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Chapter 6 Thermodynamics Questions and Answers: NCERT Solutions for Class 11th Chemistry (Partnership Chemistry )
Class 11 Chemistry (Partnership Chemistry ) Chapter 6: Thermodynamics - Questions and Answers of NCERT Book Solutions.
1. Choose the Correct Answer. a Thermodynamic State Function Is a Quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) Whose value depends on temperature only.
Ans: A thermodynamic state function is a quantity whose value is independent of the path. Functions like p, V, T etc. depend only on the state of a system and not on the path.
2. For the Process to Occur Under Adiabatic Conditions, the Correct Condition Is:
(i) Δ T=0 Δ T=0
(ii) Δ p=0 Δ p=0
(iii) q=0q=0
(iv) w=0w=0
Ans: A system is said to be under adiabatic conditions if there is zero exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q=0q=0. Therefore, alternative (iii) is correct.
3. The Enthalpies of All Elements in Their Standard States Are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Ans: The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.
4. Δ U θ Δ U θ of combustion of methane is -XkJ mol-1-XkJ mol-1. The value of Δ Hθ Δ Hθis
(i) = Δ U θ Δ U θ
(ii) > Δ U θ Δ U θ
(iii) < Δ U θ Δ U θ
(iv) = 0
Ans: Since
ΔHθ = ΔUθ + ΔngRT
ΔHθ = ΔUθ + ΔngRT
andΔUθ=−XkJ mol−1
ΔUθ=−XkJ mol−1
,
ΔHθ = (−X) + ΔngRT
ΔHθ = (−X) + ΔngRT
⇒ΔHθ<ΔUθ
5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol-1kJ mol-1 ,-393.5 kJ mol-1kJ mol-1, and -285.8 kJ mol-1kJ mol-1 respectively. Enthalpy of formation of CH4(g)CH4(g) will be
(i) -74.8 kJ mol-1kJ mol-1
(ii) -52.27 kJ mol-1kJ mol-1
(iii) +74.8 kJ mol-1kJ mol-1
(iv) +52.26 kJ mol-1
Ans: According to the question,
(i) CH4(g)+2O2(g)→CO2(g)+2H2O(l);ΔcHΘ=−890.3kJmol−1CH4(g)+2O2(g)→CO2(g)+2H2O(l);ΔcHΘ=−890.3kJmol−1
(ii) C(s)+2O2(g)→CO2(g);ΔcHΘ=−393.5kJmol−1C(s)+2O2(g)→CO2(g);ΔcHΘ=−393.5kJmol−1
(iii) 2H2(g)+O2(g)→2H2O(l);ΔcHΘ=−285.8kJmol−12H2(g)+O2(g)→2H2O(l);ΔcHΘ=−285.8kJmol−1
Thus, the desired equation is the one that represents the formation of CH4(g)CH4(g)that is as follows:
C(s)+2H2(g)→CH4(g);ΔfHCH4=ΔcHc+2ΔcHH2−ΔcHCO2
C(s)+2H2(g)→CH4(g);ΔfHCH4=ΔcHc+2ΔcHH2−ΔcHCO2
Substituting the values in the above formula :
Enthalpy of formationCH4(g)CH4(g)=(−393.5)+2×(−285.8)−(−890.3)=−74.8kJmol−1(−393.5)+2×(−285.8)−(−890.3)=−74.8kJmol−1
Therefore, alternative (i) is correct.
6. A Reaction, A + B →C + D + qA + B →C + D + q is Found to Have a Positive Entropy Change. The Reaction Will Be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Ans: The reaction of aluminum with caustic soda can be represented as:
2Al+2NaOH+2H2O→2NaAlO2+3H2
2Al+2NaOH+2H2O→2NaAlO2+3H2
At STP (273.15 K and 1 atm), 54 g (2 ×× 27 g) of Al gives 3×× 22400 mL of H2H2.
Therefore, 0.15 g Al gives3×22400×0.1554mL3×22400×0.1554mL ofH2H2 i.e., 186.67 mL of H2H2.
