RBSE Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

Rajasthan Board RBSE Class 11 Chemistry Chapter 6 Thermodynamics

RBSE Class 11 Chemistry Chapter 6 Text Book Questions

RBSE Class 11 Chemistry Chapter 6 Multiple Choice Questions

Question 1.
During isothermal expansion of ideal gas
(a) Internal energy increases
(b) Enthalpy decreases
(c) Enthalpy is unchanged
(d) Enthalpy decreases and becomes zero
Answer:
(d) Enthalpy decreases and becomes zero

Question 2.
Internal energy is
(a) Partially potential and partially kinetic
(b) Totally kinetic
(c) Totally potential
(d) None of these
Answer:
(a) Partially potential and partially kinetic

Question 3.
In which state, entropy of a substance is maximum?
(a) Solid
(b) Liquid
(c) Gas
(d) Same in all states
Answer:
(c) Gas

Question 4.
The source of a Carnot engine is at 500 K but sink is at 300 K. The efficiency of engine will be :
(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.3
Answer:
(b) 0.4

Question 5.
Which of the following reactions have maximum heat of neutralisation?
(a) NH₄ OH and CH₃ COOH
(b) NH₄ OH and HCl
(c) NaOH and CH₃ COOH
(d) NaOH and HCl
Answer:
(d) NaOH and HCl

RBSE Class 11 Chemistry Chapter 6 Very Short Answer Type Questions

Question 6.
Is entropy of universe constant?
Answer:
No, the entropy of the universe always increases in the course of every spontaneous change. It is the second law of thermodynamics.

Question 7.
Define combined form of first and second law of thermodynamics.
Answer:
In thermodynamics, the combined law of thermodynamics, also called the Gibb’s fundamental equation, is a mathematical summation of the first law of thermodynamics and the second law of thermodynamics summed into a single concise mathematical statement as
dU- TdS + PdV ≤ 0
Where dU is a variation in internal energy.
T is temperature, dS is variation in entropy
P is pressure
dV is variation in volume of a simple working body in which there is neither flows of particles out of the body nor external forces, other than gravity, acting on the body. In theoretical structure, in addition to the obvious inclusion of the first two laws, the combined law incorporates the implications of the zeroth law, via temperature T and the third law, through its use of free energy as related to the calculation of chemical affinities near absolute zero.

Question 8.
What will be the value of internal energy for one mole of an ideal gas ?
Answer:
Internal energy (E_int) of the ideal gas is defined as the,
E_int = 3/2 nRT
Where n is the number of moles, R is the gas constant and T is the temperature.
To find out the internal energy of the ideal gas, substitute 1 mole for n,
250° C for temperature T and 8.31 J/ mol K for the gas constant R in the equation.
E = 3/2 nRT
= 3/2 (1 mol) (8.31 J/ mol. K) (250°C)
= (1.5)( 1 mol) (8.31 J/ mol. K) (250 + 273) K
= (1.5)( 1 mol) (8.31 J/ mol. K) (523) K
= 6519.195 J or 6520 J

Question 9.
Energy is neither absorbed nor released during expansion of ideal gas in vacuum. Why?
Answer:
In an ideal gas, there are no intermolecular forces of attraction. Hence, no energy is required to overcome these forces. Like, when a gas expends against vacuum, work done is zero as Pext – Hence, internal energy of the system does not change, there is no absorption nor release of energy during expansion of ideal gas in vacuum.

Question 10.
When will bond energy be equal to bond dissociation energy?
Answer:
When bond energy is measured in isolated gaseous state, bond energy becomes equal to bond dissociation energy.

Question 11.
Why enthalpy changes whereas, internal energy does not change due to change in heat energy ?
Answer:
Mostly chemical reaction occurs under constant atmospheric pressure. In this situation, heat change occurs on the system is different from the change under constant volume.

Hence, heat energy change occurs under constant pressure but internal energy does not change due to this heat energy change but enthalpy changes.

