Class 11 Unit 6 Thermodynamics Full Exercise Solution 6.1 to 6.22 NCERT Solution 2022

Hi guys, This Falguni Vala from My Smart Class, in this video, I am going to teach you all about Class 11 Unit 6 Thermodynamics Full Exercise Solution 6.1 to 6.22

Time Stamp of Numerical is given below in the description.
6.1 Choose the correct answer. A thermodynamic state function is a quantity
6.2 For the process to occur under adiabatic conditions, the correct condition is:
6.3 The enthalpies of all elements in their standard states are:
6.4 ∆U0of combustion of methane is – X kJ mol–1. The value of ∆H is
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 CH4(g) will be
6.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
6.7 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?
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 mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
6.9 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–1 K–1.
6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C.
6.11 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.
6.12 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 ∆rH for the reaction:
6.13 Given N2(g) + 3H2(g) → 2NH3(g) ; ∆rH = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas?
6.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH (l) +3/2 O2(g) → CO2(g) + 2H2O(l) ; ∆rH = –726 kJ mol–1
C(graphite) + O2(g) → CO2(g) ; ∆cH = –393 kJ mol–1
H2(g) +1/2O2(g) → H2O(l) ; ∆f H = –286 kJ mol–1.
6.15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g)and calculate bond enthalpy of C – Cl in CCl4(g).∆vapH(CCl4) = 30.5 kJ mol–1.∆fH (CCl4) = –135.5 kJ mol–1.∆Ha (C) = 715.0 kJ mol–1 , where ∆Hais enthalpy of atomisation∆Ha (Cl2) = 242 kJ mol–1
6.16 For an isolated system, ∆U = 0, what will be ∆S ?
6.17 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.
6.18 For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?
6.19 For the reaction 2 A(g) + B(g) → 2D(g) ∆U  = –10.5 kJ and ∆S = –44.1 JK–1. Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously.
6.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G ? R = 8.314 JK–1 mol–1, T = 300 K.
6.21 Comment on the thermodynamic stability of NO(g), given
6.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H = –286 kJ mol–1.

Tools that I regularly use.
🎤 Used mic:

👉 Follow On Instagram: my_smartclass

My Smart Class Channel needs your support, so please SHARE & LIKE to this video for our motivation. So we can make important and useful videos for all. And don’t forget to SUBSCRIBE MY SMART CLASS YouTube channel.

🎼 Music Credit:

[nishitaverma]

I haven’t seen any teacher explaining so well in less time
Your explanation is besttest❤

[dhruvpatwal]

Thanku mam kal mera papar h aur sare question clear ho gye thanku ❤❤❤

[its_me_amit_official]

mam question no. 6.10 me ΔH 1 ki unit J/mol and ΔH 2 ki bhi unit J/mol hogi na please check🤔

[gamersworld]

Mam by this video I solve my all ncert exercise doubt

[Cherry-qj]

Thank you very much mam! Your explanation is soo good! ❤

[Xyztaken]

Q21 in the reaction exothermic heat release is more stable than endothermic reaction product
Mam a request it was quick and good for aspirants for competitive exam but it's still somewhere fast it could be of more than 7 to 10 mins more to complete it as basics would be

[manishaverma]

Mam que-6.10 me 6030 hoga apne total enthalpy change me( -) le liya hai

[DivyanshuKujur-tlgl]

Thankyou ma'am for your wonderful explanation.... ❤

[apurvapandey]

Ma'am everything is fabulous but their is one mistake ... I.e.
You are saying no.s together although after decimal also ....😅

[Mithleshkumarverma]

Hello ma'am please physics ka bhi ncert exercise banaiye

[vinamratiwari]

At 18:58 me 6030 ko minus me kyo Liye hai

[BudaramdanidiaDanidia]

Aap mast padhta ho par hindi ma pdaya karo

[rameshchandrasamal]

Thanks mam etni achhi explanation ke liye

[jaypatel-mlps]

Pura cource khatam karo
Ye kya ak ak video kyu dal rahe ho