At STP,
p1p1 = 1atm
V1V1 = 186.67 mL
T1T1 = 273.15 K
Let the volume of dihydrogen be V2V2 at p2p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2T2 = 20oC20oC = (273.15 + 20) K = 293.15 K.
p1V1T1=p2V2T2
p1V1T1=p2V2T2
V2=p1V1T2p2T1
V2=p1V1T2p2T1
=1×186.67×293.150.987×273.15
=1×186.67×293.150.987×273.15
=202.98mL
=202.98mL
=203mL=203mL
Therefore, 203 mL of dihydrogen will be released.
6. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3dm3 flask at 27oC27oC ?
Ans: For a reaction to be spontaneous, ΔGΔG should be negative ΔG = ΔH − TΔSΔG = ΔH − TΔS
According to the question, for the given reaction,
ΔS =ΔS = positive
ΔH=ΔH= negative (since heat is evolved)
That results in ΔG=ΔG= negative
Therefore, the reaction is spontaneous at any temperature.
Hence, alternative (iv) is correct
7. 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?
Ans: According to the first law of thermodynamics,
ΔU= q + W....(i)
ΔU= q + W....(i)
Where,
ΔUΔU = change in internal energy for a process
q = heat
W = work
Given,
q = + 701 J (Since heat is absorbed)
W = -394 J (Since work is done by the system)
Substituting the values in expression (i), we get
ΔU= 701 J + (−394 J)
ΔU= 701 J + (−394 J)
ΔU= 307 J
ΔU= 307 J
Hence, the change in internal energy for the given process is 307 J.
8.The reaction of cyanamide, NH2CN(s)NH2CN(s) with dioxygen was carried out in a bomb calorimeter and Δ U Δ U was found to be -742.7 KJ mol−1KJ mol−1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.
Ans: Enthalpy change for a reaction (ΔH)(ΔH) is given by the expression,
ΔH = ΔU + ΔngRT
ΔH = ΔU + ΔngRT
Where,
ΔUΔU = change in internal energy
ΔngΔng = change in number of moles
For the given reaction,
Δng=∑ngΔng=∑ng (products) - ∑ng∑ng(reactants)
Δng=(2−1.5)Δng=(2−1.5)moles
Δng=+0.5Δng=+0.5moles
And, ΔUΔU= -742.7 kJ mol−1kJ mol−1
T = 298 K
R = 8.314×10−3kJmol−1K−18.314×10−3kJmol−1K−1
Substituting the values in the expression of ΔH ΔH
ΔH = (−742.7 kJ mol−1) + (+0.5 mol) (298 K)8.314×10−3kJmol−1K−1
ΔH = (−742.7 kJ mol−1) + (+0.5 mol) (298 K)8.314×10−3kJmol−1K−1
ΔH ΔH = -742.7 + 1.2
ΔH = −741.5kJ mol−1
ΔH = −741.5kJ mol−1
9. Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from 35∘C35∘C to 55 ∘ C55 ∘ C. Molar heat capacity of Al is 24J mol-1K-124J mol-1K-1.
Ans: From the expression of heat (q),
q = m. c. ΔT
q = m. c. ΔT
Where,
c = molar heat capacity
m = mass of substance
ΔTΔT = change in temperature
Given,
m = 60 g
c = 24J mol−1K−124J mol−1K−1
DeltaT=(55−35)∘C
DeltaT=(55−35)∘C
ΔT=(328−308)K=20K
ΔT=(328−308)K=20K
Substituting the values in the expression of heat:
q=(6027mol)(24Jmol−1K−1)(20K)
q=(6027mol)(24Jmol−1K−1)(20K)
q = 1066.7 JM
q = 1.07 kJ
10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 ∘ C10.0 ∘ C to ice at -10.0 ∘ C-10.0 ∘ C, Δ fusH = 6.03 KJ mol-1 Δ fusH = 6.03 KJ mol-1 at 0 ∘ C0 ∘ C.
Cp [H2O(l) ] = 75.3 J mol-1 K-1
Cp [H2O(l) ] = 75.3 J mol-1 K-1
Cp [H2O(s) ] = 36.8 J mol-1 K-1
Cp [H2O(s) ] = 36.8 J mol-1 K-1
Ans: Total enthalpy change involved in the transformation is the sum of the following changes:
Energy change involved in the transformation of 1 mol of water at 10.0∘C10.0∘C to 1mol of water at0∘C0∘C.
