Gujarat Board GSEB Textbook Solutions Class 11 Chemistry Chapter 6 Thermodynamics Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Chemistry Chapter 6 Thermodynamics

GSEB Class 11 Chemistry Thermodynamics Text Book Questions and Answers

Question 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.
Answer:
(ii) State function does not depend upon the path followed or is independent of the path followed.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is :
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0
Answer:
(iii) No heat is allowed to enter or leave the system under adiabatic conditions.

Question 3.
The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
(ii) The enthalpies of all elements in their standard states are zero.

Question 4.
∆U^Θ of combustion of methane is – χ kJ mol^-1. The value of ∆U^Θ is
(i) = ∆U^Θ
(ii) > ∆U^Θ
(iii) < ∆U^Θ
(iv) = 0
Answer:
(iii) ∆U^Θ = ∆U^Θ – 2RT as ∆n = – 2.

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are – 890.3 kJ mol^-1 – 393.5 kJ mol^-1, and – 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be:
(i) – 74.8 kJ mol^-1
(ii) – 52.27 kJ mol^-1
(iii) + 74.8 kJ mol^-1
(iv) + 52.26 kJ mol^-1.
Answer:
Given:
(a) CH₄(g) + O₂(g) → CO₂(g) + 2H₂O(1); ∆H = – 890.3 kJ mol^-1
(b) C(graphite) + O₂(g) → CO₂(g); ∆H = – 393.5 kJ mol^-1
(c) H₂(g) +(frac { 1 }{ 2 })O₂(g) → H₂O(l); ∆H = – 285 kJ mol^-1
Our aim to get the enthalpy of formation of CH₄(g),
i. e., C(graphite) + 2H₂(g) → CH₄(g); ∆H = ?
Add b + 2c- a
C(graphite) + 2H₂(g) + O₂(g) – CH₄(g) – O₂(g) → CO₂(g) +
2H₂O(l) – CO₂(g) – 2H₂O(1); ∆H = – 393.5 + 2(- 285.8) – (- 890.3) kJ mol^-1
C(graphite) + 2H₂(g) → CH₄(g)
∆H = – 393.5 – 571.6+ 890.3 kJ mol^-1
= – 74.8 kJ mol^-1
∴ (i) answer is correct.

Question 6.
A reaction, A + 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
Answer:
The given reaction is
A + B → C + D + q
or A + B → C + D; AH = – q
i. e. The reaction is exothermic
∆G = AH – T∆S
∆S for the reaction is given to be positive
∴ The reaction is spontaneous or possible if ∆G is -ve now ∆H is already -ve
∴ The reaction is possible at any temperature or (iv) Choice is correct.

Question 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?
Answer:
∆U = q + w; where ∆U is change of internal energy.
Given q = 701 J
Since the work is done by the system; w is -ve
∆U = q – w = (701 – 394) J = 307 J
∴ Increase in internal energy = 307 J.

Question 8.
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 reactibn at 298 K.
NH₂CN(S) + (frac { 3 }{ 2 })O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
Solution:
The given reaction is
NH₂CN(s) + (frac { 3 }{ 2 })O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
∆U = – 742.7 kJ mol^-1
Enthalpy change ∆H = AU + ∆n_gRT
where ∆n_gg = change in the no. of moles (gaseous)
= n₂ – n₁
= 2 – (frac { 3 }{ 2 }) = (frac { 1 }{ 2 })
∆H = – 742.7 + (frac { 1 }{ 2 })(0.083)(298)
= – 741.5 kJ
∴ Enthalpy change for the reaction is – 741.5 kJ.

Question 9.
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^-11 K^-1
Solution:
Mass of Al given = 60.0 g
∆T = rise in temp = (55 – 35) = 20.0°C
No. of moles of Al = (frac { 60.0 }{ 27 })
Molar heat capacity of Al = 24 K mol^-1 K^-1
∴ q = Total heat required= n x C x ∆T
= (frac { 60.0 }{ 27 }) x 20.0 x 24.0 J = 1066.7 J
= 1.07 kJ

Question 10.
Calculate the enthalpy change of freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. ∆_fus = 6.03 kJ mol^-11 at 0°C.
Solution:
(i) Heat change required to lower the temperature of water from 10.0°C to 0°C .
∆H₁ = n x CP x ∆T = 1.0 x 75.3 x (- 10) = – 753 J mol^-1

(ii) Heat change required to convert 1 mol. of H₂O(l) at 0°C to H₂O at 0°C
∆H₂ = ∆H_fusion = – 6.03 kJ mol^-1 as heat is given out

(iii) Heat change reqd to change 1 mole of ice from 0°C to -10.0°C
∆H₃ = – 36.8 x 10 x 1 = – 368 J mol^-1
Total heat change = ∆H₁ + ∆H₂ + ∆H₃
= (- 0.753 – 6.03- 0.368) kJ mol^-1
∴ Total enthalpy change = – 7.151 kJ mol^-1
As in each step, heat is evolved, each step will have a negative sign with ∆H.

