KJ mol^-1

H₂_(g) + 1/2O₂_(g) → H₂O_(l) ∆ H= -285.8 KJ mol^-1

Q.7 Differentiate between spontaneous and non–spontaneous reaction?

Ans.

The process which takes place on its own without any outside help and moves from a non–equilibrium state to equilibrium state is called spontaneous or natural process. It is real, unidirectional and irreversible e.g. water flows from higher level to low leve, reaction between acid and base etc.

There are certain reactions which need energy to start but once they start they proceed their own for example burning of candle.

The process which does not take place on its own and does not occur in nature is called as non–spontaneous. It is reverse of the spontaneous process i.e. pumping of water uphill, flow of heat from colder to hotter region etc.

Q.8 Prove that change in enthalpy is equal to heat of reaction? / prove that qp =DH?

Ans.

We know that enthalpy is equal to the internal energy plus product of pressure and volume.

H = E + Pv

According to first law of the thermodynamics

q = DE + w

At constant pressure w = PDv

qp = DE + PDv

qp = (E2 – E1) + (v2 – v1)p

qp = E2 – E1 + Pv2 – Pv1

qp = (E2 + Pv2) – (E1 + Pv1) H2 = E2 + Pv2

qp =H2 – H1 H1 = E1 + Pv1

qp =DH

This shows that enthalpy change is equal to amount of heat absorbed at constant pressure.

Q.9 Briefly explain laws of thermochemistry.

Ans. First law of thermochemistry:

The enthalpy of formation of a compound to the enthalpy of decomposition of that compound. e.g.

H2(g) + O2(g) ® H2O(l) DH = – 285.5 kJ/mole

H2O(l) ® H2(g) + O2(g) DH = + 285.5 kJ/mole

Second law of thermochemistry (Hess’s law):

The amount of heat evolved or absorbed in a chemical reaction is same whether the reaction takes place in one or several steps. e.g. single step process.

C(s) + O2(g) ® CO2(g) DH1 = – 393.7 kJ

Two steps process:

C(s) + O2(g) ® CO(g) DH2 = – 110.7kJ

DH1 = DH2 + DH3

– 393.7 = – 110.7 – 283

– 393.7 = – 393.7

Q.10 Draw a complete, fully labeled Born–Haber cycle for the formation of KBr.

Reactions:

K(s) + Br2(l) ® K+ Br– D H/kJ mole–1

– 392 kJ mole–1

K(s) ® K(g) + 90 kJ mole–1

K(g) ® K+ e– + 420 kJ mole–1

Br2(l) ® Br(g) +112 kJ mole–1

Br(q) + e– ® Br–(g) – 342 kJ mole–1

Solution:

The heat of formation of KBr is equal to sum of all the enthalpies.

DHf = – 392 kJ mole–1 (Heat of formation)

DHs = + 90 kJ mole–1 (Heat of sublimation)

DHi = + 420 kJ mole–1 (Ionization pot of k)

DHD/2 = + 112 kJ mole–1 (Dissociation energy of Br2)

DHe = – 342 kJ mole (electron affinity)

DHl = ? (Lattice energy of kBr)

DHf = DHat + DHi + DHd + DHe + DHl

– 392 = + 90 + 420 + 112 + (– 342) + DHl

DHl = – 672 kJ mole–1

Q.11 Heat is evolved in exothermic reactions and absorbed in endothermic relations.

Ans.

When bond formation energy is greater than the bond breaking energy then the excess of energy is evolved making the reaction exothermic. When the bond breaking energy is greater than the bond formation energy then the difference of energy is supplied from surrounding making the reaction is endothermic.

Q.12 How would you explain that change in enthalpy is a state function?

Ans.

As H = E + PV

As E, P and v are state functions as they are independent of path and depend only on the initial and final state of the system therefore enthalpy of a system is also a state function because enthalpy depends on E, P and V.

Q.13 How can you prove that (w = – PDV)

Ans.

We know that

Work = Force x distance

Initial volume= V1

Final volume = V2

Change in volume DV = V2 – V1

In the expansion of gases, the work is expressed in terms of pressure and change in volume. So,

Work = P x DV

= w = – PDV

The work is negative because it is done by the system.

