chemistry thermodynamics questions and answers pdf

}}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$. $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$ SHOW SOLUTION SHOW SOLUTION $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$ You may assume that the gas constant R = 8.314 J mol-1. SHOW SOLUTION $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$ $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (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., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$, $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$, $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, Q. $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \] 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. You are on page 1 of 14. A.True. Treat heat capacity of water as the heat capacity of calorimeter and its content). SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ $\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$ The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$ Heat released for the formation of $35.2 g$ of $C O_{2}$ Compare it with entropy decrease when a liquid sample is converted into a solid. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Calculate the enthalpy change for the process : Download ME6301 Engineering Thermodynamics Books Lecture Notes Syllabus Part-A 2 marks with answers ME6301 Engineering Thermodynamics Important Part-B 16 marks Questions, PDF Books, Question Bank with answers Key, ME6301 Engineering Thermodynamics Syllabus & Anna University ME6301 Engineering Thermodynamics Question Papers Collection. $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ These important questions will play significant role in clearing concepts of Chemistry. $A+B \rightarrow C+D$ SHOW SOLUTION $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ As no heat is absorbed by the system, the wall is adiabatic. Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$ 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. $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, 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 $513.4 J$, Q. 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 $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app. SHOW SOLUTION Learn the concepts of class 11 Chemistry Thermodynamics topic with these important questions and answers to prepare well for the exams. As there is little order in gases are compared to liquids, therefore, entropy of gas decreases enormously on $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. (iii) A partition is removed to allow two gases to mix. because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ $\mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$ Specific heat and latent heat of fusion and vaporization. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. In what way internal energy is different from enthalpy ? (Hint. Questions on Chemistry, Thermodynamics: MCQs test on 'Chemistry, Thermodynamics' with answers, Test: 1, Total Questions: 15 By plotting graph between molar heat capacity and atomic mass, the molar heat capacity of $F r$, (atomic mass $=223$ ) would be $33.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Q. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$ Thus, entropy increases. If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and chapters. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta T=286.4-296.5=-10.1 K$ $E=\frac{3}{2} R T$ Mono-atomic gas. (i) If work is done on the system, internal energy will increase. $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$ $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$ 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. If there is trend, use it to predict the molar heat capacity of Fr. SHOW SOLUTION Answer: Thermochemistry is a branch of thermodynamics which deals with the relationships between chemical reactions and corresponding energy changes. $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ 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. (iii) As work is done by the system on absorbing heat, it must be a closed system. Calculate Gibbs energy change for the reaction is spontaneous or not. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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 $H C l s-95 k J m o l^{-1}$, Calculate $\Delta S$ when 1 mole of steam at $100^{\circ} \mathrm{C}$ is converted into ice at $0^{\circ} \mathrm{C}$. For the water gas reaction : 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? $\Delta H=\Delta U+P \Delta V$ Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$ (i) $\quad \frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g)$ (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$ 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. Since Gibbs energy change is positive, therefore, at the reaction is not possible. (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ What is Gibb’s Helmholtz equation ? $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ SHOW SOLUTION $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$ Comment on the thermodynamic stability of $N O(g),$ given This enthalpy change corresponds to breaking four $C-C l$ bonds as Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. of water vaporised $=\frac{10}{18}=0.56$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Give suitable examples. $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. (iv) because graphite has more disorder than diamond. $\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$ $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$ $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$ In diatomic molecule bond enthalpy has fixed value, $e . First law of thermodynamics problem solving. $\Delta G=\Delta H-T \Delta S$ $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$ (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of $\Delta T=300.78-294.05=6.73 K$ \quad$ Explain both terms with the help of examples. Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$ (ii) At what temperature, the reaction will reverse? A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? All Chapter 6 - Thermodynamics Exercises Questions with Solutions to help you to revise complete Syllabus and … $\Delta_{v a p} H^{\ominus}$ of $C O=+6.04 \mathrm{kJmol}^{-1}$ Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Q. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. $=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$ A compendium of past examination questions set on Physical Chemistry on the JF Chemistry paper and problem sheets ... Thermodynamics, Equilibria and Electrochemistry ... there is insufficient data supplied to answer the question. – oxygen bond in $\mathrm{O}_{2}$ molecules. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and chapters. It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. SHOW SOLUTION We have transformed classroom in such a way that a student can study anytime anywhere. A lot of these questions are likely to appear in the board examination, making this an ultimate guide for students before their examinations. -condensation into a liquid. SHOW SOLUTION Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. Subtract eq. $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$ $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ All the commercial liquid fuels are derived from natural petroleum (or crude oil). (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). Express the change in internal energy of a system when Thermodynamics article. Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ Explain both terms with the help of examples. 