Energy change involved in the transformation of 1 mol of water at 0∘C0∘Cto 1 mol of ice at 0∘C0∘C.
Energy change involved in the transformation of 1 mol of ice at 0∘C0∘C to 1 mol of ice at 10∘C10∘C.
Total ΔH= Cp [H2O(l)] ΔT+ΔHfreezing+ Cp [H2O(s)] ΔTΔH= Cp [H2O(l)] ΔT+ΔHfreezing+ Cp [H2O(s)] ΔT
ΔH=(75.3 Jmol−1K−1)(0−10)K+(−6.03×103Jmol−1)+(36.8 Jmol−1K−1)(−10−0)K
ΔH=(75.3 Jmol−1K−1)(0−10)K+(−6.03×103Jmol−1)+(36.8 Jmol−1K−1)(−10−0)K
ΔH= −753 J mol−1 − 6030 J mol−1 − 368 J mol−1
ΔH= −753 J mol−1 − 6030 J mol−1 − 368 J mol−1
ΔH= −7151 J mol−1
ΔH= −7151 J mol−1
ΔH= −7.151 kJ mol−1
ΔH= −7.151 kJ mol−1
Hence, the enthalpy change involved in the transformation is −7.151 kJ mol−1 −7.151 kJ mol−1
11. Enthalpy of combustion of carbon to carbon dioxide is −393.5 kJ mol−1−393.5 kJ mol−1 Calculate the heat released upon formation of 35.2 g of CO2CO2 from carbon and dioxygen gas.
Ans: Formation of CO2CO2 from carbon and dioxygen gas can be represented as
C(s)+O2(g) toCO2(g);ΔH=−393.5kJ,mol−1
C(s)+O2(g) toCO2(g);ΔH=−393.5kJ,mol−1
(1mole=44g)
Heat released in the formation of 44 g of CO2CO2 = 393.5 kJmoll−1kJmoll−1
Heat released in the formation of 35.2 g of
CO2=(393.5kJ)×(35.2g)(44g)=314.8kJCO2=(393.5kJ)×(35.2g)(44g)=314.8kJ
So, heat released upon formation of 35.2 g of CO2CO2from carbon and dioxygen gas is 314.8 kJ.
12. Enthalpies of formation of CO (g),CO2(g)CO2(g), N2O(g)N2O(g) and N2O4(g)N2O4(g)are -110 ,-393, 81 kJ and 9.7 kJ mol-1kJ mol-1 respectively. Find the value of Δ rH Δ rHfor the reaction:
N2O4(g)+3CO(g)→N2O(g)+3CO2(g)
N2O4(g)+3CO(g)→N2O(g)+3CO2(g)
Ans: ΔrHΔrH for a reaction is defined as the difference between ΔfHΔfHvalue of products and ΔfHΔfHvalue of reactants.
ΔrHΔrH= ∑ΔfH∑ΔfH (product) - ∑ΔfH∑ΔfH(reactant)
For the given reaction,
N2O4(g)+3CO(g)→N2O(g)+3CO2(g)
N2O4(g)+3CO(g)→N2O(g)+3CO2(g)
ΔrH=[{ΔfH(NO2)+3ΔfH(CO2)}−{ΔfH(N2O)+3ΔfH(CO)}]ΔrH=[{ΔfH(NO2)+3ΔfH(CO2)}−{ΔfH(N2O)+3ΔfH(CO)}]
Substituting the values of ΔfHΔfHfor CO (g),CO2(g)CO2(g), N2O(g)N2O(g) and N2O4(g)N2O4(g)from the question, we get:
ΔrH=[{81kJmol−1+3(−393)kJmol−1}−{9.7kJmol−1+3(−110)kJmol−1}]
ΔrH=[{81kJmol−1+3(−393)kJmol−1}−{9.7kJmol−1+3(−110)kJmol−1}]
ΔrH=−777.7kJmol−1
ΔrH=−777.7kJmol−1
Hence, the value of ΔrHΔrH for the reaction is −777.7kJmol−1
13. Given
N2(g)+3H2(g)→2NH3(g); Δ rH θ =-92.4kJmol-1
N2(g)+3H2(g)→2NH3(g); Δ rH θ =-92.4kJmol-1
What is the standard enthalpy of formation of NH3NH3 gas?