Q14. What is difference between heat and temperature? Write a mathematical relationship between these two parameters?

Ans:

Heat: The measure of total energy of a substance is called heat. It is property of a body which flows from a body at higher temperature to a body at lower temperature. It is denoted by ‘q’. It depends upon the quantity of a substance. It is measured by calorimeter. It is not a state function.

Temperature: It is measure of average K.E of the molecules in the system. It is denoted by ‘T’. It is independent of the quantity of a substance. It is measured by thermometer. It is a state function.

Realtionship: q= m x S x ∆T

Q15. What is enthalpy of neutralization? Why enthalpy of neutralization of strong acid and base is always -57.4 KJ mol^-1?

Ans:

It is the amount of heat eveolved or absorbed when one mole of H⁺ ions from an acid reacts with one mole of OH^- from a base to form one mole of H₂O. Under standard conditions it is called standard enthalpy of neutralization, and it is denoted by ∆H_n.

H⁺ + OH^- → H₂O ∆H_n = -57.4 KJ mol^-1

The heat of neutralization of strong acid or base is always -57.4 KJ mol^-1 because strong acid or base is completely ionized and when acid and base is mixed no bond has to be broken

TEXT BOOK EXERCISE

Q1. Select the suitable answer from the given choices.

(i) If an endothermic reaction is allowed to take place very rapidly in the air the temperature of the surrounding air

(a) Remains constant (b) increases

(ii) In endothermic reactions, the heat content of the

(a) Products is more than that of reactants

(b) Reactants is more than that of reactants

(d) Reactants and product are equal

(iii) Calorie is equivalent to

(a) 0.4184 J (b) 41.84J

(iv) The change in heat energy of a chemical reaction at constant temperature and pressure is called

(a) enthalpy change (b) bond energy

(v) Which of the following statements is contrary to the first law of thermodynamics?

(a) Energy can neither be created nor destroyed.

(b) One form of energy can be transferred into an equivalent amount of of other kinds of energy.

(d) Continuous production of mechanical work without supplying an equivalent amount of heat is possible

(vi) For a given process, the heat changes at constant pressure (q_p) and at constant volume (q_v) are related to each other as

(a) q_p =q_v (b) q_p < q_v

(vii) For the reaction NaOH + HC1

NaC1 + H₂O. The change in enthalpy is called

(a) Heat of reaction (b) heat of formation

(viii) The net heat change in a chemical reaction is same whether it is brought about in two or more different ways in one or several steps. It is known as

(a) Henry’s law (b) Hess’s Law

(ix) Enthalpy of neutralization of all the strong acids and strong bases has the same value because

(a) Neutralization leads to the formation of salts and water

(b) Strong acids and bases are ionic substances

(d) The net chemical change involve the combination of H+ and OH- ions to form water

Ans. i)c ii)a iii)c iv)a v)d vi)c vii)c viii)b ix)d

Q2. Fill in the blanks with suitable words.

(i) The substance undergoing a physical or a chemical change forms a chemical ________.

(ii) The change in internal energy _________be measured.

(iii) Solids which have more than one crystalline forms possess_________values of heats of formation.

(iv) A process is called___________if it takes place on its own without any outside assistance.

(v) A __________is a macroscopic property of a system which is __________of the path adopted to bring about that change.

Ans. i)system ii)can iii)different iv)spontaneous v)state function : independent

Q3. Indicate the true or false as the case may be.

(i) It is necessary that a spontaneous reaction should be exothermic.

(ii) Amount of heat absorbed at constant volume is internal energy change.

(iii) The work done by the system is given the positive sign.

(iv) Enthalpy is a state function but internal energy is not.

(v) Total heat content of a system is called enthalpy of the system.

Ans. i) False ii) True iii) False iv) False v) True

Q4. Define the following terms and give three examples of each

i) System ii) Surroundings

iii) State function iv) Units of energy

v) Exothermic reaction vi) Endothermic reaction

vii) Internal energy of the system viii)Enthalpy of the system.

