Under what conditions will the reaction occur spontaneously? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various $T_{b}=35+273=308 K$ Zeroth law of thermodynamics. Questions and Answers in Thermodynamics. $q=m \times s \times\left(t_{2}-t_{1}\right)$ Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$ Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero SHOW SOLUTION Thus, entropy increases. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Read online Class 11 Chemistry Thermodynamics Questions And Answers book pdf free download link book now. the condensation of diethyl ether is the reverse process, therefore, $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. $\Delta H=\Delta E$ during a process which is carried out in a closed vessel $(\Delta v=0)$ or number of moles of gaseous products $=$ number of moles of gaseous reactants or the reaction does not involve any gaseous reactant or product. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $g$ of $C O_{2}$ from carbon and dioxygen gas. Red phosphorus reacts with liquid bromine as: Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$ Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. If there is trend, use it to predict the molar heat capacity of Fr. What type of system would it be ? For $\Delta G$ to be negative, $T \Delta S$ must be $>\Delta H$, For $\Delta G$ to be negative, $T \Delta S$ must be $>\Delta H$, Q. Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Comment on the thermodynamic stability of $N O(g),$ given Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION (i) $\quad H g(l) \rightarrow H g(g)$ Compare it with entropy decrease when a liquid sample is converted into a solid. $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$ $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$, (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. For an ideal gas, from kinetic theory of gases, the average kinetic energy per mole $\left(E_{k}\right)$ of the gas at any temperature $T K,$ is given by $E_{k}=3 / 2 R T$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) A liquid substance crystallises into a solid. The standard Gibbs energies of formation of $S i H_{4}(g), S i O_{2}(s)$ and $H_{2} O(l)$ are $+52.3,-805.0$ and Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ (i) $\frac{1}{2} N_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow N O(g) ; \Delta_{r} H^{\circ}=90 k J \operatorname{mol}^{-1}$, (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. $\Delta T=300.78-294.05=6.73 K$ (iii) by 2 and add to eqn. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. It is based on 1st law of thermodynamics. How many times is molar heat capacity than specific heat capacity of water ? SHOW SOLUTION $\Delta_{v a p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}, T_{b}=373 \mathrm{K}$ Practice: Thermodynamics questions. SHOW SOLUTION The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. 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. What will be sign of for backward reaction? $\Delta G^{\circ}=-2.303 R T \log K .$ Hence, $\log k=0$ or $K=1$. (i) $\quad \Delta H=\Delta U+\Delta n R T$ Thus, entropy increases. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$, (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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. This is the currently selected item. $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$ No, there is no enthalpy change in a cyclic process because the system returns to the initial state. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ First method: by using the relation Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - … Class 11 Important Questions for Chemistry – Thermodynamics NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. A lot of these questions are likely to appear in the board examination, making this an ultimate guide for students before their examinations. SHOW SOLUTION The enthalpy of vaporisation of liquid diethyl ether $\left[\left(C_{2} H_{5}\right)_{2} O\right]$ is $26.0 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point $\left(35^{\circ} \mathrm{C}\right)$ SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 1 Mole of solid $x<1$ mole of liquid $x<1$ mole of gas, 1 Mole of solid $x<1$ mole of liquid $x<1$ mole of gas. SHOW SOLUTION to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. SHOW SOLUTION Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. The process consists of the following reversible steps : Welcome to 5.1 THERMODYNAMICS. $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is, $\Delta U$ at $298 K ? CBSE Class 11 Chapter 6 Thermodynamics Chemistry Marks Wise Question with Answers All books are in clear copy here, and all files are secure so don't worry about it. The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ $-C l$ bond in $C C l_{4}(g)$ $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$ Isolated system : (vi) Coffee in thermos flask, Q. 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. $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$ $A g_{2} O(s) \rightarrow 2 A g(s)+\frac{1}{2} O_{2}(g)$ Treat heat capacity of water as the heat capacity of calorimeter and its content). constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$ Calculate the bond enthalpy of $H C l .$ Given that the bond enthalpies of $H_{2}$ and $C l_{2}$ are $430 \mathrm{kJ} \mathrm{mol}^{-1}$ and $242 \mathrm{kJ} \mathrm{mol}^{-1}$ respectively and $\Delta \mathrm{H}_{f}^{\circ}$ for $H C l s-95 k J m o l^{-1}$ The enthalpy change for the reaction: $R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ SHOW SOLUTION combustion of $C$ to $C O_{2} .$ The net free energy change is calculated Jump to Page . $\Delta G=\Delta H-T \Delta S$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). which is very easy to understand and improve your skill. What is its equilibrium constant. Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: A piston exerting a pressure of 1 atm rests on the surface of water at $100^{\circ} \mathrm{C} .$ The pressure is reduced to smaller extent when $10 g$ of water evaporates and $22.2 \mathrm{kJ}$ of heat is absorbed. $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$ [NCERT] $E=\frac{3}{2} R T$ Mono-atomic gas. Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2 R. $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$ Treat heat capacity of water as the heat capacity of calorimeter and its content). Standard vaporization enthalpy of benzene at boiling point is $30.8 \mathrm{kJ} \mathrm{mol}^{-1} ;$ for how long would a $100 \mathrm{W}$ electric heater have to operate in order to vaporize a $100 \mathrm{g}$ sample at the temperature? (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$ [NCERT] Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? chemical thermodynamics problems and solutions, chemical thermodynamics problems and solutions pdf, class 11 chemistry thermodynamics questions and answers pdf, JEE Main Previous Year Questions Topicwise. Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$ Revision : Electricity. Q. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$ SHOW SOLUTION $\Delta U$ at $298 K ?$ Since Gibbs energy change is positive, therefore, at the reaction is not possible. Q. Calculate $\Delta S$ when 1 mole of steam at $100^{\circ} \mathrm{C}$ is converted into ice at $0^{\circ} \mathrm{C}$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Calculate the temperature at which the Gibbs energy change for the reaction will be zero. $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Q. Therefore $\Delta E=0$ under isothermal conditions. This question bank is designed, keeping NCERT in mind and the questions are updated with respect to … Red phosphorus reacts with liquid bromine as: Standard enthalpy of vapourisation of benzene at its boiling point is $30.8 \mathrm{kJ} \mathrm{mol}^{-1} .$ For how long would a $100 \mathrm{Welectric}$ heater have to operate in order to vapourise $100 \mathrm{g}$ of benzene at its boiling point. $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ Jump to Page . (iii) Calculation of $\Delta U$ SHOW SOLUTION Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ $-228.6 \mathrm{kJmol}^{-1}$ respectively. Also calculate the enthalpy of combustion of octane. Thermodynamics The normal boiling point of ethanol, C 2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ/mol. What is meant by average bond enthalpy ? (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ $\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)$ Power $\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$ $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. Our online thermodynamics trivia quizzes can be adapted to suit your requirements for taking some of the top thermodynamics quizzes. Predict the sign of entropy change for each of the following changes of state: (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$ SHOW SOLUTION 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. the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. We won’t spam you. Explain. Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. $\Delta n=2-4=-2$ $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. $-92380=\Delta U-2 \times 8.314 \times 298$ Free PDF download of NCERT Solutions for Class 11 Chemistry Chapter 6 - Thermodynamics solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$, $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$. Q. (iii) w amount of work is done by the system and q amount of heat is supplied to the system. Question from very important topics are covered by NCERT Exemplar Class 11.You also get idea about the type of questions and method to answer in your Class 11th … In this unit we shall focus our You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. The heat released in the above two reactions will be different. SHOW SOLUTION Thermodynamics key facts (4/9) ... • Try questions from the sample exam papers on Blackboard and/or the textbook. $20.0 \mathrm{g}$ 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. $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ For the water gas reaction : What is the value of internal energy for 1 mole of a mono-atomic gas ? $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION SHOW SOLUTION \quad$, $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$, $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, Q. [NCERT] B. – For a spontaneous process, $\Delta G<0 .$ Also entropy change (\DeltaS) during polymerization is negative. Learn the concepts of class 11 Chemistry Thermodynamics topic with these important questions and answers to prepare well for the exams. (i) $\quad C H_{4}(g)+2 O_{2}(g) \rightarrow C O_{2}(g)+2 H_{2} O(g)$ $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$ $\Delta G_{f}^{o} \mathrm{CO}_{2}(g)=-394.36 \mathrm{kJ} \mathrm{mol}^{-1}$ Thus, enthalpy change $=513.4 \mathrm{J}$, $\Rightarrow \Delta H$ 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 $=513.4 \mathrm{J}$, Q. since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$ (i) Entropy increases due to more freedom of movement of Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Please enter your email address. What type of wall does the system have ? $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app. For a reaction both $\Delta H$ and $\Delta S$ are positive. \[ \Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ} \] $\Delta H$ and $\Delta S$ for the reaction: Calculate the standard Gibb’s energy change, $\Delta G_{f}^{\circ},$ for the following reactions at $298 \mathrm{Kusing}$ standard Gibb’s energy of formation. $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ or $\log K=\frac{-\Delta_{r} G}{2.303 R T}$, $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$, $R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, $\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$ or $K=2.64$, Q.