Ans: 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.
Re-writing the given equation for 1 mole of NH3(g)NH3(g)is as follows:
12N2(g)+32H2(g)→2NH3(g)
12N2(g)+32H2(g)→2NH3(g)
Therefore, standard enthalpy of formation of NH3(g)NH3(g)
= 1/2ΔrHθ1/2ΔrHθ
1/2 (−92.4 kJ mol−1)
1/2 (−92.4 kJ mol−1)
−46.2 kJ mol−1
−46.2 kJ mol−1
14. Calculate the standard enthalpy of formation of CH3OH(ℓ)CH3OH(ℓ) from the following data:
CH3OH(l)+32O2(g)→CO2(g)+2H2O(l), Δ rH θ =-726 kJ mol-1
CH3OH(l)+32O2(g)→CO2(g)+2H2O(l), Δ rH θ =-726 kJ mol-1
C(g)+O2(g)→CO2(g); Δ cH θ =-393 kJ mol-1
C(g)+O2(g)→CO2(g); Δ cH θ =-393 kJ mol-1
H2(g)+12O2(g)→H2O(l); Δ fH θ =-286kJ mol-1
H2(g)+12O2(g)→H2O(l); Δ fH θ =-286kJ mol-1
Ans: The reaction that takes place during the formation of CH3OH(ℓ)CH3OH(ℓ)can be written as:
C(s)+2H2O(g)+12O2(g)→CH3OH(ℓ)(1)
C(s)+2H2O(g)+12O2(g)→CH3OH(ℓ)(1)
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 ×× equation (iii) - equation (i)
ΔfHθ[CH3OH(ℓ)]=ΔcHθ+2ΔfHθ[H2O(l)]−ΔrHθ
ΔfHθ[CH3OH(ℓ)]=ΔcHθ+2ΔfHθ[H2O(l)]−ΔrHθ
= (-393kJmol−1kJmol−1 ) + 2(-286 kJmol−1kJmol−1) - (-726 kJmol−1kJmol−1)
= (-393 - 572 + 726) kJmol−1kJmol−1
Therefore, ΔfHθ[CH3OH(ℓ)]=−239 kJmol−1ΔfHθ[CH3OH(ℓ)]=−239 kJmol−1
15. Calculate the Enthalpy Change for the Process
Ans: Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself. The electronegativity of any given element is not constant. It varies according to the element to which it is bonded. It is not a measurable quantity. It is only a relative number.
On the other hand, electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed.
An element has a constant value of the electron gain enthalpy that can be measured experimentally.
16. For an isolated system, Δ U=0 Δ U=0 , what will be Δ S Δ S?
Ans: ΔSΔSwill be positive i.e., greater than zero.
Since for an isolated system, ΔU=0ΔU=0, hence ΔSΔSwill be positive and the reaction will be spontaneous.
17. For the reaction at 298 K, 2A+B→C2A+B→C
Δ H= 400 kJ mol-1 Δ H= 400 kJ mol-1 and Δ S= 0.2 kJ K-1 mol-1 Δ S= 0.2 kJ K-1 mol-1 At what temperature will the reaction become spontaneous considering Δ H Δ H and Δ S Δ S to be constant over the temperature range?
Ans: From the expression,
ΔG= ΔH−TΔS
ΔG= ΔH−TΔS
Assuming the reaction at equilibrium, ΔTΔTfor the reaction would be:
T=(ΔH−ΔG)1ΔS
T=(ΔH−ΔG)1ΔS
=ΔHΔS
=ΔHΔS
(ΔGΔG = 0 at equilibrium)
=400 kJ mol−10.2 kJ K−1 mol−1
=400 kJ mol−10.2 kJ K−1 mol−1
T = 2000 K
For the reaction to be spontaneous, ΔGΔGmust be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.
