MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

In this article, we will share MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics Pdf, These solutions are solved by subject experts from the latest edition books.

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

MP Board Class 11 Chemistry Thermodynamics Textbook 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.
Solution.
As per the definition, the correct answer is (ii).

Question 2.
For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0
Solution.
The correct answer is (iii) because in an adiabatic process no heat change occurs.

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

Question 4.
ΔU° of combustion of methane is – X kJ mol-1. The value of ΔH° is
(i) = ΔU°
(ii) > ΔU°
(iii) < ΔU°
(iv) 0
Solution.
The correct answer is (iii). The combustion of methane is represented as
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Here, = Δn(g) = 1 – 3 = – 2
ΔH° < ΔU°

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
Solution.
The correct answer is (i) ΔfH° (CH4) = 2 ΔcH° (H2) + ΔcH° (C) – ΔcH° (CH4)
= 2 (- 285.8) + (- 393.5) – (- 890.3)
= – 74.8 kJ mol-1

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

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
Solution.
The correct answer is (iv). Since the reaction is exothermic (ΔH < 0) and has ΔS > 0, therefore, it is feasible at all temperatures because both the factors favour spontaneity.

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?
Solution.
Heat absorbed by the system, (q) = + 701 J
Work done by the system (W) = – 304 J
Change in internal energy (AU) = q + w
= 701 – 394
= 307 J.

Question 8.
The reaction of cyanamide, NH2CN(s) with oxygen was affected in a bomb calorimeter and ΔU was found to be – 742.7 kJ mol-1 of cyanamide at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH2CN(S) + 3/2O2(g)→ N2(g) + CO2(g) + H2O(l)
Solution.
ΔU = – 742.7 kJ mol-1; Δn(g) = 2 – 3/2 = + 0.5
R = 8.314 × 10-3 kJ K-1 mol-1; T = 298.15 K
According to the relation,
ΔH = ΔU + Δn(g) RT
or ΔH = – 742.7 (kJ) + 0.5 (mol) × 8.314 × 10-3 (kJ K-1 mol-1) × 298.15 (K)
= – 742.7 kJ + 1.239 kJ
= – 741.5 kJ.

Question 9.
Calculate the number of kJ necessary to raise the temperature of 60 g of aluminium from 35 to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.
Solution.
Moles of Al =\(\frac{60(\mathrm{~g})}{27\left(\mathrm{~g} \mathrm{~mol}^{-1}\right)}\) = 2.22 mol
Molar heat capacity (Cm) = 24.0 J mol-1 K-1
ΔT = 55° – 35° = 20° or 20 K
q = Cm × n × ΔT
= 24.0 (J mol-1 K-1) × 2.22 (mol) × 20 (K)
= 1065.6 J
= 1.067 kJ.

Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at – 10.0°C to ice at – 10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C. Cp [H2O(l) ]= 75.3 J mol-1m K-1; Cp [H2O(l) ] = 36.8 J mol-1m K-1
MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics 1
Heat released in Step I
q = – nCpΔT
= – 1 × 75.3 × 10 = – 753 J
Heat released in Step II
q = -nΔHfreezing
= – 1 ×6.03 = – 6.03 kJ or – 6030 J
Heat released in Step III
q = – nCpΔT
= – 1 × 36.8 × 10 = – 368 J
Total heat released q = – 753 – 6030 – 368
= – 7151 J.

Question 11.
Enthalpy of combustion of carbon to carbon dioxide is – 393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of C02 from carbon to oxygen gas.
Solution.
The combustion equation is:
C(s) + O2(g) → CO2(g)(44 g) ; ΔcH = – 393.5 kJ mol-1
Heat released in the formation of 44 g of CO2 = – 393.5 kJ
Heat released in the formation of 35.2 g of CO2
= \(\frac{-393.5(\mathrm{~kJ}) \times 35.2(\mathrm{~g})}{44(\mathrm{~g})}\)
= – 314.8 kJ.

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

Question 12.
The enthalpy of formation of CO (g), CO2(g), N2O(g), N2O4 (g) are – 110, – 393,81 and 9.7 kJ mol-1 respectively. Find the value of ΔrH° for the reaction,
N2O4(g) + 3CO (g) → N2O(g) + 3CO2 (g).
Solution.
ΔrH° for the given reaction is given as
ΔrH° = Σ ΔfH° (Products) – ΣΔfH°(reactants)
= ΔfH° (CO2) + ΔfH° (N2O) – 3 ΔfH° (CO) – ΔfH° (N2O4)
3(-393) + 81.0-31- 110J-97
= – 1179 + 81.0 + 330-9.7
= – 777.7 kJ mol-1.

Question 13.
Given: N2(g) + 3H2(g) → 2NH3(g); ΔrH°= – 92.4 kJ mol-1. What is the standard enthalpy of formation of NH3 gas?
Solution.
ΔfH° of NH3(g) = \(\frac{\Delta_{r} \mathrm{H}^{\circ}}{2}\)
= –\(\left(\frac{92.4}{2}\right)\) = -46.2 KJ mol-1.

