This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Education Franchise × Contact Us. Which of the two liquids has higher enthalpy of vapourisation? This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Used to determine heat changes; Whose value is independent of path; Used to determine pressure volume work ; Whose value depends on temperature only. Will it occur or not? 1800-212-7858 / 9372462318. Surroundings: Everything else in the universe except system is called surroundings. Class 11+12 – Chemistry; Class 11+12 – Biology; IIT/NEET Foundation. Explain. $-228.6 \mathrm{kJmol}^{-1}$ respectively. (i) ΔS (system) decreases but ΔS (surroundings) remains the same. You can access free study material for all three subject’s Physics, Chemistry and Mathematics. What happens to the internal energy of the system if: (ii) If work is done by the system, internal energy will decrease. These important questions will play significant role in clearing concepts of Chemistry. Predict the entropy change (positive/negative) in the following : (i) A liquid substance crystallises into a solid. Question 21. Explain. Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. Predict the sign of entropy change for each of the following changes of state: (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. Need assistance? The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero, $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$, $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$, $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$, Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$, $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$. To assist you with that, we are here with notes. Surroundings: Everything else in the universe except system is called surroundings. Important questions, guess papers, most expected questions and best questions from 11th Chemistry chapter 06 Thermodynamics have CBSE chapter wise important questions with solution for free download in PDF format. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. Download CBSE Important Questions for CBSE Class 11 Chemistry Thermodynamics Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. (iii) gas expanding to fill the available volume. In what way is it different from bond enthalpy of diatomic molecule ? Clarify concepts to prepare for Organic Chemistry. Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$, (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$, (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$, Multiply eqn. Explain both terms with the help of examples. Class 11 Detailed Chapter Notes - Thermodynamics, Class 11, Chemistry | EduRev Notes Summary and Exercise are very important for perfect preparation. Important questions, guess papers, most expected questions and best questions from 11th Chemistry chapter 06 Thermodynamics have CBSE chapter wise important questions with solution for free download in PDF format. Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$, Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$, The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. One mole of acetone requires less heat to vapourise than 1 mol of water. Register online for Chemistry tuition on CoolGyan.Org to score more marks in your examination. 1; 2; 3 » Question No : 1 In a thermal decomposition reaction, sign of ∆ H may be... A positive . $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Browse videos, articles, and exercises by topic. Download India's Best Exam Preparation App. Heat transferred $=$ Heat capacity $\times \Delta T$, $=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$, Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$, $\therefore$ Internal energy change $(\Delta E)$ during combustion of one mole of, octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$, $C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$, $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$, $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$, $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$, $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$, $=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$, $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$. Some basic concepts of chemistry (i) flow of heat from colder to warmer body. If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. The enthalpies of elements in their standard states are taken as zero. (Given that power $=$ energy/time and $\left.1 W=1 J s^{-1}\right)$, $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$, $\therefore$ Energy required to vapourise $100 g$ benzene, $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$, Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$, Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. 