Question 12.
How non-spontaneous process be converted to spontaneous process?
Answer:
AG = ∆H + TAS; when the signs for ∆H and ∆S are both the same, then temperature determines that the process is spontaneous. When ∆H is positive (unfavourable) and ∆S is positive (favourable), high temperatures are needed so the favourable ∆S term dominates making the process spontaneous (∆G < 0). When ∆H is negative (favourable) and ∆S is negative (unfavourable), low temperatures are needed so the favourable ∆H term dominates making the process spontaneous (∆G < 0).

Question 13.
Differentiate between isothermal and adiabatic process?
Answer:
Process in which temperature remains constant throughout the process is called isothermal process. When such a process occurs, heat can flow from the system to the surrounding and vice versa in order to keep the temperature constant. In an isothermal process, dT= 0.

When a process is carried out in such a way that no heat can flow from the system to the surroundings or vice versa i.e., system is completely insulated from the surroundings. In such a system, temperature of the system always changes. For an adiabatic process dQ = 0

Question 14.
Internal energy change is state function but work is not a state function. Why?
Answer:
Internal energy is state function because it describes quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state. In contrast, work is process quantity or path function, because its value depends on the specific transition (or path) between two equilibrium states.

Question 15.
Which of the two, diamond or graphite, has more enthalpy?
Answer:
Graphite has greater enthalpy since it is loosely packed.

Question 16.
What is the relation between E and H?
Answer:
E = Internal energy, H = Enthalpy,
∆H = ∆E + P∆V or ∆H = ∆E + ∆n_gRT
here ∆E = Change in internal energy
∆H = Enthalpy change,
∆ n_g = Number of moles of gaseous products – Number of moles of geseous reactants Where R = Gas constant
T = Temperature

Question 17.
In which state a substance has maximum entropy?
Answer:
Entropy by definition is the degree of randomness in a system. Out of three states of matter: Solid, Liquid and Gas, the gaseous particles move freely and therefore, the degree of randomness is the highest. Solids have highly ordered arrangement and has low entropy. The entropy of liquids is more than solids because molecules of liquid shows less ordered arrangement than solids. The molecules of gas are free to move and not constrained to be adjacent to each other. Therefore, S_gas >> S_liquid > S_solid. In gaseous state, entropy is maximum.

Question 18.
Give examples of endothermic and exothermic reactions.
Answer:
Reaction in which heat is given out along with the products is called exothermic reaction.
Example:
1. Burning of methane gas releases a large amount of energy. Hence it is an exothermic reaction.
2. Reaction of calcium oxide and water
Reactions that absorb energy or require energy in order to proceed are called endothermic reactions.
Example :
(a) When ammonium chloride is dissolved in water in a test tube, the test tube becomes cold.
(b) Reaction of barium hydroxide with ammonium chloride.

Question 19.
What is the heat of neutralisation of NH₄ OH and HCl?
Answer:
Enthalpy of neutralization of ammonium hydroxide (weak base) and hydrochloric acid (strong acid) is -51.5 kJ.

Question 20.
Write Gibb’s Helmholtz Equation.
Answer:
Gibb’s Helmholtz Equation can be given as :
∆G = ∆H – T∆S
Where; ∆G = Free energy change
∆H = Enthalpy change
∆S = Entropy change
T = Absolute temperature

RBSE Class 11 Chemistry Chapter 6 Short Answer Type Questions

Question 21.
What will be the efficiency of Carnot engine if its source is at 500 K and sink at 300 K?
Answer:
Efficiency, η = (T₂-T₁)/T₂
T₂ = 500 K (temperature of the source)
T₁ = 300 K (temperature of the sink)
Therefore,
η = (T₂ – T₂])/r₂
= (500 – 300) / 500
= 200/500 .
= 2/5
= 0.4

Question 22.
What are the necessary conditions for adiabatic process?
Answer:
The necessary conditions for adiabatic process are—
(a) There should not be any exchange of heat between the system and its surrounding. All walls of the container and the piston must be perfectly insulating.
(b) The system should be compressed or allowed to expand suddenly so that there is no time for the exchange of heat between the system and its surroundings.

