NCERT Solutions for Class 11 Chemistry Chemistry Part-1 Chapter 6

Thermodynamics Class 11

Chapter 6 Thermodynamics Exercise Solutions

<< Previous Chapter 5 : States of Matter Next Chapter 7 : Equilibrium >>

Exercise : Solutions of Questions on Page Number : 182

Q1 :  

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 :

A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, Tetc. depend only on the state of a system and not on the path.

Hence, alternative (ii) is correct.

Answer needs Correction? Click Here

Q2 :  

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

(i) ΔT = 0

(ii) Δp = 0

(iii) q = 0

(iv) w= 0


Answer :

A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.

Therefore, alternative (iii) is correct.

Answer needs Correction? Click Here

Q3 :  

The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q4 :  

ΔUθof combustion of methane is - XkJ mol-1. The value of ΔHθis

(i) = ΔUθ

(ii) > ΔUθ

(iii) < ΔUθ

(iv) = 0


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q5 :  

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 :

Please Register/Login to get access to all solutions Facebook Login
Q6 :  

A reaction, A + B → C + D + qis 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 :

Please Register/Login to get access to all solutions Facebook Login
Q7 :  

In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process?


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q8 :  

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


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q9 :  

Calculate the number of kJ of heat necessary to raise the temperatureof 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q10 :  

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-1 K-1

Cp[H2O(s)] = 36.8 J mol-1 K-1


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q11 :  

Enthalpy of combustion of carbon to CO2is -393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2from carbon and dioxygen gas.


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q12 :  

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

N2O4(g)+ 3CO(g) N2O(g)+ 3CO2(g)


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q13 :  

Given

; ΔrHθ= –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3gas?


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q14 :  

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

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

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

H2(g) +


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q15 :  

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


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q16 :  

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


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q17 :  

For the reaction at 298 K,

2A + B → C

ΔH = 400 kJ mol-1and ΔS = 0.2 kJ K-1mol-1

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q18 :  

For the reaction,

2Cl(g) → Cl2(g),what are the signs of ΔH and ΔS ?


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q19 :  

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.


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q20 :  

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


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q21 :  

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

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

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


Answer :

Please Register/Login to get access to all solutions Facebook Login
Q22 :  

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


Answer :

Please Register/Login to get access to all solutions Facebook Login
<< Previous Chapter 5 : States of Matter Next Chapter 7 : Equilibrium >>

Chemistry Part-1 - Chemistry : CBSE NCERT Exercise Solutions for Class 11th for Thermodynamics will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

Popular Articles