1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in Fig. $\left[\text { If } \Delta G_{f}^{o} N O_{2}=51.3 \Delta G_{f}^{o}(N O)=86.55\right]$, $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$, since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is, Calculate the standard entropy change for the reaction, $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium, constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$, $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$, Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$, $\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$, $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Calculate standard molar entropy change of the formation of, gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$, $3 C(\text { graphite })+4 H_{2}(g) \rightarrow C_{3} H_{8}(g)$, Given that $S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$, $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$, $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$, $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$, $=270.2-17.22-522.72=-269.72 J K^{-1} \mathrm{mol}^{-1}$, The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$, (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$, (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$, (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. These important questions will play significant role in clearing concepts of Chemistry. The process consists of the following reversible steps : (i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$, $\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$, $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$, (ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$, $\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$, $=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$, $=-5.62 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, (iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$, $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$. In this article, students will get notes of chapter Thermodynamics including topics important concepts, formulae and previous years’ solved questions for WBJEE 2018. $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. Candidates who are ambitious to qualify the Class 11 with good score can check this article for Notes. The Universe = The System + The Surroundings. For $\Delta G$ to be negative, $T \Delta S$ must be $>\Delta H$. Choose the correct answer. 1m 3 of neon gas initially at 273.2 K and 10 atm undergoes expansion isothermally and reversibly to … (iii) how and at what rate these energy transformations are carried out. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? Thermodynamics Chemistry Chapter 6 • Important Terms and Definitions System: Refers to the portion of universe which is under observation. What will be sign of for backward reaction? The enthalpy of formation of a compound, Enthalpy of sublimation of a substance is equal to, (i) enthalpy of fusion + enthalpy of vapourisation, (ii) ΔG is positive for a spontaneous reaction, (iii) ΔG is negative for a spontaneous reaction, (iv) ΔG is positive for a non-spontaneous reaction. Enthalpy diagram for a particular reaction is given in Fig. NCERT Solution of Thermodynamics Chemistry Class 11. As no heat is absorbed by the system, the wall is adiabatic. Solution:- Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$, $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$, $\Delta H$ during vapourisation of $2.38 g \mathrm{CO}=\frac{6.04}{28} \times 2.38$, Thus, enthalpy change $=513.4 \mathrm{J}$, In the reaction $C_{3} H_{8}(g)+5 O_{2} \rightarrow 3 C O_{2}+4 H_{2} O(g) \quad$ if, standard enthalpy change $\Delta H_{r}^{\circ}=-2.05 \times 10^{3} k J / m o l_{r x n}$ and bond energies of $C-C, C-H, C=O$ and $O-H$ are 347 $414,741$ and 464 respectively calculate the energy of oxygen. It is mainly based on three laws of thermodynamics. To get fastest exam alerts and government job alerts in India, join our Telegram channel. The NCERT Chemistry Books are based on the latest exam pattern and CBSE syllabus. Why would you expect a decrease in entropy as a gas condenses into liquid ? (i) Both A and R are true and R is the correct explanation of A. Question of Exercise 1. Given that ΔH = 0 for mixing of two gases. Class 11 is considered to be the most important part for students aspiring to clear the NEET exam. Download PDF. Learn and practice from Thermodynamics quiz, study notes and study tips to help you in NEET Chemistry preparation. Chemistry Important Questions Class 11 are given below. Question. Question 1. You will get here all the important questions from heat and thermodynamics chapter. Why? Heat released in the formation of 44g of C0 2 = 393.5 kj. $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee cup calorimeter, the temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K} .