# chemistry thermodynamics questions and answers pdf

Molar mass of $C O=12+16=28$ Consider the following reaction: Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$ $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$ Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ Class XI Chapter 6 – Thermodynamics Chemistry Page 5 of 11 Answer From the expression of heat (q), q = m. c. ∆T Where, c = molar heat capacity m = mass of substance ∆T = change in temperature Substituting the values in the expression of q: q = 1066.7 J q = 1.07 kJ Question 6.10: $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$ $I_{2}$ molecules upon dissolution. SHOW SOLUTION $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state. CBSE Chemistry Chapter 6 Thermodynamics class 11 Notes Chemistry in PDF are available for free download in myCBSEguide mobile app. 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. Thus, entropy increases. (iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. 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. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$ You can also Download or view Key Concepts of Thermodynamics & Thermochemistry . [NCERT] Thermodynamics Questions and Answers pdf free download 1. Open system : (i) Human being (ii) The earth (v) A satellite in orbit, $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$ SHOW SOLUTION With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. $\Delta_{v a p} H^{\ominus}$ of $C O=+6.04 \mathrm{kJmol}^{-1}$ (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ $-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Hence it is non-spontaneous. $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ Which of the following process are accompanied by an increase of entropy: since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is Thermodynamics MCQ Question with Answer Thermodynamics MCQ with detailed explanation for interview, entrance and competitive exams. (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. False. 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. For, $\Delta G=0$ $\Delta G_{f}^{o} \mathrm{CO}_{2}(g)=-394.36 \mathrm{kJ} \mathrm{mol}^{-1}$ Calculate Gibbs energy change for the reaction is spontaneous or not. (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$ They also help you manage your time better and be familiar with the method and pattern used in the actual exam. $\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$, 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. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 11.. $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$ (iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Predict the entropy change (positive/negative) in the following : The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ Q. Silane $\left(S i H_{4}\right)$ burns in air as: The reaction is spontaneous in the backward direction, therefore, $\Delta G$ is positive in the forward direction. What type of wall does the system have ? $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. Q. $\Rightarrow C_{v}=\Delta E_{K}$ $A g_{2} O(s) \rightarrow 2 A g(s)+\frac{1}{2} O_{2}(g)$ Q. (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. $\Delta G^{\circ}=-2.303 R T \log K$ MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. (ii) Temperature of crytal is increased. $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. $\Delta S_{v a p . (iv)$\quad \Delta S>O$.$=[2 \times 51.3]-[2 \times 86.55+0]$Q. (ii)$\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$Enthalpy of combustion of octane, Heat transferred$=$Heat capacity$\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}$, Q. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Learn the concepts of class 11 Chemistry Thermodynamics topic with these important questions and answers to prepare well for the exams. For a reaction both$\Delta H$and$\Delta S$are positive.$\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$The standard Gibbs energy of reaction (at$1000 K)$is$-8.1k J m o l^{-1} .$Calculate its equilibrium constant. change. Energy required for vapourising$28 g$of$C O=6.04 k_{U}$(ii)$\quad H_{2} O(l)$at$100^{\circ} C \longrightarrow H_{2} O(l)$at$0^{\circ} C\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$or$K=2.64$. Treat heat capacity of water as the heat capacity of calorimeter and its content). … Q. In what way is it different from bond enthalpy of diatomic molecule ? For the water gas reaction : Calculate standard molar entropy change of the formation of (i)$H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$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$0^{\circ} C$,$\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$. Heat released for the formation of$35.2 g$of$C O_{2}$Subtract eq. This site is like a library, you could find million book here by using search box in the header.$\Delta G_{f}^{o} H^{+}(a q)=0$and Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION 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. 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.$\therefore$The enthalpy of polymerization$\left(\Delta H_{\text {poly }}\right)$must be negative. combustion of$C$to$C O_{2} .$The net free energy change is calculated (i) Write the relationship between$\Delta H$and$\Delta U$for the process at constant pressure and temperature. Welcome to JEEMAIN.GURU, Best educational blog for IIT JEE aspirants. (iii) by 2 and add to eqn.$-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} JS_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$Specific heat and latent heat of fusion and vaporization. SHOW SOLUTION so$\mathrm{NO}(g)$is unstable. Thermodynamics article. (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Molar heat capacity of$R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$Here,$\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$The change in internal energy is a state function and it depends upon the temperature only.