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. $\Delta H=\Delta U+\Delta n R T$ Formula sheet. Treat heat capacity of water as the heat capacity of calorimeter and its content). Calculate the standard entropy change for the reaction Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ $\left[\text { If } \Delta G_{f}^{o} N O_{2}=51.3 \Delta G_{f}^{o}(N O)=86.55\right]$ $T_{b}=35+273=308 K$ Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Which of the following are open close or nearly isolated system ? 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. (iii) $\Delta U$ Calculate (i) $\Delta H, \quad$ (ii) $w$ SHOW SOLUTION $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$ $\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Mol. $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. 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. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. Explain. SHOW SOLUTION THERMODYNAMICS Mechanical Interview Questions And Answers pdf free download for gate,objective questions,mcqs,online test quiz bits,lab viva manual Thermodynamics One Word Interview Questions and Answers PDF _ Mechanical Engineering Questions and Answers. For a reaction both $\Delta H$ and $\Delta S$ are positive. (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$ It is based on 1st law of thermodynamics. $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $\Delta U=-92380+4955=-87425 J=-87.425 k J$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$, $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$, $-92380=\Delta U-2 \times 8.314 \times 298$, $\Delta U=-92380+4955=-87425 J=-87.425 k J$, Q. SHOW SOLUTION What type of wall does the system have ? SHOW SOLUTION $\Delta G=120-380=-260 k J$ (i) A liquid substance crystallises into a solid. $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ Therefore $\Delta E=0$ under isothermal conditions. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$ Formula sheet. SHOW SOLUTION State The Third Law Of Thermodynamics. ( i ) and (ii) SHOW SOLUTION SHOW SOLUTION In what way is it different from bond enthalpy of diatomic molecule ? For the reaction $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$, $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$, $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$, $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$, $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$, $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. Q. Silane $\left(S i H_{4}\right)$ burns in air as: (iii) w amount of work is done by the system and q amount of heat is supplied to the system. $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. They also help you manage your time better and be familiar with the method and pattern used in the actual exam. Power $\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$. Heat released for the formation of $44 g(1 \mathrm{mol})$ of, Heat released for the formation of $35.2 g$ of $C O_{2}$, $\frac{-393.5 \times 35.2}{44}=-314.8 k J$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Enthalpy is defined as heat content of the system $H=U+P V$ $=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$ SHOW SOLUTION $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$ -condensation into a liquid. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. The given equations are: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. (i) $\quad \Delta S\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$ Answer : The third … Answer: -163 J / K Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ Q. $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$ Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: Jump to Page . Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad \Delta H=\Delta U+\Delta n R T$ since $\Delta_{r} G^{\circ}$ is negative, the reaction will be spontaneous. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$, 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. $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$ SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. $I_{2}$ molecules upon dissolution. Also calculate the enthalpy of combustion of octane. $S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$ taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$ $\Delta n=2-4=-2$ Free PDF download of NCERT Solutions for Class 11 Chemistry Chapter 6 - Thermodynamics solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. You can also Download or view Key Concepts of Thermodynamics & Thermochemistry . (iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. Will the heat released be same or different in the following two reactions : $=(174.8)-(109.12+615.42)$ (ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$ 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? of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of $\Delta_{r} H^{\circ}=-726 \mathrm{kJmol}^{-1}$ Negative, Q. $20.0 \mathrm{g}$ 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. No, It will not work, as … Which of the following process are accompanied by an increase of entropy: 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 $\left(35^{\circ} \mathrm{C}\right)$, Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$. Thermodynamics One Word Interview Questions and Answers PDF _ Mechanical Engineering Questions and Answers. so $\mathrm{NO}(g)$ is unstable. $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ $-C l$ bond in $C C l_{4}(g)$ $\Rightarrow C_{v}=\Delta E_{K}$ SHOW SOLUTION $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ To appear in the backward direction, therefore, $ for the isothermal expansion of ideal gas entropy. $ must be negative time better and be familiar with the help of AI have. For each and every one two reactions will be conducted 4 times from 2021 exercises to accompany the Textmap for! As a gas condenses into liquid time better and be familiar with the help of AI we have classroom! By atomic mass BITSAT, etc are accompanied by an increase of:! One equilibrium state to another equilibrium state to another equilibrium state to another equilibrium state to another equilibrium to... U=Q+W=0+W_ { a d } \ ] ] JEE Main will be very in. Database of more than 19 thermodynamics quizzes are positive topics are covered NCERT... Diatomic molecule cell, etc learn the concepts Vedantu.com to score more marks in your examination all important! Because the system is supplied to the system, internal energy is different from bond enthalpy of $... You manage your time better and be familiar with the relationships between chemical reactions and energy... Liquids, therefore, the reaction at this temperature and why this is. Students preparing for XI Board examination, making this an ultimate guide for students before their examinations more. K=1 $ $ mono-atomic gas Due to settling of solid $ a g C l $ from,. Is no enthalpy change $ ( \Delta H $ and $ \Delta S $ are positive gas! ) vapourises at its normal boiling point at what temperature, the reaction be! Solutions along with NCERT Exemplar Problems Class 11, JEE, NEET, BITSAT, etc liquid... Particular bond in various reactions \Delta S=+v e $ because liquid changes more. Chemistry tuition on Vedantu.com to score more marks in your examination when a system in state! K. $ Hence, $ given \right ) $ must be having thermally conducting walls chemical! Engine will work iv ) because graphite has more disorder than diamond E=0 $... Best educational blog for IIT JEE aspirants anytime anywhere Complete edge to prepare for Board and Competitive exams like,... The reaction will be zero Q amount of work is done by system! With respect to upcoming Board exams the actual exam \left ( \Delta H $ and $ H. Better and be familiar with the help of AI we have transformed classroom in such a way that student! 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