18. For the reaction, 2Cl(g)→Cl2(g)2Cl(g)→Cl2(g) What are the signs of ∆H and ∆S ?
Ans: ΔHΔH and ΔSΔS are negative.
The given reaction represents the formation of chlorine molecules from chlorine atoms. Here, bond formation is
occurring. Therefore, energy is being released. Hence, ΔHΔHis negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔSΔSis negative for the given reaction.
19. For the reaction
2A(g) + B(g) → 2D(g)2A(g) + B(g) → 2D(g)
Δ U θ Δ U θ = -10.5 kJ and Δ S θ Δ S θ = -44.1JK-1JK-1 .
Calculate Δ G θ Δ G θ for the reaction, and predict whether there action may occur spontaneously.
Ans: For the given reaction,
2A(g) + B(g) → 2D(g)
2A(g) + B(g) → 2D(g)
Δng=2−(3)Δng=2−(3) = -1 mole
Substituting the value of ΔUθΔUθ in the expression ofΔHΔH:M
ΔHθ=ΔUθ+ΔngRT
ΔHθ=ΔUθ+ΔngRT
=(−10.5 kJ)−(−1)(8.314×10−3 kJ K−1 mol−1)(298 K)
=(−10.5 kJ)−(−1)(8.314×10−3 kJ K−1 mol−1)(298 K)
−10.5kJ−2.48kJ
−10.5kJ−2.48kJ
ΔHθ=−12.98kJ
ΔHθ=−12.98kJ
Substituting the values of ΔHθΔHθ and ΔSθΔSθ in the expression of ΔGθΔGθ:
ΔGθ = ΔHθ −TΔSθ
ΔGθ = ΔHθ −TΔSθ
=−12.98kJ−(298K)(−44.1)J K−1
=−12.98kJ−(298K)(−44.1)J K−1
=−12.98kJ+13.14kJ
=−12.98kJ+13.14kJ
ΔGθΔGθ= + 0.16 kJ
Since ΔGθΔGθ for the reaction is positive, the reaction will not occur spontaneously.
20. The equilibrium constant for a reaction is 10. What will be the value of Δ G θ Δ G θ ? R = 8.314 text JK-1 mol-1text JK-1 mol-1 , T = 300 K.
Ans: From the expression,
ΔGθ = −2.303 RTlogKeq
ΔGθ = −2.303 RTlogKeq
ΔGθΔGθ for the reaction,
=(2.303)(8.314 JK−1 mol−1)(300K)log10
=(2.303)(8.314 JK−1 mol−1)(300K)log10
=−5744.14 Jmol−1
=−5744.14 Jmol−1
=−5.744k Jmol−1
21. Comment on the thermodynamic stability of NO(g), given
12NO(g)+12O2(g)→NO2(g): Δ rH θ =90kJmol-1
12NO(g)+12O2(g)→NO2(g): Δ rH θ =90kJmol-1
NO(g)+12O2(g)→O2(g): Δ rH θ =-74 kJ mol-1
NO(g)+12O2(g)→O2(g): Δ rH θ =-74 kJ mol-1
Ans: The positive value of ΔrHΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N2N2 andO2O2). Hence, NO(g) is unstable. The negative value of ΔrHΔrHindicates that heat is evolved during the formation of NO2(g)NO2(g) from NO(g) and O2O2(g). The product, NO2(g)NO2(g)is stabilized with minimum energy.
Hence, unstable NO(g) changes to unstable NO2(g)NO2(g).
22. Calculate the entropy change in surroundings when 1.00 mol of H2O(l)H2O(l) is formed under standard conditions. Δ fH θ Δ fH θ = -286 kJ mol-1kJ mol-1 .
Ans: It is given that 286 kJ mol−1kJ mol−1 of heat is evolved on the formation of 1 mol of H2O(l)H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.
qsurr=+286kJkJ mol−1
qsurr=+286kJkJ mol−1
Entropy change (ΔSsurr)(ΔSsurr) for the surroundings =qsurr7
=qsurr7
=286kJmol−1298K
=286kJmol−1298K
Therefore, (ΔSsurr)(ΔSsurr)= 959.73Jmol−1K−1959.73Jmol−1K−1.
Last Updated on: Mar 28, 2024