Question 23.
Heat of neutralisation of strong acid and strong base is constant. Why?
Answer:
The enthalpy of neutralization of all strong acids and strong bases is always constant, i.e., -57.1 kJ. The explanation to this generalization can be provided on the basis of Arrhenius theory of ionization. The strong acids and strong bases are almost completely ionized in dilute aqueous solutions. The neutralization of strong acid and strong bases simply involve the combination of H⁺ ions (from acid) and OH^– ions (from base) to form water molecules. It has been found experimentally that when 1 mole of water is formed by the neutralization of 1 mole of H⁺ ions and 1 mol of OH^– in aqueous solutions, 57.1 kJ of energy is released.

Question 24.
Though dissolution of NaCl in water is an endothermic reaction, it is soluble in water. Explain.
Answer:
The enthalpy of solution of NaCl in water (that is, the energy change associated with the dissolution of sodium chloride crystals in water) at standard conditions is very slightly positive, it is an endothermic process. A positive enthalpy change indicates that the system acquires heat from the surroundings in order for the reaction (dissolution) to proceed forward. When something dissolves in water, some of these O-H bonds are broken. This requires heat energy. Entropy increases when solute dissolves in a solvent. Also, if H is positive, G will be negative and process is spontaneous.

Question 25.
Given N_2(g) + 3H_2(g) ➝ 2NH_3(g);
∆H° = -92.4 kJ mol^-1
What is the standard enthalpy of formation of ammonia gas?
Answer:
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
For 1 mole of NH₃(g),

Question 26.
Enthalpy of combustion of carbon to CO₂ is -393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Answer:
Formation of CO₂ from carbon and dioxygen gas is represented as:
2C(s) + 2O₂ (g) ➝ 2CO₂ ∆_fH = -393.5 kJ mol^-1
(1 mole = 44 g)
Heat released on formation of 44g CO₂ = -393.5 kJ mol
Therefore, Heat released on formation of 35.2 g CO₂ = 393.5 × 35.2 / 44 = 3.14 kJ mol^-1

Question 27.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Answer:
According to the first law of thermodynamics,
∆U = q + W …(i)
Where,
∆U = change in internal energy
q = heat = 701 J (Heat is absorbed so, it is positive) (given)
W = work done = (-394 J) (Work is done by the system so, it is negative) given
Substituting the values in expression (i), we get
∆U = 701 + (-394)
∆U = 307 J
Hence, the change in internal energy for the given process is 307 J.

Question 28.
For an isolated process, U = 0, what will be the value of S for this?
Answer:
Since ∆U = 0, there is no exchange of energy with the surroundings. In the process, AS will be positive and the reaction will be spontaneous. That means AS is greater than zero.

Question 29.
In a process, 5 kJ of heat is absorbed by a system and 1 kJ of work is done by the system. What is the change in internal energy for the process?
Answer:
According to first law of thermodynamics,
U = Q+W
Where U = change in internal energy
Q = heat
W = work
According to this, there are two kinds of processes that can lead to a change in the internal energy of the system – they are heat and work.
Q = + 5kJ
W = -1 kJ

Question 30.
For the reaction, 2Cl (g) ➝ Cl₂ (g), what are the signs of H and S?
Answer:
The given reaction represents the formation of chlorine molecule form two chlorine atoms. Here bond formation is taking place. Therefore energy is being released. Hence H is negative. Also two moles of atoms have more randomness than one mole of molecule. Since spontaneity is decreased; S is negative for the given reaction.

Question 31.
Give an example of isolated system.
Answer:
Thermos flask is the best example of isolated system. A thermos flask is used to keep things either cold or hot. Thus, a thermos does not allow energy for transfer. Additionally, the thermos, like any other closed container, does not allow matter transfer because it has a lid that does not allow anything to enter or leave the container. An isolated system does not exchange energy or matter with its surroundings.

Question 32.
In which process of system, temperature decreases?
Answer:
Adiabatic process is the process in which change in pressure, volume and temperature takes place without any heat entering or leaving the system is called adiabatic process. So the total heat of the system, undergoing an adiabatic process always remains constant. In an adiabatic expansion since no heat is supplied from outside, therefore the energy required for the expansion of the gas is taken from the gas itself. This signifies that, the internal energy of an ideal gas undergoing in an adiabatic expansion decreases, and because the internal energy of an ideal gas depends only on the temperature, therefore its temperature must decreases. That is why the temperature of a gas drops in an adiabatic expansion.

