Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. Q. $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ Welcome to 5.1 THERMODYNAMICS. Calculate the standard molar entropy change for the following reactions at $298 K$ 2000 AP CHEMISTRY FREE RESPONSE QUESTIONS. $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ We know, Q. SHOW SOLUTION SHOW SOLUTION eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. (ii) Temperature of crytal is increased. $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$ (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What happens to the internal energy of the system if: Why would you expect a decrease in entropy as a gas condenses into liquid ? (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$, Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$, $\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$, (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is, (ii) Below this temperature, $\Delta G$ will be $+$ve because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$. (iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$ $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ First law of thermodynamics problem solving. SHOW SOLUTION $\therefore \Delta U=q+w=0+w_{a d}=w_{a d}$. $\therefore$ Energy required to vapourise $100 g$ benzene Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $44 g(1 \mathrm{mol})$ of SHOW SOLUTION $\therefore \Delta U=q+w=0+w_{a d}=w_{a d}$ The First Law of Thermodynamics Work and heat are two ways of transfering energy between a system and the environment, causing the system’s energy to change. Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ SHOW SOLUTION Q. Predict the entropy change (positive/negative) in the following : $\Delta_{f} H^{\circ}\left(C C l_{4}\right)=-135.5 k J \mathrm{mol}^{-1}$ Also calculate the enthalpy of combustion of octane. [NCERT] $5 B_{O=O}=-2050+4152=+2102$ Also calculate enthalpy of solution of ammonium nitrate. (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. Questions Answer PDF Download. 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. These important questions will play significant role in clearing concepts of Chemistry. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). (iii) by 2 and add to eqn. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 11. What kind of system is the coffee held in a cup ? $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$ (ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$ Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$ Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - вЂ¦. 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. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. (i) At what temperature the reaction will occur spontaneously from left to right? Therefore, $C_{v}=3 / 2 R$, 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 $\Delta T=1^{\circ} \mathrm{C}$, Q. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. Compute the molar heat capacity of these elements and identify any periodic trend. What is the change in entropy in the system when 68.3 g of C 2H5OH( g) at 1 atm condenses to liquid at the normal boiling point? If not at what temperature, the reaction becomes spontaneous. (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ What will be the direction of the reaction at this temperature and below this temperature and why? NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. change. $\Delta_{r} G^{\circ}=-2.303 R T \log K$ $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium $C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. $g$ of $C O_{2}$ from carbon and dioxygen gas. Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: Calculate the enthalpy change for the process : Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. 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. Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$ $=\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1}$ A comprehensive database of more than 19 thermodynamics quizzes online, test your knowledge with thermodynamics quiz questions. A-Level Chemistry. What is its equilibrium constant. of water vaporised $=\frac{10}{18}=0.56$ You may assume that the gas constant R = 8.314 J mol-1. (i) Work is done on the system, 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? since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$ Visit eSaral Website to download or view free study material for JEE & NEET. (i) $\quad \Delta G_{f}^{\circ}=\left[\Delta G_{f}^{\circ} C O_{2}(g)+2 \times \Delta G_{f}^{\circ} H_{2} O(g)\right]$ $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$, $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$. SHOW SOLUTION $=[-394.36+\{2 \times(-228.57)\}-[-50.72+0]$ Thermodynamics Questions and Answers pdf free download 1. $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$ Also calculate the enthalpy of combustion of octane. – For a spontaneous process, $\Delta G<0 .$ Also entropy change (\DeltaS) during polymerization is negative. $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. 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 $\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)$ ( i ) and (ii) (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$ Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. (iii) A partition is removed to allow two gases to mix. State The Third Law Of Thermodynamics. (i) $\quad \Delta H=\Delta U+\Delta n R T$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. -condensation into a liquid. We know Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. (ii) Work is done by the system? Under what condition $\Delta H$ becomes equal to $\Delta E ?