Q5. (a) Differentiate between the following:

(i) Internal energy and enthalpy

(ii) Internal energy change and enthalpy change

(iii) Exothermic and endothermic reactions

(b) Define the following

(i) Standard enthalpy of reaction

(ii) Standard enthalpy of combustion

(iii) Standard enthalpy of atomization

(iv) Standard enthalpy of solution

Q6. (a) What are spontaneous and non-spontaneous processes? Give examples.

(b) Explain that burning of a candle is a spontaneous process.

Explain it.

Q.7 (a) What is the first law of thermodynamics? How does it explain that

(i) q_v =

E ii) q_p=

(b) Wow will you differentiate between

E and

H ? Is it true that

H and

E have the same values for the reaction taking place in the solution state.

Hint: For reactions taking place in the solution state. Since the change in volume is insignificant, i.e .

V=0, so

E + P

H =

E +Px0

H =

Hence,

H and

E have the same values for the reaction taking place in the solution state.

Q.8. (a) What is the difference between heat and temperature ? Write a mathematical relationship between these two parameters.

Ans. Difference between heat and temperature

Temperature: “ A measure of the average kinetic energy of all the particles in a system is called temperature .”

Temperature is a state function. If we transfer energy to a system, the kinetic energy of the particles in the system increases. Therefore, the temperature rises.

Heat: “The transfer of energy caused by a difference in temperature between a system and its surroundings or between a system and another system is called heat.”

Heat is not a state function. It depends on the path of the system. A mathematical relationship between heat and temperature is :

q= ms

(b) How do you measure the heat of combustion of a substance by bomb calorimeter.

Q9. Define heat of neutralization. When a dilute solution of a strong acid is neutralized by a dilute solution of a strong base, the heat of neutralization is found to be nearly the same in all the cases. How do you account for this?

Q10. (a) State the laws of thermo-chemistry and show how are they based on the firs law of thermodynameics.

Ans. Laws of Thermochemistry

There are two thermochmical laws. They are based on the law of conservation of energy. These law are:

1. First Thermo chemical Law (Lavoisier and Laplace-1780)

“The quantity of heat required to decompose a compound into its elements is equal to the heat evolved when that compound is formed from its elements, but with opposite sign.”

In other words the heat of decomposition of a binary compound is numerically equal to the heat of formation of a compound, but of opposite sign.

Examples:

(1) H_{2(g )}+

O₂ _(g)

H ₂O_(g)

H= -285.58kJ mol^-1

H₂O_{(l )}

H ₂O_(g)+

O₂ _(g)

H= +285.58kJ mol^-1

(2) C_{(s )}

O_2(g)+CO₂ _(g)

H= -393.7kJ mol^-1

CO_{(g )}

C_(s)+ O₂ _(g)

H= +393.7kJ mol^-1

(3)

H₂ _(g)+

1₂ _(s)

H1_(g)

H= +26.48kJ mol^-1

H1_(g)

H₂ _(g)+

1₂ _(s)

H= -26.48kJ mol^-1

2. Second Thermaochemical law (Hess’s Law-1840)

“ The amount of heat evolved or absorbed in a chemical reaction is the same whether the reaction takes place in one step or in several steps.”

Mathematically:

H =

H₁+

H₂+ ………….

Where

H is the enthalpy change for the reaction when the reactants are directly converted into products in one step and

H₁and

H₂are the enthalpies of different steps when the reactants are converted into products in two steps.

(b) What us a thermochmical equation? Give three examples. What information do they convey?

Ans. Thermochmical Equation

“A chemical equation which gives an idea about the heat evolved or absorbed during the reaction is called a thermochmical equation.”

Examples: (i) C_(s) +O₂ _(g)

CO_3(g)

H= -393.7kJ mol^-1

(ii) H_{2(g )}+

O₂ _(g)

H ₂O_(l)

H= -285.58kJ mol^-1(iii) N_2(s) +O₂ _(g)

2NO_(g)

H= +180.51kJ mol^-1

(iv) N_2(s) +3H₂ _(g)

2NH_3(g)

H= -41.6kJ mol^-1

Information conveyed by a thermochmical equation

A thermochmical equation give the following information.

