$\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$, Molar mass of $C O=28 g \mathrm{mol}^{-1}$, Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$, Energy required for vapourising $2.38 g$ of $\mathrm{CO}$, $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Enthalpy of combustion of carbon to $\mathrm{CO}_{2}(g)$ is $-393.5 \mathrm{kJ}$ $m o l^{-1} .$ Calculate the heat released upon the formation of 35.2. Class 9+10 – Foundation for IIT Physics; Class 9+10 – Foundation for IIT Chemistry; View All Courses; Live Tutoring; Study Material; Test Series; Watch Video Lectures; LIVE Class Login; Doubts; Contact. Solve MCQs for Physical Chemistry and learn concepts of Inorganic Chemistry. These important questions will play significant role in clearing concepts of Chemistry. strategies to Crack Exam in limited time period. What happens to the internal energy of the system if: (ii) If work is done by the system, internal energy will decrease. Our mission is to provide a free, world-class … Answer. (i) Human being (ii) The earth (iii) Cane of tomato soup, (iv) Ice-cube tray filled with water, (v) A satellite in orbit, (vi) Coffee in a thermos flask, (vii) Helium filled balloon. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$, \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \], $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$, $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$, The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be, taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$, $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$, $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$. Vectors; Class-XI. 11th Chemistry chapter 06 Thermodynamics have many topics. In what way is it different from bond enthalpy of diatomic molecule ? In an adiabatic process, no transfer of heat takes place between system and surroundings. 1m 3 of neon gas initially at 273.2 K and 10 atm undergoes expansion isothermally and reversibly to … Predict the sign of entropy change in the following reactions: (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$, (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$, (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$, (iii) $\quad \Delta \mathrm{S}=-\mathrm{ve}$. A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? $q=125 g \times 4.18 J / g \times(286.4-296.5)$, $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$. Solution - Here ∆Uᶱ of combustion methane is – x kJ mol-1, students have to determine the value of ∆Hᶱ. Surroundings: Everything else in the universe except system is called surroundings. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. Multiple Choice Questions (Type-I) Units and Measurement; 04. During complete combustion of one mole of butane, 2658 kJ of heat is released. (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. Last Updated on March 20, 2019 By Mrs Shilpi Nagpal 1 Comment. The specific heat will be ______. 11th Chemistry chapter 06 Thermodynamics have many topics. Class 11+12 – Chemistry; Class 11+12 – Biology; IIT/NEET Foundation. Learn and practice from Thermodynamics quiz, study notes and study tips to help you in NEET Chemistry preparation. (Hint. (i) At what temperature the reaction will occur spontaneously from left to right? These CBSE NCERT Class 11 Chemistry MCQs have been developed by experienced teachers of StudiesToday.com for benefit of Class 11 students. (ii) ΔS (system) increases but ΔS (surroundings) decreases. For such system, The spontaneity means, having the potential to proceed without the assistance of external agency. $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$, $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$, $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$, $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$, $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$, $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. Therefore, the decrease in entropy when a gas condenses into a liquid is much more as compared to decrease in entropy when a liquid solidifies. Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2, At $(T+1) K,$ the kinetic energy per mole $\left(E_{k}\right)=3 / 2 R(T+1)$ Therefore, increase in the average kinetic energy of the gas for $1^{\circ} \mathrm{C}(\text { or } 1 \mathrm{K})$ rise in temperature $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$, When the gas is heated to raise its temperature by $1^{\circ} \mathrm{C},$ the increase in its internal energy is equal to the increase in kinetic energy, i.e., $\Delta U=\Delta E_{K}$, Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$, Calculate the number of $k J$ necessary to raise the temperature of $60.0 \mathrm{g}$ of aluminium from $35-55^{\circ} \mathrm{C} .$ Molarheat capacity of $A l$ is $24 J m o l^{-1} K^{-1}$, Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$, \[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \], Now, $q=53.28 \mathrm{JK}^{-1} \times \Delta T$, $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ or $1.065 \mathrm{kJ}$. This unit is part of the Chemistry library. The state of a gas can be described by quoting the relationship between___. Download CBSE Class 11 Chemistry Thermodynamics MCQs in pdf, Chemistry chapter wise Multiple Choice Questions free, Question: Thermodynamics is not concerned abouta) the rate at which a reaction proceedsb) the feasibility of a chemical reactionc) the extent to which a chemical reaction proceedsd) energy changes involved in a chemical reactionAnswer: the rate at which a 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 $=\frac{1304.0}{4}$. When liquid benzene is oxidised at constant pressure at 300 K, the change in enthalpy is -3728 kJ. CBSE Class 11 - Chemistry - Thermodynamics Thermodynamics (Short Q & A) Q1: Define Thermodynamics Answer: It is a physical science that deals with quantitative relation between heat and mechanical energy. Used to determine heat changes; Whose value is independent of path; Used to determine pressure volume work ; Whose value depends on temperature only. The standard free energy of a reaction is found to be zero. What is the sign of $\Delta S$ for the forward direction? What will be the direction of the reaction at this temperature and below this temperature and why? (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$, (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$. Find the unit of ΔG. $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. (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. Enter OTP. Free PDF download of Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics prepared by expert Chemistry teachers from latest edition of CBSE(NCERT) books. 