$ $\Delta H$ 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. Answer: -163 J / K (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. April 15th, 2018 - Thermodynamics Multiple Choice Questions Has 100 MCQs Thermodynamics Quiz Questions And Answers Pdf MCQs On Applied Thermodynamics First Law Of Thermodynamics MCQs With Answers Second Law Of Thermodynamics Reversible And Irreversible Processes And Working Fluid MCQs And Quizzes To Practice Exam Prep Tests' Predict the sign of entropy change for each of the following changes of state: $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ This will be so if, $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. They also help you manage your time better and be familiar with the method and pattern used in the actual exam. $q=125 g \times 4.18 J / g \times(286.4-296.5)$ Will the heat released be same or different in the following two reactions : g . Ans: d 2. $-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$ (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$ $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state. $S^{\circ}(F e(s))=27.28 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$ (ii) Calculate the value of $\Delta n$ in the following reaction: 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., $\Delta U=\Delta E_{K}$ At equilibrium $\Delta G=0$ so that The heat released in the above two reactions will be different. SHOW SOLUTION $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$ What type of system would it be ? $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. No, there is no enthalpy change in a cyclic process because the system returns to the initial state. Predict the sign of entropy change in the following reactions: Predict the entropy change (positive/negative) in the following : A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? $\Delta U$ is measured in bomb calorimeter. Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$, $\therefore$ 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 $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of 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 $\left(35^{\circ} \mathrm{C}\right)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. Negative, Q. Bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ bond $=\frac{1304.0}{4}$, $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ 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 $=\frac{1304.0}{4}$, Q. Comment on the thermodynamic stability of $N O(g),$ given $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$ strategies to Crack Exam in limited time period. because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$, 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. Q. (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. SHOW SOLUTION (i) Liquid to vapours and What will be sign of for backward reaction? Multiply eqn. (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$ $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ Treat heat capacity of water as the heat capacity of calorimeter and its content). (i) A liquid substance crystallises into a solid. $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ SHOW SOLUTION Compare it with entropy decrease when a liquid sample is converted into a solid. 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 $H C l s-95 k J m o l^{-1}$ An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. These important questions will play significant role in clearing concepts of Chemistry. For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$ $\Delta_{a} H^{o}(C l)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$ $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual … $\therefore \Delta H^{\circ}=715.0+4 \times 121-(-105.0)=1304.0 \mathrm{kJ} \mathrm{mol}^{-1}$ Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. In the reaction $C_{3} H_{8}(g)+5 O_{2} \rightarrow 3 C O_{2}+4 H_{2} O(g) \quad$ if Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. $\frac{109}{18} \times 36=218 \mathrm{JK}^{-1}$, $\Delta_{v a p} S=\frac{\Delta_{v a p} H}{T_{b}}=\frac{40.63 \times 1000 \mathrm{J} \mathrm{mol}^{-1}}{373 \mathrm{K}}=109 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Entropy change for evaporation of $36 g$ of wate, $\frac{109}{18} \times 36=218 \mathrm{JK}^{-1}$. First method: by using the relation $\mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. ( } i v)$, (i)$\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(l)$,$\Delta_{r} H^{\circ}=-726 \mathrm{kJmol}^{-1}$,$C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$,$\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, Q. Isolated system : (vi) Coffee in thermos flask, Q. For an ideal gas, from kinetic theory of gases, the average kinetic energy per mole$\left(E_{k}\right)$of the gas at any temperature$T K,$is given by$E_{k}=3 / 2 R T$Consider the following reaction: The temperature of calorimeter rises from$294.05 K$to$300.78 K .$If the heat capacity of calorimeter is$8.93 \mathrm{kJK}^{-1}$, calculate the heat transferred to the calorimeter. It is based on 1st law of thermodynamics. Discuss the possibility of reducing$A l_{2} O_{3}$and$P b O$with carbon at this temperature. 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. Q. Silane$\left(S i H_{4}\right)$burns in air as: The change in internal energy is a state function and it depends upon the temperature only. What is meant by average bond enthalpy ? An exothermic reaction$X \rightarrow Y$is spontaneous in the back direction.$\Delta H=\Delta U+P \Delta V$Download India's Best Exam Preparation App. Given that$S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$As there is little order in gases are compared to liquids, therefore, entropy of gas decreases enormously on$=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}\Delta H$and$\Delta U$are related as (iv)$\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$What happens if gasoline is used in a Diesel Engine, Siesel Engine will work? (ii) Calculation of$w$Thermodynamics Multiple choice Questions and answers pdf,mcqs,objective type questions,lab viva manual technical basic interview questions download Skip to content Engineering interview questions,Mcqs,Objective Questions,Class Notes,Seminor topics,Lab Viva Pdf free download.