(i) The heat evolved or absorbed in a reaction.

H is always negative for an exothermic reaction where as it is positive for an endothermic reaction.

(ii) The physical state of reaction and products because they will effect the heat of reaction.

(iii) It is treated as a standard equation. The coefficients of reactants and products always represent their mole.

(iv) If the conditions of the reaction are not given, then it is assumed that the reaction is taking place at 1 atm pressure and 25^oC.

(v) If the termochemical equation is reversed, the sign of the heat of reaction will be reversed but the magnitude remains the same.

(c) Why is it necessary to mention the physical states of reactants and products in a thermochemical reaction? Apply Hess’s law to justify your answer.

Ans. Since the heat of reaction depends upon the physical state of the reactants and products, therefore, while writing a thermochemical equation, the physical state of the reactants and products must be mentioned.

If the reaction A

B is exothermic, than B

A will be endothermic.

Q11. (a) Sate and explain Hess’s law of constant heat summation. Explain it with examples and give its applications.

(b) Hess’s law helps us to calculate the heats of those reactions, which cannot be normally carried out in a laboratory. Explain it.

Q12. (a) What is lattice energy ? How does Bron-Haber Cycle help to calculate the lattice energy of NaC1?

(b) Justify that heat of formation of a compound is the sum of all the other enthalpies.

Q13. 50 cm³of 1.0 M HC1 is mixed with 50 cm3 of 1.00 M NaOH in a glass calorimeter. The temperature of the resultant mixture increases from 21.0^oC to enthalpy change mole^-1for the reactions. The density of solution to be considered is 1 g cm^-3and specific heat is 4.18 Jg^-1k^-1.

Solution:

Volume of 1.0 MHC1 =50cm³

Volume of 1.0 MNaOH =50cm³

Initial temperature of HC1 =21.0^oC

Initial temperature of NaOH =21.0^oC

Final temperature of reaction mixture =27.5^oC

Temperature rise for reaction mixture =27.5 – 21.0 =6.5^oC

Specific heat capacity of reaction mixture =4.10 jg^-1k^-1

=4.18 jg ^-1oC^-1

Density of solution =1 gcm^-3

Volume of reaction mixture =50 cm ³+ 50 cm³=100 cm³

Mass of reaction mixture =volume x density

=100 cm³x 1 g cm^-3

=100 g

Amount of total heat evolved, q =m x s x

=100g x 4.18 Jg ^-1oC^-1x 6.5^oC

=2717 J , =2.717 KJ

Since the reaction is exothermic, so, q = - 2.717 kJ

Vol. Of NaOH =

=0.05 dm³

No. of moles of HC1 =Molarity x vol. of soln. In dm3

=1.0 x 0.05

0.05 mole

Vol. Of NaOH =

=0.05 dm³

No. of moles of HC1 =Molarity x vol. of soln. In dm3

=1.0 x 0.05

0.05 mole

Equation of reaction: HC1_(aq) +NaOH _(aq)

NaC1_(aq)+ H₂O_(l)

^1mole ^1mole ^1mole

^{0.05 mole} ^{0.05mole 0.05mole}

Now, 0.1 mole of formation of H₂O liberates heat =-2.717 KJ

1 mole of formation of H2 O liberates heat =

=- 54.34 kJ

Enthalpy of neutralization, H_n= -54.34 kJ Answer

Q14. Hyrazine N₂H₄ is a rocket fuel. It burns in o₂to N₂and H₂O.

N₂H4_(l) +O_2(g)

N_2(g)+ 2H₂O_(g)

1.00 g of N₂H₄is burned in a bomb calorimeter. An increase of temperature 3.51^oC is recorded. The heat capacity of calorimeter is 5.5 kJK^-1. Calculate the quantity of heat evolved. Also calculate the heat of combustion of 1 mole of N₂H₄.

Solution:

Mass of N₂H₄= 1g

Rise of temperature =3.51 ^oC

Sp. Heat capacity of calorimeter =5.5 kJK^-1=5.5 kJ ^oC^-1

q =?