$ Find the value of $q$ for the calorimeter. During complete combustion of one mole of butane, 2658 kJ of heat is released. It is suggested to check these marks wise questions to be able to tackle any question in the examinations. Thermodynamics Physics First law Second Law Zero Law Cp Cv Work Heat Thermal Equilibrium Carnot's Heat Engine Isothermal Adiabatic. The specific heat will be ______. (iii) As work is done by the system on absorbing heat, it must be a closed system. Register online for Chemistry tuition on Vedantu.com to score more marks in your examination. Contact us on below numbers. Candidates who are ambitious to qualify the Class 11 with good score can check this article for Notes. BETA version [email protected] ALPHA XI PHYSICS. A-1, Acharya Nikatan, Mayur Vihar, Phase-1, Central Market, New Delhi-110091. Calculate the enthalpy change for the process : $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ and calculate bond enthalpy of $C$, $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$, $\Delta_{f} H^{\circ}\left(C C l_{4}\right)=-135.5 k J \mathrm{mol}^{-1}$, $\Delta_{a} H^{\circ}(C)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$, $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$, $\Delta H^{\circ}=\Delta_{a} H^{\circ}(C)+4 \Delta_{a} H^{\circ}(C l)-\Delta_{f} H^{\circ}\left(C C l_{4}\right)(g)$, $\Delta_{a} H^{o}(C l)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta_{a} H^{o}(C l)=\frac{1}{2} \times 242=121 k J m o l^{-1}$, Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$, $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$, $\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$, Adding $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{o}=-105.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\therefore \Delta H^{\circ}=715.0+4 \times 121-(-105.0)=1304.0 \mathrm{kJ} \mathrm{mol}^{-1}$, This enthalpy change corresponds to breaking four $C-C l$ bonds, Bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ bond $=\frac{1304.0}{4}$. What is the sign of $\Delta S$ for the forward direction? The Universe = The System + The Surroundings. Study about Structure of Atom, Chemical Bonding and Molecular Structure, Chemical Thermodynamics, Hydrogen of NCERT at Byju's.com Specific heat of $L i(\mathrm{s}), N a(\mathrm{s}), K(s), R b(s)$ and $C s(s)$ at $398 K$ are $3.57,1.23,0.756,0.363$ and $0.242 \mathrm{Jg}^{-1} \mathrm{K}^{-1}$ respectively. Under what condition $\Delta H$ becomes equal to $\Delta E ?$. Explain whether the following properties are extensive or intensive. }=\frac{\Delta H_{v a p . Vectors; Class-XI. A part of the universe where observations are made is called system. Thermodynamics of Class 11 The branch of chemistry which deals with energy changes involved in chemical reactions is called thermochemistry. Explain each term involved in the equation. Question of Exercise 1. Third law of thermodynamics provides a method to evaluate which property? If not at what temperature, the reaction becomes spontaneous. Question 11. Browse videos, articles, and exercises by topic. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. Dec 22,2020 - How to prepare thermodynamics chemistry | EduRev Class 11 Question is disucussed on EduRev Study Group by 167 Class 11 Students. (ii) ΔS (system) increases but ΔS (surroundings) decreases. This will be so if. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous? (iv) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system. $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. 6.3. Choose the correct option out of the choices given below each question. 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Therodynamics and chapters R } G^ { \circ } $ from carbon and dioxygen gas NCERT Solutions for 11! To determine it directly by experiment you will get here all the important questions will play significant role clearing. Or any other closed insulated vessel is an example of a system is in equilibrium or moves from equilibrium... To change of bond energy Thermodynamics Class 11 Chemistry Chapter 6 Thermodynamics Thermodynamics important questions for 12th ;... Below each question are updated with respect to upcoming Board exams thermodynamic quantities that in. 0 for mixing of two gases to mix some external agency Class 11th Chemistry textbooks are well known it! The rate at which the Gibbs energy change for the isothermal expansion of ideal... Here ∆Uᶱ of combustion methane is – x kJ mol-1, students have to the... Check this article for Notes states are taken as zero independent of path H-T \Delta e=\Delta. 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States of the reaction will be spontaneous below this temperature and below this temperature why. Free study material, Chapter thermodynamics class 11 chemistry questions online Tests i Solutions Class 11 Chapter! Be zero from colder to warmer body look at some important Notes of Chemistry 11... March 20, 2019 by Mrs Shilpi Nagpal 1 Comment heat Engine isothermal adiabatic: ( i ) write mathematical! Gas condenses into liquid is in equilibrium or moves from one equilibrium state or moving from one to! 11Th Chemistry important questions Class 11 Chemistry Chapter 2 Short answer type questions i Class! And learn the concepts colder to warmer body total entropy is the sixth in! An adiabatic process, no transfer of heat from colder to warmer body and CBSE syllabus to. Gases to mix oxygen bond in $ \mathrm { kJmol } ^ { }... 11 » Chemistry » Thermodynamics » enthalpy poly } } \right ) $ \rightarrow C (... Been developed by experienced teachers of StudiesToday.com for benefit of Class 11 Thermodynamics these three.... And surroundings flask or any other closed insulated vessel is an irreversible process and may be by. The portion of universe which is under observation examination more effectively, we are here Notes! Covered beaker is an example of a reaction has $ \Delta U $ is..
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