NCERT Solutions For Class 11 Chemistry Chapter 6 Thermodynamics Laws

NCERT Solutions For Class 11 Chemistry Chapter 6 Thermodynamics Laws

Class 11 Chemistry Chapter 6 Thermodynamics Laws

NCERT Solutions For Class 11 Chemistry Chapter 6 Thermodynamics Laws, (Chemistry) exam are Students are taught thru NCERT books in some of the state board and CBSE Schools.  As the chapter involves an end, there is an exercise provided to assist students to prepare for evaluation.  Students need to clear up those exercises very well because the questions withinside the very last asked from those.

Sometimes, students get stuck withinside the exercises and are not able to clear up all of the questions.  To assist students to solve all of the questions and maintain their studies without a doubt, we have provided step-by-step NCERT Solutions for the students for all classes.  These answers will similarly help students in scoring better marks with the assist of properly illustrated solutions as a way to similarly assist the students and answering the questions right.

NCERT Solutions For Class 11 Chemistry Chapter 6 Thermodynamics Laws

Class 11 Chemistry Chapter 6 Thermodynamics Laws

 

Question 1. Choose the correct answer:
A thermodynamic state junction 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) whose value is independent of path

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) q = 0

6.3. The enthalpies of all elements in their standard states are : ‘

(i) unity (ii) zero
(iii) < 0 (iv) different for each element

Answer: 

(ii) zero

6.4.

Answer:

6.5. 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 CHJg) 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: 

As per the available data :

6.6. 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: 

(iv) possible at any temperature

6.7. In a process, 701 ] 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: 

Heat absorbed by the system, q = 701 J Work done by the system = – 394 J Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

6.8. The reaction of cyanamide,NH2CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742,7 KJ-1  mol-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.NH2CN (S) + 3/202(g) —–>N2(g) + CO2(g) + H20(Z)

Answer: 

∆U = – 742.7 KJ-1  mol-1 ; ∆ng = 2 – 3/2 = + 1/2 mol.
R = 8.314 x 10-3KJ-1  mol-1 ; T = 298 K
According to the relation,∆H = ∆U+∆ngRT
∆H = (- 742.7 kj) + (1/2 mol) x (8.314 x10-3 KJ-1  mol-1 ) x (298 K)
= – 742.7 kj + 1.239 kj = – 741.5 kj.

6.9. Calculate the number of kj of heat necessary to raise the temperature of 60 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.

Answer: 

No. of moles of Al (m) = (60g)/(27 g mol-1) = 2.22 mol
Molar heat capacity (C) = 24  J mol-1 K-1.
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C x m x T = (24 J mol-1 K-1) x (2.22 mol) x (20 K)
= 1065.6 J = 1.067 kj

6.10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. A, H = 6.03 KJ mot1 at 0°C. Cp [H20(l)J = 75.3 J mol-1 K-1; Cp [H20(s)J = 36.8 J mol-1 K-1.

Answer: 

The change may be represented as:

6.11. Enthalpy of combustion of carbon to carbon dioxide is – 393.5 J mol-1   .Calculate the heat released upon formation of 35.2 g of C0from carbon and oxygen gas.

Answer: 

The combustion equation is:
C(s) + 0(g) —–> C02(g); AcH = – 393.5 KJ mol-1
Heat released in the formation of 44g of C02 = 393.5 kj
Heat released in the formation of 35.2 g of C02=(393.5 KJ) x (35.2g)/(44g) = 314.8 kj

6.12. Calculate the enthalpy of the reaction:
N204(g) + 3CO(g) ———->N20(g) + 3CO2(g)
Given that;∆fHCO(g) = – 110 kj mot-1; ∆fHC02(g) = – 393 kj mol-1
fHN20(g) = 81 kj mot-1; ∆fN2O4(g) = 9.7 kj mol-1

Answer: 

Enthalpy of reaction (∆r,H) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kj mol-1

6.13. Given : N2(g) + 3H2(g) ————> 2NH3(g); ∆r H = -92.4 kj mot-1 What is the standard enthalpy of formation of NH3 gas?

Answer: 

∆H NH3 (g) = – (92.4)/2 = – 46.2 kj mol-1

6.14. Calculate the standard enthalpy of formation of CH3OH. from the following data:
(i) CH3OH(l) + 3/2 02 (g) ———-> CO2 (g) + 2H20 (l); ∆rH = – 726kj mol-1
(ii) C(s) + 02(g) —————>C02 (g); ∆cH = -393 kj mol-1
(iii) H2(g) + 1/202(g) —————->H20 (l); ∆fH = -286 kj mol-1

Answer: 

The equation we aim at;
C(s) + 2H2(g) + l/202(g) ———> CH3OH (l);∆fH = ±? … (iv)
Multiply eqn. (iii) by 2 and add to eqn. (ii)
C(s) + 2H2(g) + 202(g) ————->C02(g) + 2H20(Z)
∆H = – (393 + 522) = – 965 kj moH Subtract eqn. (iv) from eqn. (i)
CH3OH(Z) + 3/202(g) ————> C02(y) + 2H20(Z); ∆H = – 726 kj mol-1
Subtract: C(s) + 2H2(y) + l/202(g) ———-> CH3OH(Z); ∆fHe = – 239 kj mol-1

6.15.

Answer:

6.16.For an isolated system∆U = 0; what will be ∆S? 

Answer: 

Change in internal energy (∆U) for an isolated system is zero for it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.

6.17. For a reaction at 298 K
2A + B————->C
∆H = 40Q kj mot1 and AS = 0.2 kj Kr-1 mol-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

Answer:  

As per the Gibbs Helmholtz equation:
ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS
T = (400 KJ mol-1)/(0.2 KJ K-1 mol-1) = 2000 k
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

6.18. For the reaction; 2Cl(g) ———-> Cl2(g); what will be the signs of ∆H and ∆S?

Answer: 

∆H : negative (- ve) because energy is released in bond formation
∆S : negative (- ve) because entropy decreases when atoms combine to form molecules.

6.19.

Answer:

6.20.

Answer:

6.21.

Answer:

6.22.

Answer:

Leave a Comment