Question 33.
What will be the sign of G for melting of ice at 267 K and 276 K? (Melting point of ice = 272 K).
Answer:
At 267 K, the process is non-spontaneous ∆G = + ve While at 276 K, it become spontaneous ∆G = -ve

Question 34.
At 25°C, the heat required to dissociate 4 g hydrogen gas into free gaseous atoms is 208 kCal. Then what will be the bond energy of H-H bond?
Answer:
Bond enthalpy is defined as amount of energy required to break one mole of bonds of a particular type between atoms in gaseous state.
For 4 g = 2 mole of H₂
E = 208 kcal
as bond energy is always for 1 mole thus bond energy of H-H will be 208/2= 104 kcal

Question 35.
Give relation between ∆H and ∆E.
Answer:
In only pressure-volume work is done at constant pressure, then according to first law of thermodynamics,
∆E = q_p – P∆V
q_p = ∆E + P∆V
q_p = (E₂ – E₁) – (PV₂ + PV₁)
For constant pressure, q is represented by q_p.
Here, E, P and V, all are state functions, E + PV is also a state function. This is known as enthalpy or heat content.
Enthalpy of a system is equal to sum of its internal energy and pressure – volume energy. The enthalpy change for initial and final state are:
H₂ – H₁ = (E₂ -E₁ + PV₂ – PV₁)
∆H = ∆E + P∆V = q_p
Thus, heat absorbed by the system at constant pressure is equal to its enthalpy change. H is a state function. Hence, q_p is also a state function.
If V₂ = V₁ then, ∆H = ∆E
In gaseous state, there is an appreciable distance between H and E, then according to ideal gas equation,
PV_A = n_ART
PV_B = n_BRT
PV_B – PV_A – n_BRT – n_ART = (n_B – n_A )RT
P(V_B – V_A) = (n_B – n_A)RT
P∆V = n_gRT
Here, ∆n_g refers to the number of moles of gaseous products minus the number of moles of gaseous reactants. Substituting the value of P∆V from equation, we get
∆H = ∆E + ∆n_gRT

RBSE Class 11 Chemistry Chapter 6 Long Answer Type Questions

Question 36.
Explain first law of Thermodynamics and give its limitations.
Answer:
This law is given by Helmholtz and Robert Mayer. This law is also known as “Law of conservation of energy”. According to this law, “Energy can neither be. created nor destroyed although it can be converted from one form into another. OR “Total energy of the universe remains constant”, or “Energy of an isolated system is constant.” The internal energy of the system can be changed in two ways :

1. By allowing the heat to flow into the system or out of the system,
2. By doing work on the system or by the system. Consider a system whose internal energy is U₁ Now if the system absorbs q amount of heat, then the internal energy of the system increases and becomes U₁ + q.

If work is done on the system then its internal energy further increases and becomes U₂. Thus,
U₂ = U₁ + q + w
or U₂ – U₁ = q + w
or ∆U = q + w
The above relationship is mathematical statement of first law of thermodynamics

Limitations:

• The first law of thermodynamics merely indicates that in any process there is a transformation between the various forms of energy involved in the process but provides no information regarding the feasibility of such transformation.
• First law does not provide any information regarding the direction. A process will take place whether it is a spontaneous or a non-spontaneous process.
• Practically it is not possible to convert the heat energy into an equivalent amount of work.
• No restriction on the direction of the flow of heat : The first law establishes definite relationship between the heat absorbed and the work performed by a system. The first law does not indicate whether heat can flow from a cold end to a hot end or not. For example, we cannot extract heat from the ice by cooling it to a low temperature. Some external work has to be done.

Question 37.
Give Hess’s Law of constant Heat summetion and write its uses.
Answer:
This law is given by a Russian Chemist, G. H. Hess, in 1840. The law is known after his name as Hess’s Law. It states that:

“If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature”.

