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NCERT solutions Chemistry Class 11 Part 1 chapter 6 Thermodynamics

Chapters

NCERT Solutions for Chemistry Class 11 Part 2

NCERT Chemistry Class 11 Part 1

Chemistry Textbook for Class 11 Part 1

Chapter 6 - Thermodynamics

Pages 182 - 184

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.

Q 1 | Page 182

For the process to occur under adiabatic conditions, the correct condition is:

(i) Δ= 0

(ii) Δ= 0

(iii) = 0

(iv) w = 0

Q 2 | Page 182

The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

Q 3 | Page 182

ΔUθof combustion of methane is – X kJ mol–1. The value of ΔHθ is

(i) = ΔUθ

(ii) > ΔUθ

(iii) < ΔUθ

(iv) = 0

Q 4 | Page 182

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.

Q 5 | Page 182

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

Q 6 | Page 182

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?

Q 7 | Page 182

The reaction of cyanamide, NH2CN(s),with dioxygen was carried out in a bomb calorimeter, and ΔUwas found to be –742.7 kJ mol–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

`NH_2 CN(g) + 3/2 O_2(g) -> N_2(g) + CO_2(g) + H_2O(1)`

Q 8 | Page 182

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.

Q 9 | Page 183

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. Δfus= 6.03 kJ mol–1 at 0°C.

Cp[H2O(l)] = 75.3 J mol–1 K–1

Cp[H2O(s)] = 36.8 J mol–1 K–1

Q 10 | Page 183

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.

Q 11 | Page 183

Enthalpies of formation of CO(g), CO2(g), N2O(gand N2O4(gare –110 kJ mol–1, – 393 kJ mol–1, 81 kJ mol–1 and 9.7 kJ mol–1 respectively. Find the value of Δrfor the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Q 12 | Page 183

Given

`N_(2(g)) + 3H_(2(g)) -> 2NH_(3(g))`; ΔrHθ = –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?

 
Q 13 | Page 183

Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

CH3OH(l) + `3/2` O2(g) →CO2(g) + 2H2O(l) ; ΔrHθ = –726 kJ mol–1

C(g) + O2(g) →CO2(g) ; ΔcHθ = –393 kJ mol–1

H2(g) +`1/2` O2(g) → H2O(l) ; ΔfHθ = –286 kJ mol–1.

Q 14 | Page 183

Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4Cl(g)

and calculate bond enthalpy of C–Cl in CCl4(g).

ΔvapHθ (CCl4) = 30.5 kJ mol–1.

ΔfHθ (CCl4) = –135.5 kJ mol–1.

ΔaHθ (C) = 715.0 kJ mol–1, where ΔaHθ is enthalpy of atomisation

ΔaHθ (Cl2) = 242 kJ mol–1

Q 15 | Page 183

For an isolated system, Δ= 0, what will be ΔS?

Q 16 | Page 183

For the reaction at 298 K,

2A + B → C

Δ= 400 kJ mol–1 and Δ= 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering Δand Δto be constant over the temperature range?

Q 17 | Page 183

For the reaction,2Cl(g) → Cl2(g), what are the signs of Δand ΔS?

Q 18 | Page 183

For the reaction

2A(g) + B(g) → 2D(g)

ΔUθ = –10.5 kJ and ΔSθ= –44.1 JK–1.

Calculate ΔGθ for the reaction, and predict whether the reaction may occur spontaneously.

Q 19 | Page 183

The equilibrium constant for a reaction is 10. What will be the value of ΔGθ? R = 8.314 JK–1 mol–1,T = 300 K.

Q 20 | Page 184

Comment on the thermodynamic stability of NO(g), given

`1/2` N2(g+ `1/2` O2(g→ NO(g) ; ΔrHθ = 90 kJ mol–1

NO(g) +`1/2` O2(g→ NO2(g: ΔrHθ= –74 kJ mol–1

Q 21 | Page 184

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfHθ = –286 kJ mol–1.

Q 22 | Page 184

NCERT Chemistry Class 11 Part 1

Chemistry Textbook for Class 11 Part 1
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