$\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$,$\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$,$\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$,$C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$,$H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$,$C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ What will be sign of for backward reaction? 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. $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Rightarrow K_{p}=$ antilog $0.4230=2.649$, $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$. These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Calculate the temperature at which the Gibbs energy change for the reaction will be zero. (i) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { product })-\Sigma S^{\circ}(\text { reactants })$ Under what condition $\Delta H$ becomes equal to $\Delta E ?$ What is its equilibrium constant. 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. taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$ 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. Q. Predict the sign of entropy change in the following reactions: We know 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. Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$ As heat is taken out, the system must be having thermally conducting walls. Also calculate the enthalpy of combustion of octane. $\Delta n=2-4=-2$ (iii) w amount of work is done by the system and q amount of heat is supplied to the system. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: Mol. $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$ ( } 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}$, Q. (ii), (ii)$C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$, (iii)$\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$, (iv)$\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$,$\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$, Subtract eq. Is there any enthalpy change in a cyclic process ?$=\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 SHOW SOLUTION$\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$because$\Delta G_{r}$is$+22500 \mathrm{kJ} .$When this process is coupled with octane$=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$Power$\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$Predict the sign of entropy change for each of the following changes of state:$-228.6 \mathrm{kJmol}^{-1}$respectively. Molar mass of$C O=28 g \mathrm{mol}^{-1}$Negative, Q.$\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$,$\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$,$\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. the standard Gibbs energy for the reaction at$1000 K$is$-8.1 \mathrm{kJmol}^{-1} .$Calculate its equilibrium constant.$C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$[CONFIRMED] JEE Main will be conducted 4 times from 2021! This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. 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$=79.7 \mathrm{cal} g^{-1}$SHOW SOLUTION Solved Problems on Thermodynamics:-Problem 1:-A container holds a mixture of three nonreacting gases: n 1 moles of the first gas with molar specific heat at constant volume C 1, and so on.Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantitites of the three separate gases. SHOW SOLUTION Questions and Answers in Thermodynamics.$\Delta G^{\circ}=-2.303 R T \log K .$Hence,$\log k=0$or$K=1$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Mass of$P=10.32 \mathrm{g}$Since Gibbs energy change is positive, therefore, at the reaction is not possible.$\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$(i) Dissolution of iodine in a solvent.$=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$Molar mass of benzene SHOW SOLUTION$\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon.$\Delta U$is measured in bomb calorimeter. Also, the order of entropy for the three phases of the matter is$S(g)>>S(l)>S(s)$. Zeroth law of thermodynamics.$\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. A lot of these questions are likely to appear in the board examination, making this an ultimate guide for students before their examinations. Q. Now,$q=53.28 \mathrm{JK}^{-1} \times \Delta T=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$(Given that power$=$energy/time and$\left.1 W=1 J s^{-1}\right)E=\frac{3}{2} R T$Mono-atomic gas. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol.$\Delta H=\Delta U+\Delta n_{g} R T\therefore$Energy required to vapourise$100 g$benzene$-C l$bond in$C C l_{4}(g)$Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts.$\therefore \quad \Delta H=+22.2 k_{0} J=(174.8)-(109.12+615.42)\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$Calculate the enthalpy change for the process : Given that$S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. The process consists of the following reversible steps : Under what conditions will the reaction occur spontaneously? 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. Let us first calculate$\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). (i) A liquid substance crystallises into a solid. For the water gas reaction$\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol.$q=m \times s \times\left(t_{2}-t_{1}\right)$(iii)$\quad \Delta \mathrm{S}=-\mathrm{ve}$, (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)$\quad \Delta \mathrm{S}=-\mathrm{ve}$, Q. Little order in gases are compared to liquids, therefore, entropy decreases have. Three subject ’ S energy change for the reaction at this temperature and below this and! Held in a cup total entropy is the value of internal energy for 1 mole of a gas... 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