In general, if enthalpy of an overall reaction A ➝ B along one route is ∆H and H₁, H₂, H₃……

Representing enthalpies of reactions leading to same product, B along another route, if the reaction is taking place in three steps, then we have
∆H = H₁ + H₂ + H₃
It can be represented as : ,

Example : Combustion of carbon dioxide can be done in two ways :
Path I: C_(s) + O_2(g) ➝ CO_2(g) ∆H = -393.5 kJ

Thus, the heat of reaction in single step and that in two steps is same. This proves Hess’s Law.

Uses of Hess’s Law of constant Heat:

• Determination of transition : It helps in the determination of enthalpy of transition during allotropic modification.
• Determination of enthalpy of formation : It helps in the determination of enthalpy of formation which cannot be determined experimentally e.g. it is not possible to calculate enthalpy of formation of CO experimentally, but can be calculated by Hess’s law.
• Bond Energy : It may be defined as, “The quantity of heat evolved when a bond is formed between two free atoms in a gaseous state to form a molecular product in a gaseous state”. It is also known as enthalpy of formation of the bond. It may also be defined as, “The average quantity of heat required to break (dissociate) bonds of that type present in one mole of the compound”.

Question 38.
Explain second law of thermodynamics with the help of efficiency of Carnot engine. How free energy is a measure of spontaneity of a reaction?
Answer:
According to second law of thermodynamics, :
“All spontaneous processes are thermodynamically irreversible”.

Or

“Without the help of an external agency, a spontaneous process cannot be reversed”.

The change of heat into mechanical work is done by Carnot heat engine. In real engines, it is not possible to convert heat energy completely into work because heat is lost in friction, as radiation etc. Hence, Carnot calculated efficiency and reversible work by ideal engine.

In this engine, at standard conditions, source of heat (at T₂K) and after work, to store rest of heat,a sink (at T₂K) is required. As a source of heat, working substance, ideal gas is taken in such a cylinder whose walls are insulators but bottom is conducting. A frictionless piston is required in insulators. The Carnot cycle when acting as a heat engine consists of the following four steps: two isothermal and two adiabatic processes. Four stages are A, B, C and D, In whole Carnot cycle, at temperature T₁, q₁ heat is absorbed from the source and w work is done and rest part q₂ is given in the sink at temperature T₂. The work done by Carnot engine is given by the following formula,
w = q₁ – q₂ = q₁(T₂ -T₁)/T₁

In this way, in a cyclic process, heat absorbed by the source cannot be converted to equivalent amount of work without any loss of heat. The efficiency of Carnot Engine can be calculated by this. “That absorbed heat which can be converted to work is called efficiency of Carnot engine.
n = w/q₁ = (T₂ -T₁)/T₁.
(T₂ – T₁)/T₁ < 1

Hence, efficiency of Carnot engine is always less than one. This proves second law of thermodynamics, according to it, “heat can neither be converted to equivalent amount of work”.

The efficiency of Carnot engine explains the entropy concept of second law of thermodynamics.
Free energy (∆G) gives a criteria for spontaneity at constant pressure and temperature,

• If ∆G is negative (<0), the process is spontaneous.
• If ∆G is positive (>0), the process is non spontaneous.

Question 39.
Explain the following:
1. Enthalpy of formation
2. Phase Transition enthalpy
3. Entropy
4. Enthalpy of solution
Answer:
(1) Enthalpy of Formation : It is a form of standard enthalpy of reaction in which one mole of compound is formed from its elements. The standard enthalpy change for the formation of one mole of a compound from its elements. The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation is called Standard Molar Enthalpy of Formation. Its symbol is ∆H°.
N_2(g) + O_2(g) ➝ 2NO_(l) ; ∆_fH° =+180 kJ

(ii) Phase Transition Enthalpy: The change in phase of a substance takes place at a very small rate. Thus, it is not possible to measure the heat change for phase transition experimentally. But, phase transition enthalpy can be calculated with the help of Hess’s Law. The heat of transition of diamond to graphite can be calculated from the heat of combustion data. For diamond and graphite, which is -395.4 kJ and -393.5 kJ respectively.
The thermochemical equations showing the combustion reaction of diamond and graphite are:
C_(diamond) + O_2(g) ➝ CO_2(g) ∆H = -395.4 kJ …(i)
C_(graphite) + O_2(g) ➝ CO_2(g) ∆H = -393.4 kJ …(ii)
The conversion that is required is :
C_(diamond) ➝ C_(graphite) ∆H = ?
This can be obtained by subtracting the second equation from the first one.