Formula Used: q = m x s x

=1 g x 5.5 kJ ^oC^-1x 3.51^oC

=19.31 kJ

Since combustion is an exothermic reaction, so

q =- 19.31 kJ Answer

Now, No of moles of N₂H₄ =1 mole

Mass of N₂H₄ =No. of moles x Molar mass

=1 mol x 32 g mol^-1

=32 g

Heat of combustion of 1 g of N₂H₄ = -19.31 kJ

Heat of combustion of 32 g of N₂H₄ =- 19.31 kJ x 32

Hence, heat of combustion of N₂H₄ = -618 kJ mol^-1Answer

Q15. Octane C₈H₁₈is a motor fuel. 1.80 g of a sample of octane is burned in a bomb calorimeter having heat capacity 11.66 kJk^-1. The temperature of the calorimeter increases from 31.36^oC to 28.78^oC. Calculate the heat of combustion for 1 g of octane. Also calculate the heat for 1 mole of octane.

Solution:

Mass of octane, C₈H₁₈ =1.80g

Sp. Heat capacity of calorimeter =11.66 kJK^-1=11.66 kJ ^oC^-1

Rise in temperature,

T =(28.78 – 21.36 ) ^oC =7.42 ^oC

Heat of combustion, q =?

Formula Used: q=m x s x

=1.80 g x 11.66 kJ ^oC^-1x 7.42 ^oC

=155.73 kJ Answer

Since combustion reaction is an exothermic, so

Heat of combustion = -155.73 kJ

Now, heat o combustion of 1.80 g of octane =- 155.73 kJ

Heat of combustion of 1 mole of octane = -

x 1 g

= - 86.52 kJ

Heat of combustion of 1 mole of octane = -86.52 kJ

No, of moles of octane =1 mole

Molar mass of octane, C₈H₁₈ =96+18=114 g mol^-1

Mass of octane = moles of octane x molar mass of octane

= 1 mol x 114 g mol^-1

=114 g

Now, heat of combustion of 1 g of octane = -86.52 kJ

Heat of combustion of 114 g of octane =

x 114 g

= - 9863 kJ

Hence, heat of combustion of 1 mole of octane=- 9863 kJ mol^-1Answer

Q16. By applying Hess’s law calculate the enthalpy change for the formation of an aqueous solution of NH₄C1 from NH₃gas and HC1. The results for the various reactions and pressure are as follows.

(i) NH_3(s) +aq

NH_3(aq)

H= -35.16kJ mol^-1

(ii) HC1_(g) +aq

HC1_(aq)

H= -72.41kJ mol^-1

(iii) NH_3(s) +HC1_(aq)

NH₄ C1_(aq)

H= -35.16kJ mol^-1

Solution:

NH_3(g) +HC1_(g)+ aq

NH₄C1_(aq)

H= ?

(i) NH_3(g) +aq

NH_3(aq)

H= -35.16kJ mol^-1

(ii) NH_3(g) +aq

HC1_(aq)

H= -72.41kJ mol^-1

(iii) NH_3(aq) +HC1_(aq)

NH₄ C1_(aq)

H= -51.48kJ mol^-1

Add Eq (i), Eq (ii)

(i) NH_3(g) +aq

NH_3(aq)

H= -35.16kJ mol^-1

(ii) NH_3(g) +aq

HC1_(aq)

H= -72.41kJ mol^-1

____________________________________________________________

(iv) NH_3(g) +HC1_(g)+2aq

NH_3(aq)+HC1_(aq)

H= -107.57kJ mol^-1

Now, add Eq (iii)and Eq(iv)

(iii) NH_3(aq) +HC1_(aq)

NH₄ C1_(aq)

H= -51.48kJ mol^-1

(iv) NH_3(g) +HC1_(g)+2aq

NH_3(aq)+HC1_(aq)

H= -107.57kJ mol^-1

____________________________________________________________

(v) NH_3(g) +HC1_(g)+2aq

NH_3(aq)+HC1_(aq)