$\Delta_{v a p} H^{\ominus}$of$C O=+6.04 \mathrm{kJmol}^{-1}$SHOW SOLUTION (ii) Calculate the value of$\Delta n$in the following reaction:$\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$.$=0.5134 \mathrm{kJ}=(2 \times 14)+(4 \times 1)+(3 \times 16)=80S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$SHOW SOLUTION Give Its Limitations And Importance.? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol.$=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$(iii) w amount of work is done by the system and q amount of heat is supplied to the system. (i) If work is done on the system, internal energy will increase.$\Delta H=\Delta U+P \Delta V$. [CONFIRMED] JEE Main will be conducted 4 times from 2021!$=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$, (i)$\quad C H_{4}(g)+2 O_{2}(g) \rightarrow C O_{2}(g)+2 H_{2} O(g)$,$\Delta G_{f}^{o} \mathrm{CO}_{2}(g)=-394.36 \mathrm{kJ} \mathrm{mol}^{-1}$,$\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$,$\Delta G_{f}^{\circ} C H_{4}(g)=-50.72 \mathrm{kJ} \mathrm{mol}^{-1}$and$\Delta \mathrm{G}_{f}^{\circ} \mathrm{O}_{2}(g)=0$, (ii)$\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$,$\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$,$\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}$,$\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$,$\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$,$-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$,$=[-394.36+\{2 \times(-228.57)\}-[-50.72+0]$, (ii)$\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$,$\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]$,$=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$, (ii)$\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$,$-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(s)+2 \Delta G_{f}^{\circ} H^{+}(a q)\right]$, Q. Average bond enthalpy$\Delta H(O-H)=\frac{498+430}{2}=464 k J$,$H-O-H \rightarrow H(g)+O H(g) ; \Delta H=498 k_{\mathrm{d}}$,$O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$, Average bond enthalpy$\Delta H(O-H)=\frac{498+430}{2}=464 k J$, Q.$A+B \rightarrow C+D$As per the second law of thermodynamics god exist, some person have believe that the world is so beautiful and only god can construct such beautiful world.Some scientist believe that the DNA structure is very much complicated and only god can create it.But as per the one scientist that god exist out of the our knowledge boundary.$=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Therefore, the reaction will not be spontaneous below this temperature. SHOW SOLUTION 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. (ii) At what temperature, the reaction will reverse? Q. (i) At$463 K,$the reaction will be at equilibrium because$\Delta G$is$\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$Red phosphorus reacts with liquid bromine as:$\Delta G^{\circ}=-2.303 R T \log K .$Hence,$\log k=0$or$K=1$. Last updated on 12.08.2020: Previous question papers are a great way to revise and prepare for Higher secondary exams. SHOW SOLUTION SHOW SOLUTION We know SHOW SOLUTION Q. 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.$=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$or$1.065 \mathrm{kJ}$, 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$1.065 \mathrm{kJ}$, Q. (ii)$\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$Enthalpy change for 2 mole of$P=-243 \mathrm{kJ}-92380=\Delta U-2 \times 8.314 \times 298\Delta H=\Delta U+\Delta n R T$Calculate the standard molar entropy change for the following reactions at$298 K$. JEEMAIN.GURU is a free educational site for students, we started jeemain.guru as a passion now we hope that this site would help students to find their required study materials for free. Explain both terms with the help of examples. (ii)$C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}\Delta U$is measured in bomb calorimeter. (ii) SHOW SOLUTION 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. For,$\Delta G=0\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$SHOW SOLUTION$\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol.$g$of$C O_{2}$from carbon and dioxygen gas. Welcome to JEEMAIN.GURU, Best educational blog for IIT JEE aspirants.$B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$, 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.$\left[\text { If } \Delta G_{f}^{o} N O_{2}=51.3 \Delta G_{f}^{o}(N O)=86.55\right]\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$,$M g O(s)+C(s) \rightarrow M g(s)+C O(g)$,$\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$and$\Delta_{r} S^{\circ}$,$=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. (ii) Due to settling of solid$A g C l$from solution, entropy decreases. Place the following systems in order of increasing randomness : The equilibrium constant for a reactions is$10 .$What will be the value of$\Delta G^{\circ} ?$.$\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}f$(Products)$-\Sigma \Delta H^{\circ}$f$(\text {Reactants})$Reaction of combustion of octane: PV diagrams - …$2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} .$=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$of$20 g$of$N H_{4} N O_{3}$in$125 g$water Molecular mass of What type of wall does the system have?$N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$is$-92.38 k J$at$298 K .$What is (iii)$\quad \Delta S=-v e$because gas is changing to less disorder solid. Q. SHOW SOLUTION A$1.250 \mathrm{g}$sample of octane$\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$is burned in excess of oxygen in a bomb calorimeter. 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