(3) Entropy : Entropy is a measure of the molecular disorder, or randomness, of a system. The concept of entropy provides deep insight into the direction of spontaneous change for many everyday phenomena. Its introduction by the German physicist Rudolf Clausius in 1850 is a highlight of 19th-century physics.

There is increase in entropy during melting of a solid and vaporisation of a liquid, in solids, there is a definite crystal lattice whereas in gases and liquids, this arrangement is very less. Therefore, with increase in entropy in gases and liquids, disorderlness also increases. Thus, entropy increases in spontaneous processes. Therefore, entropy of a system is a measure of orderlness. In this way, entropy changes related to molecules take place.

• More the number of molecules, more will be the entropy.
• More the number of energy levels available to molecules on heating, more will be the entropy.

4. Enthalpy of solution : Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions (or solute molecules) are negligible.

When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice. These are now more free in solution. But solvation of these ions (hydration in case solvent is water) also occurs at the same time. Thus, in dissolution of ionic solid, contributing of both the energies is necessary.

RBSE Class 11 Chemistry Chapter 6 Numerical Problems:

Question 40.
For a reaction, X—Y at 298 K, what is standard entropy when equilibrium constant is 1.8 × 10^-7?
Answer:
Generic reaction
Reactants ➝ Products
q_reaction = 1.8 × 10^-7 kJ
Then, for the surroundings, one has,
∆S_{(surroundings)} = 0.8 × 10^-7)/ 298 = 33.8 J/K mol .

Question 41.
At 60°C, dinitrogen tetroxide dissociates to 50%. Calculate standard free energy change at 1 atm and same temperature.
Answer:
Dissociation of N²O⁴ proceed as
N²O⁴ ➝ 2NO²
Mole fraction, dissociation is 50% is 0.5
= (1- 0.5)/(1 + 0.5) = 0.5/1.5 = 0.33
Therefore, P = 0.33 × 1 atm. = 0.33
Similarly for NO₂ = 2 × 0.5/(1.0.5) = 1/1.5 = p = 0.66
Equilibrium constant Kp = P(NO₂)²/P(N²O⁴)
= (0.66)²/(0.33) = 1.33
G = -RT In Kp
= – 8.314 × 333 × 2.303 × (0.1239)
= -768.3 kJ/mol

Question 42.
The volume of a gas at STP is 2 L. 300 J heat is given to it, as a result of which volume changes to 2.5 L at 1 atm. Calculate the change in internal energy of the system.
Answer:
∆U = q + w
W = -P∆V
W = -1(2.5 – 2) = 0.5 L atm
= -0.5 × 101.3 J
= – 50.65 J
Q = 330J
∆U = 300J + (-50.65J)
= 249.35 J

Question 43.
At 0°C and 1 atm, heat absorbed by the system by melting of one mole of ice is 6.05 KJ/mol. The molar volume of ice and water are 0.0196 L and 0.0180 L respectively. Calculate ∆H and ∆U.
Answer:
Heat is absorbed here at constant pressure so ∆H = 1440 cal
∆U can be calculated by : ∆U – ∆H – P∆V
P = 1 atm, ∆V = 0.196L – 0.180L
= 0.0016 L
P∆V = 1 × 0.0016 = 0.0016 L atm
= 0.0016 × 24.217 calories
1 l atm = 24.217 calories
So ∆U = 1440cal – 0.0016 × 24.217 cal
= 1439.96 calories
So ∆H = ∆U