H= -159.05kJ mol^-1

Since any integer multiplied by ‘aq’ is equal to ‘q’

So, 2 x aq =aq

Hence, Eq(v)becomes,

NH_3(g) +HC1_(g)+2aq

NH_3(aq)+HC1_(aq)

H= -159.05kJ mol^-1

H= .159.05 kJ mol^-1Answer

Q17. Calculate the heat of formation of ethyl alcohol from the following information

(i) Heat of combustion of ethyl alcohol is – 1367 kJ mol-1

(ii) Heat of formation of ethyl alcohol is – 393.7 kJ mol-1

(iii) Heat of formation of water is - 285.8 kJ mol-1

Solution:

2C_(s) +3H _2(g)+

O_2(g)

C₂H 5OH_(l)

H= ?

(i) C₂H₅OH_(l) + 3O_2(g)

2CO_2(g)+ 3H₂O_(g)

H= -1367kJ

(ii) C_(s) +O_2(g)

CO_2(g)

H= -393.7kJ

(iii) H_2(s)+

O_2(g)

H₂O_(l)

H= -285.8 kJ

Multiply Eq(ii) by 2 and Eq (iii) by 3 and then add the resultant equation

(iv) 2C_(s) +2O_2(g)

2CO_2(g)

H= -787.4kJ

(v) 3H_2(g)+

O_2(g)

3H₂O_(l)

H= -857.4 kJ

____________________________________________________________

(iv) 2C_(s)+ 3H_2(g)+

O_2(g)

2CO_2(g)+3H₂O_(l)

H= -1644.8 kJ

Now, subtract Eq(i) from Eq(iv),

(vi) 2C_(s)+ 3H_2(g)+

O_2(g)

2CO_2(g)+3H₂O_(l)

H= -1644.8 kJ

(i)

C₂H₅OH_(l)

3O_2(g

2CO_2(g)

3H₂O_(l)

1367 kJ

____________________________________________________________

2C_(s)+ 3H_2(g)+

O_2(g) – C₂H₅OH_(l)

H= -277.8 kJ

2C_(s)+ 3H_2(g)+

O_2(g)

C₂H₅OH_(l)

H= -277.8 kJ

Hence, the heat of formation of ethyl alcohol is - 277.8 kJ Answer

Q18. If the heats of combustion of C₂H₂, H₂and C₂H₆ are – 337.2 , - 68.3 and – 372.8 k calories respectively , then calculate the heat of the following reaction.

C₂H_2(g)+ 2H_2(g)

C₂H_{6 (g)}

Solution:

C₂H_2(g)+ 2H_2(g)

C₂H_{6 (g)}

H= ?

(i) C₂H_2(s)+

O_2(g)

2CO_2(g)+ C₂H_6(g)

H= -337.2 kJ

(ii) H_2(s)+

O_2(g)

H₂O_(l)

H= -68.3 kJ

(iii) C₂ H_6(s)+

O_2(g)

2CO_2(g) +3H₂O_(l)

H= -372.8 kJ

Multiply Eq (ii) by2

(iv) 2H_2(g)+ O_2(g)

2H₂O_(l)

H= -136.3kJ

Add Eq (i) , Eq (iv)

(i) C₂H_2(s)+

O_2(g)

2CO_2(g)+ H₂O_(l)

H= -337.2 kJ

(ii) 2H_2(g)+ O_2(g)

2H₂O_(l)

H= -136.3kJ

(v) C₂ H_2(s)+ 2H_2(g)+

O_2(g)

2CO_2(g) +3H₂O_(l)

H= -473.8 kJ

From Eq(v) subtract Eq(iii)

(v) C₂ H_2(s)+ 2H_2(g)+

O_2(g)

2CO_2(g) +3H₂O_(l)

H= -473.8 kJ

(iii)

C₂H_6(g)

O_2(g)

2CO_2(g)

3H₂O_(l)

372.8 kJ

C₂H_2(g)+ 2H_2(g) - C₂H_{6 (g)}

H= -101.0 k cal

C₂H_2(g)+ 2H_2(g)

- C₂H_{6 (g)}

H= -101.0 k cal

H= -101.0 k cal Answer

Q19. Graphite and diamond are two forms of carbon. The enthalpy of combustion of graphite at 25^oC is – 393.51 kJ mol^-1and that of diamond is – 395.41 kJ mol^-1.