Question 11.
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.
Solution:

For formation of 44.0 g of CO₂, heat released is = – 393.5 kJ
For formation of 35.2 g of CO₂, heat released is
= (frac { -393.5 }{ 44 }) x 35.2 = – 314.8 kJ
∴ Heat released = – 314.8 kJ or – 315 kJ

Question 12.
Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are – 110, – 393, 81 and 9.7 kJ mol^-1 respectively. Find the value of ∆H for the reaction:
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Solution:
∆_fH(CO) = – 110 kJ mol^-1
∆_fH(CO₂) = – 393 kJ mol^-1
∆_fH(N₂O) = 81.0 kJ mol^-1
∆_fH(N₂O₄) = 9.7 kJ mol^-1
The given reaction is
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g); ∆_rH = ?

Question 13.
Given
N₂(g) + 3H₂(g) → 2NH₃(g); ∆_rH^Θ = – 92.4 kJ mol^-1 What is the standard enthalpy of formation of NH₃ gas?
Solution:
Given : N₂(g) + 3H₂(g) → 2NH₃(g); ∆_rH^Θ = – 92.4 kJ mol^-1
This is the heat evolved for 2 moles of NH₃(g)
∴Heat evolved for 1 mole of NH₃(g) = (frac { -92.4 }{ 2 }) = – 46.2 kJ
Hence ∆_rH^Θ of NH₃ gas = – 46.2 kJ mol^-1

Question 14.
Calculate the standard enthalpy of formation of CH₃ OH(l) from the following data :
CH₃OH(l) + (frac { 3 }{ 2 })O₂(g) → CO₂(g) + 2H₂O(l); ∆_rH^Θ = – 726 kJ mol^-1
C(graphite) + O₂(g) → CO₂(g); ∆_cH^Θ= – 393 kJ mol^-1
H₂(g) + (frac { 1 }{ 2 })O₂(g) → H₂O(l); ∆_fH^Θ = – 286 kJ mol^-1
Solution:
Given data is:
(i) CH₄OH(l) + (frac { 3 }{ 2 })O₂(g) → CO₂(g) + 2H₂O(l); ∆_rH^Θ = – 726 kJ mol^-1

(ii) C(s) + O₂(g) → CO₂(g); ∆_cH^Θ = – 393 kJ mol^-1

(iii) H₂(g) + (frac { 1 }{ 2 }) O₂(g) → H₂O(l); ∆_fH^Θ = – 286 kJ mol^-1
Our aim is to get
C(s) + 2H₂(g) + 2O₂(g) – CH₃OH(l) – (frac { 3 }{ 2 })O₂(g) → CO₂(g) + 2H₂O(l) – CO₂(g) – 2H₂O(l)
or C(s) + 2H₂(g) + (frac { 1 }{ 2 })O₂(g) → CH₃OH(l); ∆_fH^Θ [- 393 + 2(- 286) – (- 726)]
= – 239 kJ mol^-1
Thus the enthalpy of formation of CH₃OH(l) = – 239 kJ mol^-1.

Question 15.
Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4Cl(g) and calculate bond enthalpy change of C-Cl in CCl₄(g)
∆_vapH^Θ(CCl₄) = 30.5 kJ mol^-1,
∆_fH^Θ(CCl₄) = – 135.5 kJ mol^-1
∆_aH^Θ(C) = 715.0 kJ mol^-1 where ∆_aH^Θ is enthalpy of atomisation
∆_aH^Θ(C) = 242 kJ mol^-1.
Solution:
Here we are given
(i) CCl₄(g) → C(g) + 4Cl(g); ∆_vapH^Θ(CCl₄) = 30.5 kJ mol^-1

(ii) C(s) + 2Cl₂ = CCl₄(1); ∆_fH^Θ(CCl₄) = – 135.5 kJ mol^-1

(iii) C(s) → C(g); ∆_gH^Θ(C) = 715.0 kJ mol^-1
where ∆_aH^Θ is the enthalpy of atomisation