Question 14.
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
(i) CH3OH(l) + 3/2O2(g) →CO2(g) + 2H2O(1); ΔrH°= – 726 kJ mol-1
(ii) C(s) + O2(g) → CO2(g); ΔrH°= – 393 kJ mol-1
(iii) H2(g) + 1/2O2(g) > H2O(l); ΔrH° = – 286 kJ mol-1
Solution.
The equation we require is:
C(s) + 2H2(g) + 1/2O2(g) → CH3OH(l); ΔfH°= ±?
Multiply eqn. (iii) by 2 and add to eqn. (ii)
C(s) + 2H2(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = – (393 + 522) = – 965 kJ mol-1
Subtract eqn. (i) from eqn. (iv)
MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics 2
or C(s) + 2H2 + \(\frac{1}{2}\)O2(g) →
ΔH = – 239 kJ mol-1.

Question 15.
Calculate the enthalpy change for the process CCl4 (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C-Cl in CCl4 (g) Given: Δvap H° (CCl4) = 30.5 kJ mol-1 ΔfH° (CCl4) = – 135.5 kJ mol-1
ΔaH° (C) = 715.0 kJ mol-1 where ΔaH° is enthalpy of atomisation 3 ΔaH° (Cl2) = 242 kJ mol-1
Solution.
The thermochemical equations for given data are as under:
(i) CCl4 (l) → CCl4(g); ΔH1 = 30.5 kJ mol-1
(ii) C (s) + 2 Cl2 (g) → CCl4 (l); ΔH2 = – 135.5 kJ mol-1
(iii) C (s) → C (g); ΔH3 = 715.0 kJ mol-1
(iv) Cl2 (g) → 2 Cl (g); ΔH4 = 242 kJ mol-1
Required equation is:
CCl4 (g) → C (g) + 4 Cl (g), ΔH =?

Now carry out the following operations to get the required equation.
Eqn. (iii) + 2 × Eqn. (iv) – Eqn. (i) – Eqn. (ii)
∴ ΔH = ΔH3 + 2ΔH4 – ΔH1 – ΔH2
or ΔH = 715.0 + 2 (242) – 30.5 – (- 135.5) kJ mol-1
= 1304 kJ mol-1
Bond enthalpy of C-Cl in CCl4 (average value)
= \(\frac{1304}{4}\)
= 326 kJ mol-1.

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

Question 16.
For an isolated system, ΔU = 0, what will be ΔS?
Solution.
For an isolated system with ΔU = 0, the spontaneous change will occur if ΔS >0. For example, in an isolated system involving intermixing of gases ΔU = 0 but. ΔS > 0 because of increase of disorder.

Question 17.
For a reaction at 298 K 2 A + B →C
ΔH = 400 kJ mol1 and ΔS = 2.0 kJ K1 mol-1.
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.
Solution.
According to Gibb’s Helmholtz equation:
ΔG = ΔH – TΔS

At equilibrium, ΔG = 0 ΔH = TΔS or T = \(\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}\)
T = \(\frac{400\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)}{2\left(\mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)}\) = 200 K
Thus, reaction will be in a state of equilibrium at 200 K and will become spontaneous as the temperature becomes > 200 K.

Question 18.
For the reaction; 2Cl (g) → Cl2(g); what will be the signs of ΔH and ΔS?
Solution.
ΔH: negative (- ve) because energy is released in bond formation . ΔS: negative (- ve) because entropy decreases when atoms combine to form molecules.

Question 19.
For a reaction; 2A (g) + B(g) → 2D(g)
ΔU298 = -10.5 kJ and ΔS° = – 44.1 J
Calculate ΔG0298 for the reaction and predict whether the reaction is spontaneous or not.
Solution.
ΔH° = ΔU° + Δn(g)RT
Here, ΔU° = – 10.5 kJ; Δn(g) = 2 – 3 = – 1 mol;
T = 298 K
R = 8.314 × 10-3 kJ K-1 mob-1
∴ ΔH° = – 10.5 (kJ) + [- l(mol) × 8.314 x 10-3 (kJ K-1 mol-1) × 298 (K)]
= – 10.5 kJ – 2.478 kJ = – 12.978 kJ

According to Gibb’s Helmholtz equation:
ΔG° = ΔH° – TΔS°
ΔG° = – 12.978 (kJ) – 298 (K) x – 0.044 (kJ K1)
= – 12.978 + 13.112
= + 0.134 kJ
Since ΔG° is positive, the reaction is non-spontaneous.

Question 20.
The equilibrium constant for the reaction is 10. Calculate the value of ΔG°; Given R = 8 JK-1 mol-1; T = 300 K.
Solution. ΔG° = – 2.303 RT log K.
Here, R = 8.0 JK-1 mol-1; T = 300 K; K = 10
ΔG° = – 2.303 × 8(JK-1 mol-1) × 300(K) × log 10
= – 5527 J mol-1 = – 5.527 kJ mol-1.