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. (iv) burning carbon in oxygen to give carbon dioxide. Thermodynamics - Chemistry Notes, Questions and Answers, Free Study Material, Chapter wise Online Tests. Internal energy change is measure at constant volume. (ii) At what temperature, the reaction will reverse? $-228.6 \mathrm{kJmol}^{-1}$ respectively. Solution - This question requires students to find the enthalpy of combustion of methane, graphite and hydrogen at 298k, -890.3 kJ mol-1, -393.5 kJ mol-1 and -285.8 kJ mol-1. The enthalpy of reaction for the reaction : What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, p. How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps? Register online for Chemistry tuition on CoolGyan.Org to score more marks in your examination. (iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system. Explain each term involved in the equation. Download PDF. Thermodynamics of Class 11. CBSE Class 11 Chemistry , CBSE Class 11 Physics. A part of the universe where observations are made is called system. (iv) ΔS (system) decreases and ΔS (surroundings) also decreases. NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. 10:00 AM to 7:00 PM IST all days. What will be sign of for backward reaction? (i) flow of heat from colder to warmer body. (iii) the rate at which a reaction proceeds. Calculate Gibbs energy change for the reaction is spontaneous or not. Standard molar enthalpy of formation, Δ, Enthalpy is an extensive property. Benefits of CBSE … Browse videos, articles, and exercises by topic. The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. Chemistry Important Questions Class 11 are given below. Question 1. Suggest and explain an indirect method to measure lattice enthalpy of NaCl(s). Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of. Enthalpy diagram for a particular reaction is given in Fig. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K. An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Match the following processes with entropy change: Match the following parameters with description for spontaneity : Assertion (A): Combustion of all organic compounds is an exothermic reaction. JEE NEET Study Material : Notes , Assignment . (ii) gas in a container contracting into one corner. Represent the potential energy/enthalpy change in the following processes graphically. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. $H-O-H \rightarrow H(g)+O H(g) ; \Delta H=498 k_{\mathrm{d}}$, $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$, Average bond enthalpy $\Delta H(O-H)=\frac{498+430}{2}=464 k J$, Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. (ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ. Assertion (A) : Spontaneous process is an irreversible process and may be reversed by some external agency. Question from very important topics are covered by NCERT Exemplar Class 11. This question bank is designed by expert faculties keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p . Free PDF download of Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics prepared by expert Chemistry teachers from latest edition of CBSE(NCERT) books. You can access free study material for all three subject’s Physics, Chemistry and Mathematics. 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. ? Nice and helpful for practicing the sum on thermodynamics, What is there in pdf about thermodynamics, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app, Class 11 Chemistry Thermodynamics Questions and Answers-Important for Exams. 1; 2; 3 » Question No : 1 In a thermal decomposition reaction, sign of ∆ H may be... A positive . Dec 22,2020 - How to prepare thermodynamics chemistry | EduRev Class 11 Question is disucussed on EduRev Study Group by 167 Class 11 Students. Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. We have transformed classroom in such a way that a student can study anytime anywhere. Power $\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$, Express the change in internal energy of a system when, (i) No heat is absorbed by the system from the surroundings, but work (, \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \]. If there is trend, use it to predict the molar heat capacity of Fr. so $\mathrm{NO}(g)$ is unstable. -condensation into a liquid. Thermodynamics MCQ Question with Answer Thermodynamics MCQ with detailed explanation for interview, entrance and competitive exams. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct? Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. Third law of thermodynamics provides a method to evaluate which property? Also calculate the enthalpy of combustion of octane. The best app for CBSE students now provides Thermodynamics class 11 Notes Chemistry latest chapter wise notes for quick preparation of CBSE exams and school based annual examinations. In the following questions two or more options may be correct. A spherical constant temperature heat source of radius r 1 is at the center of a uniform solid sphere of radius r 2.The rate at which heat is transferred through the surface of the sphere is propertional to (A) r 2 2 – r 1 2 (B) r 2 – r 1 (C) ln r 1 – ln r 2 (D) 1/r 2 – 1/r 1 (E) (1/r 2 – 1/r 1)-1. Given that ΔH = 0 for mixing of two gases. Question 21. Why? For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w = – nRT ln(V. (i) Work done at 600 K is 20 times the work done at 300 K. (ii) Work done at 300 K is twice the work done at 600 K. (iii) Work done at 600 K is twice the work done at 300 K. Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below : (i) The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ. 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