Question 44.
Calculate the entropy change in surroundings when 1.00 mol of water is formed under standard conditions.
Answer:
Formation of water is an exothermic process So, H = -286 kJ/mol
Hence, heat is transferred to the surroundings
Thus Q_srr = + 286 kJ/mol
Now, S = q/t
= (286 * 1000)/298
= 959 JK^-1 mol^-1

Question 45.
The reaction of cyanamide, NH²CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be -742.7 kJ mol^-1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN(g) + 3/2O₂(g) ➝ N₂(g) + CO₂(g) + H₂O(l)
Answer:
Enthalpy change for a reaction (∆H) is given by the expression,
∆H = ∆U + ∆n_gRT
Where,
∆U = change in internal energy
∆n_g = change in number of moles
For the given reaction,
∆n_g = Σn_g (products) – Σ n_g (reactants)
= (2 – 2.5) moles
∆U = – 0.5 moles
And,
∆U = -742.7 kJ mol^-1
T = 298 K
R = 8.314 × 10^-3 kJ mol^-1
Substituting the values in the expression of ∆H :
∆H = (-742.7 kJ mol^-1) + (- 0.5 mol) (298 K) (8.314 × 10^-3 mol^-1 K^-1) = -742.7 – 1.2.
∆H = – 743.9 kJ mol^-1

Question 46.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Answer:
The expression of heat (q) is written as follows :
q = m.c. ∆T ….(i)
Where, c = molar heat capacity = 24 J mol^-1 K^-1
m = mass of substance = 60 g = 60/27 mol
∆T = change in temperature = 20°C (35°C to 55°C)
Substituting the values in equation (i), we get
q = m.c.∆T
q = (60/27) × 24 × 20
= 1066.7 J q
= 1.07 kJ

Question 47.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C.
∆_fusH = 6.03 kJ mol^-1 at 0^oC
C_p (H₂O_(l)] = 75.3 J mol^-1K^-1
C_p [H₂O_(s)] = 36.8 J mol^-1K^-1
Answer:
Total enthalpy change involved in the transformation is the sum of the following changes and can be represented as :

According to Hess’ law:
(∆H) = (∆H₁) + (∆H₂) + (∆H₃)
(∆H₁) = (75.3 J mol^-1K^-1) (0 – 10) K = – 753 J mol^-1
(∆H₂) (solidification) = (-6.03 × 10³ J mol^-1) or 60630 J mol^-1
(∆H₃) = (36.8 J mol^-1K^-1) (-10 – 0) K = – 368 J mol^-1
Substituting the values we get,
= -753 J mol^-1 – 6030 J mol^-1 – 368 J mol^-1
= -56451 J mol^-1
= -5.645 kJ mol^-1
Hence, the enthalpy change involved in the transformation is – 5.645 kJ mol^-1.

Question 48.
Enthalpies of formation of CO(g), CO² (g), NaO (g) andN²O⁴ (g) are -110, -393.81 and 9.7 kJ mol^-1 respectively. Find the value of ∆H for the reaction.
N²O^3(g) + 3CO^(g) ➝ N²O^(g) + 3CO^(g)
Answer:
∆rH for a reaction is calculated as the difference between ∆_fH (Heat of formation) value of products and ∆_fH (Heat of formation) value of reactants. For the given reaction,
N₂O_4(g) + 3CO_(g) ➝ N₂O_(g) + 3CO_2(g)

Question 49.
For the reaction
2A_(g) + B_(g) ➝ 2D_(g)
∆U° = -10.5 kJ and ∆S°= – 44.1 JK_-1.
Calculate ∆G° for the reaction and predict whether the reaction may occur spontaneously.
Answer:
For the given reaction,

When we substitute the values of ∆H^θ and ∆S^θ in the expression of ∆G^θ, then according to Gibbs’ Helmholtz equation;
∆G^θ = ∆H^θ – T∆S^θ
= -12.98 kJ – (298 K) (-44.1 J K^-1)
=-12.98 kJ-(-13.14 kJ)
∆G^θ = 0.16 kJ
Since ∆G^θ for the reaction is positive, the reaction will not occur spontaneously.

RBSE Solutions for Class 11 Chemistry