What is the enthalpy change of the process? Graphite

Diamond at the same temperature?

Solution: C(graphite)

C(diamond)

H=?

(i) C_(graphite)+O_2(g)

CO_2(g)

H= -393.51 kJmol^-1

(ii) C_{(diamond )} +O_2(g)

CO_2(g)

H= -395.41 kJmol^-1

From Eq(i) Subtract Eq(ii)

(i) C_(graphite)+O_2(g)

CO_2(g)

H= -393.51 kJmol^-1

(ii) C_(diamond)

O_2(g)

CO_2(g)

393.41 kJmol^-____________________________________________________________

C_(graphite)- C_(diamond)

H=1.9 kJ mol^-1

C_(graphite)

C_(diamond)

H=1.9 kJ mol^-1

H=1.9 kJ mol^-1Answer

Q20. What is the meaning of the term enthalpy ionization? If the heat of neutralization of HC1 and NaOH is – 57.3 KJ mol^-1and heat of neutralization of CH₃COOH with NaOH is - 55.2 kJ mol-1 , calculate the enthalpy of ionization of CH₃COOH.

Solution:

CH₃COOH_(l)

CH₃COO^-_(aq)+ H^-_(aq)

H=?

(i)H⁺_(aq)+ C1^-_(aq)+ Na⁺_(aq)+OH^-_(aq)Na⁺_(aq)+H₂O_(l)

H=57.3 kJ mol^-1

(ii)CH₃COOH_(l)+ Na⁺_(aq)+OH^-_(aq)CH₃COO ^–_(aq)+ Na⁺_(aq)+H₂O_(l)

H=55.2kJ mol^-1

Now, from Eq(ii), Subtract Eq(i)

(ii) CH₃COOH_(l)+ Na⁺_(aq)+OH^-_(aq)CH₃COO ^–_(aq)+ Na⁺_(aq)+H₂O_(l)

H=-55.2kJ mol^-1

(i) H⁺_(aq)

C1^-_(aq)

Na⁺_(aq)

OH^-_(aq)Na⁺_(aq)

H₂O_(l)

57.3 kJ mol^-1

CH₃COOH_(l)

CH₃COO^-_(aq)+ H^-_(aq)

H=2.1 kJ mol^-1

CH₃COOH_(l)

CH₃COO^-_(aq)+ H⁺_(aq)

H=2.1 kJ mol^-1

H=2.1 kJ mol^-1Answer

Q21. (a) Explain what is meant by the following terms.

(i) Atomization energy

(ii) Lattice energy

(b) Draw a complete, fully labeled Born-Haber Cycle for the formation of potassium bromide.

Reactions:

H / kJ mol^-1

K_(s)+ Br_2(l) -392

K_(s)

K_(g)+90

K_(g) +420

Br_2(l)

+112

Br_(g)+-342

Solution:

H_f(KBr)=-392 kJ mol^-1

H_{at (K)}=90kJ mol^-1

H_i(K)=420 kJ mol^-1

H_{D/2 (Br}²)=112 kJ mol^-1

H_{e (Br)}= -342 kJ mol^-1

H_l=?

According to Born-Haber cycle

H_f(KBr)=

H_f+

H_at(k)+

H_i(k)+

H_{D/2 (Br}²₎+

H_e(Br)

On putting the values,

-392 kJ mol^-1=

H_l+ 90 kJ mol^-1+ 420 kJ mol^-1+ 112 kJ mol^-1– 342 kJ mol^-1

-392 kJ mol^-1=

H_l + 280 kJ mol^-1

H_l= -392 kJ mol^-1– 280 kJ mol^-1

H_l=-672 kJ mol^-1

Hence, lattice energy of KBr = - 672 kJ mol^-1Answer

Chemistry For F.S.C

Chapter 7th

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