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

Question 21.
Comment on the thermodynamic stability of NO(g) and N02(g) given:
\(\frac{1}{2}\) N2(g) +\(\frac{1}{2}\) O2(g) → NO (g); ΔfH° = 90 kJ mol-1
NO(g) +\(\frac{1}{2}\)O2(g) → NO2(g); ΔfH° = – 74 kJ mol-1
Solution.
For NO(g); ΔfH° = + ve: Unstable in nature
For NO2(g); ΔfH° = – ve: Stable in nature

Question 22.
Calculate the entropy change in surroundings when 1.0 mol of H2O (l) is formed under standard conditions. Given ΔfH° = – 286 kJ mol-1.
Solution.
ΔfH° of H2O(l) = – 286 kJ mol-1 which refers to heat lost to surroundings.
∴ q(surr)=-ΔfH°0f H2O(l)
= — (- 286 kJ mol-1)
= + 286 kJ mol-1
ΔSsurr = \(\frac{q_{\text {(surr) }}}{\mathrm{T}}\)
= \(\frac{+286 \times 10^{3}\left(\mathrm{~J} \mathrm{~mol}^{-1}\right)}{298.15(\mathrm{~K})}\)
= 959.24 JK1 mol-1.

MP Board Class 11 Chemistry Thermodynamics Important Questions and Answers

Question 1.
What are the units of standard enthalpy of a substance?
Answer:
kJ mol-1.

Question 2.
What type of system is constituted by the coffee placed in a thermos flask?
Answer:
Isolated system.

Question 3.
Under what conditions is the heat absorbed or evolved by the system representing internal energy change?
Answer:
At constant temperature and constant volume.

Question 4.
Name the law given by the mathematical expression q + w = ΔU.
Answer:
Law of conservation of energy.

Question 5.
Why is it necessary to define the standard state?
Answer:
In the form of work.

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

Question 6.
If ΔfH0 of Freon, CHClF2 is – 480.0 kJ mol-1, what is the thermochemical equation for this ?
Answer:
C(s) + \(\frac{1}{2}\)H2(g) + \(\frac{1}{2}\)Cl2(g) + F2(g) → CHClF2; ΔH = – 480.0 kJ.

Question 7.
If ΔfH0 of H(g) is 218 kJ mol-1, what is the value of enthalpy of atomisation of dihydrogen?
Answer:
+ 436 kJ mol-1.

Question 8.
What type of enthalpy change is represented by the following thermochemical equation –
\(\frac{1}{2}\) Cl2(g) + e- + aq → Cl(aq)? ΔH = – x kJ.
Answer:
Enthalpy of formation of Cl (aq) ion.

Question 9.
Although diamond is an elementary substance, yet its ΔfH0 ≠ 0. Why?
Answer:
By convention ΔfH0 of graphite is taken to be zero.

Question 10.
What is the difference in the two processes?
(i) Burning of petrol in the engine of a ear
(ii) Burning of petrol in a bucket in open. Will the energy change in two processes be same or different?
Answer:
In the first case the energy liberated during combustion is converted into mechanical work. In the second case it appears as heat energy.

Question 11.
Will the process N2-(g) + e→ N3-(g) be exothermic or endothermic?
Answer:
The overall process will be endothermic because 2nd and successive electron gain enthalpies are positive.

Question 12.
Make necessary corrections in following relations
(i) ΔS = q/T
(ii) U = H-TS,
(iii) ΔGT V < 0.
Answer:
ΔS =\(\frac{q_{\text {(rev })}}{\mathrm{T}}\) ; G = H – TS; ΔGTp < 0.

Question 13.
At what temperature the entropy of a perfectly crystalline substance is zero?
Answer:
At absolute zero.

MP Board Class 11th Chemistry Solutions Chapter 6 Thermodynamics

Question 14.
Should the following reaction be favourable at high temperature or at low temperature?
SO2(g) + \(\frac{1}{2}\)O2(g) → SO3(g).
Answer:
Low temperature, because it is an exothermic process.

Question 15.
For a reaction, both ΔH and ΔS are positive, under what conditions will the reaction occur spontaneously?
Answer:
When TΔS > ΔH.

Question 16.
Give the relation between ΔSuniverse and ΔG.
Answer:
– ΔG = TΔSuniverse.

Question 17.
Will melting of ice be spontaneous below 273 K at one atmospheric pressure?
Answer:
Non-spontaneous.

Question 18.
Give the relationship between Gibb’s energy change and equilibrium constant.
Answer:
ΔG° = – 2.303 RT log K.

Question 19.
What happens to work, when
(i) gas expands against an external pressure.
(ii) gas is compressed.
(iii) gas expands into vacuum.
(iv) an ideal gas expands reversibly and isothermally.
Answer:
(i) When a gas expands against the external pressure pex in an irreversible manner, work is done by the gas and it is given by the expression: Wpv = -pexΔV.
(ii) When gas is c ompressed, work is done on the gas.
(iii) When the gas expands into vaccum, no work is done because external pressure is zero.
(iv) In case the expansion is done under reversible conditions, the work done by the gas is maximum because the opposing force is infinitessimally smaller than driving force.