(iv) Cl₂(g) → 2Cl(g); ∆_aH^Θ(Cl₂) = 242 kJ mol^-1
Out aim is CCl₄(g) → C(g) + 4(Cl(g); ∆_rH^Θ = ?
Add (i) and (ii), subtract (iii) and (iv) x 2
CCl₄(l) + C(s) + 2Cl₂(g) – C(s) – 2Cl₂(g) → CCl₄(g) + CCl₄(l) – C(g) – 4Cl(g)
or ∆_rH^Θ = 30.5 – 135.5 – 715 – 484 = – 1304 kJ
0 = CCl₄(g) – C(g) + 4 Cl(g); ∆_rH^Θ = – 1304 kJ
or CCl₄(g) → C(g) + 4Cl(g); ∆_rH^Θ = 1304 kJ
There are 4 bonds of C – Cl in CCL₄
∴ Bond enthalpy, of C – Cl bond = (frac { 1304 }{ 4 }) = 326 kJ mol^-1

Question 16.
For an isolated system, ∆U = 0, what will be ∆S?
Solution:

Question 17.
For the reaction at 298 K,
2A + B → C
∆H = 400 kJ mol^-1 and ∆S = 0.2 kJ K^-1 mol^-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.
Solution:
For the reaction 2A + B → C;
∆H = 400 kJ K^-1 mol^-1 and ∆S = 0.2 kJ K^-1 mol^-1
∆G = ∆H-T∆S
For a spontaneous reaction ∆G has to be negative
∆H is 400 kJ mol^-1 (positive)
∴ T∆S has to be > ∆H
or T > (frac { ∆H }{ ∆S }) > (frac { 400 }{ 0.2 })
or T > 2000 K
Thus for the reaction to be spontaneous, temperature should be > 2000 K.

Question 18.
For the reaction,
2Cl(g) → Cl₂ → (g), what are the signs of ∆H and ∆S?
Solution:
2Cl(g) → Cl₂(g) indicates the formation of bonds between the two chlorine atoms to form Cl₂ molecule.
During bond formation energy is released.
∴ ∆H is negative
Randomness decreases as Cl atoms combine to form Cl₂.
∴ ∆S is also negative [ ∵ There is less randomness in molecules than in atoms ]

Question 19.
For the reaction 2A(g) + B(g) → 2D(g)
∆U^Θ = – 10.5 kj and ∆S^Θ = – 44.1 JK^-1
Calculate ∆G^Θ for the reaction, and predict whether the reaction may occur spontaneously.
Solution:
For the given reaction

Since ∆G^Θ is positive
∴ reaction is not spontaneous.

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of ∆G^Θ? R = 8.314 JK^-1 mol^-1 , T = 300 K.
Solution:
The equilibrium given for a reaction is 10
i.e. K = 10
∆G = – 2.303 RT log K
= – 2.303 x 8.314 x 300 log 10 J
= – 2.303 x 8.314 x 300 x 1 [As log 10 = 1]
= – 5744.42 J
= – 5.744 kJ mol^-1.

Question 21.
Comment on the thermodynamic stability of NO(g), given
(frac { 1 }{ 2 }) N₂(g) + (frac { 1 }{ 2 })O₂(g) → NO(g); ∆_rH^Θ = 90 kJ mol^-1
NO(g) + (frac { 1 }{ 2 })O₂(g) → NO₂(g); ∆_rH^Θ = – 74 kJ mol^-1
Solution:
(frac { 1 }{ 2 }) N₂(g) + (frac { 1 }{ 2 }) O₂(g) → NO(g); ∆_rH^Θ = 90 kJ mol^-1
NO(g) + (frac { 1 }{ 2 })O₂(g) → NO₂(g); ∆_rH^Θ = – 74 kJ mol^-1
As is evident from the above, NO(g) is unstable but NO₂(g) is formed since energy is absorbed in the first case and it is released in the second case.

Question 22.
Calculate the entropy change is surrounding when 1.00 mol of H₂O(l) is formed under standard conditions. ∆_fH^Θ = – 286 kJ mol^-1.
Solution:
H₂(g) + O₂→ H₂O(l); ∆H^Θ = – 286 kJ mol^-1
Since heat (enthalpy) is evolved in the above reaction.
∴ It is absorbed by the surroundings .

∴ Entropy change in surroundings = 959.73 JK^-1 mol^-1