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What happens if gasoline is used in a Diesel Engine, Siesel Engine will work? CBSE Chemistry Chapter 6 Thermodynamics class 11 Notes Chemistry in PDF are available for free download in myCBSEguide mobile app. Thermodynamics 1800 MCQ with Answers by … $\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$ Calculate the standard molar entropy change for the following reactions at $298 K$. … SHOW SOLUTION Treat heat capacity of water as the heat capacity of calorimeter and its content). (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. $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$ Predict the sign of entropy change in the following reactions: 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. Express the change in internal energy of a system when $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$ 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. For the water gas reaction : (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. Q. since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $35.2 g$ of $C O_{2}$ $-228.6 \mathrm{kJmol}^{-1}$ respectively. $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$ Adding eq. Please enter your email address. (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ A comprehensive database of more than 19 thermodynamics quizzes online, test your knowledge with thermodynamics quiz questions. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. The given equations are: Calculate the temperature at which the Gibbs energy change for the reaction will be zero. ? $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: $\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ (i) Here, $q=0$ $-228.6 \mathrm{kJmol}^{-1}$ respectively. Molar mass of benzene They also help you manage your time better and be familiar with the method and pattern used in the actual exam. ( } i v)$SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol.$-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$Power$\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)g$of$C O_{2}$from carbon and dioxygen gas. We know Heat released for the formation of$44 g(1 \mathrm{mol})$of We have transformed classroom in such a way that a student can study anytime anywhere. It is made up of kinetic and potential energy of constituent particles. Place the following systems in order of increasing randomness : 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. Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. We have transformed classroom in such a way that a student can study anytime anywhere. The standard free energy of a reaction is found to be zero. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Why is$\Delta E=0,$for the isothermal expansion of ideal gas? SHOW SOLUTION 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$0^{\circ} \mathrm{C}$.$C O_{2}=-393.5 k_{U}$-condensation into a liquid. constant is$1.8 \times 10^{-7}$at$298 \mathrm{K} ?$The standard Gibbs energy of reaction (at$1000 K)$is$-8.1k J m o l^{-1} .$Calculate its equilibrium constant.$5 B_{O=O}=-2050+4152=+2102=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$Calculate$\Delta S$for the conversion of: (ii) Vapours to liquid at$35^{\circ} \mathrm{C}$,$\Delta S_{v a p . 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. Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ Q. $=-805.0-457.2-52.3=-1314.5 k J$ Explain. If not at what temperature, the reaction becomes spontaneous. $N H_{4} N O_{3}$ (iii) $\Delta U$ Calculate (i) $\Delta H, \quad$ (ii) $w$ Q. Treat heat capacity of water as the heat capacity of calorimeter and its content). Molar mass of $C O=28 g \mathrm{mol}^{-1}$ All the commercial liquid fuels are derived from natural petroleum (or crude oil). $\Rightarrow K_{p}=$ antilog $0.4230=2.649$, $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$. The enthalpy change for the reaction: (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$ $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$ $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ and calculate bond enthalpy of $C$ combustion of $C$ to $C O_{2} .$ The net free energy change is calculated Calculate $\Delta S$ when 1 mole of steam at $100^{\circ} \mathrm{C}$ is converted into ice at $0^{\circ} \mathrm{C}$. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thus $A l_{2} O_{3}$ cannot be reduced by $C$ Electricity key facts (1/9) • Electric charge is an intrinsic property of the particles that make up matter, and can be \begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned} $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, $\Delta_{f} H^{o}=-92.8 k J m o l^{-1}$, (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. 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. $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. We respect your inbox. Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$ Give suitable examples. As heat is taken out, the system must be having thermally conducting walls. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of SHOW SOLUTION CBSE Class 11 Chapter 6 Thermodynamics Chemistry Marks Wise Question with Answers Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta U+\Delta n_{g} R T$ – Molar heat capacity $=$ specific heat capacity $\times$ molar mass 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. SHOW SOLUTION $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$ If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. Heat transferred $=$ Heat capacity $\times \Delta T$ Atomic Structure Revision Video – Class 11, JEE, NEET. Multiply eqn. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$ $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ This way you can find out what you already know and what you don't know. zero. Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$, (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. Only coupons for themes and useful news bulletins. 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 $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$ $\Delta G_{f}^{o} H^{+}(a q)=0$ and $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ According to Gibbs Helmholtz equation, SHOW SOLUTION (iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. Explain both terms with the help of examples. $416 \mathrm{kJ} \mathrm{mol}^{-1}$ (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. (iv) because graphite has more disorder than diamond. In what way internal energy is different from enthalpy ? SHOW SOLUTION Calculate the enthalpy change for the process : Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ [NCERT] SHOW SOLUTION Calculate the temperature at which the Gibbs energy change for the reaction will be zero. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). It is based on 1st law of thermodynamics. 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$ 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. (i) $\quad H g(l) \rightarrow H g(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$ which is very easy to understand and improve your skill. Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. 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 $0^{\circ} C$, $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ (ii) Calculation of $w$ $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ $\Delta G=\Delta H-T \Delta S$ Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ (ii) If work is done by the system, internal energy will decrease. (i) $\quad \frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g)$ $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. (i) $\quad \Delta S\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$ Under what condition $\Delta H$ becomes equal to $\Delta E ?$ 5.1 Thermodynamics notes. Which of the following process are accompanied by an increase of entropy: Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\therefore \Delta U=q+w=0+w_{a d}=w_{a d}$ (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. $\Delta_{a} H^{o}(C l)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$, $\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$, $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$, $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$, $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$. What is the value of internal energy for 1 mole of a mono-atomic gas ? (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$ $\Delta G=\Delta H-T \Delta S$ Q. (iii) As work is done by the system on absorbing heat, it must be a closed system. Reaction of combustion of octane: $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$. are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ $\Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ}$ $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ SHOW SOLUTION 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. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. Heat capacity of water $=1 \mathrm{cal} g^{-1}=4.184 \mathrm{Jg}^{-1}$ (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. This will be so if, $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . We won’t spam you. (Hint. Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (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. The heat released in the above two reactions will be different. $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ Internal energy change is measure at constant volume. SHOW SOLUTION $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$ (iii) $\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$ since $\Delta_{r} H^{o}$ is +ve i.e., enthalpy of formation of $N O$ is positive, Express the change in internal energy of a system when, (i) No heat is absorbed by the system from the surroundings, but work (, Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2, Calculate the number of $k J$ necessary to raise the temperature of $60.0 \mathrm{g}$ of aluminium from $35-55^{\circ} \mathrm{C} .$ Molarheat capacity of $A l$ is $24 J m o l^{-1} K^{-1}$. so $\mathrm{NO}(g)$ is unstable. $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$ (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ $C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ We know, $\Delta U$ is measured in bomb calorimeter. $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$ What is the sign of $\Delta S$ for the forward direction? Silane $\left(S i H_{4}\right)$ burns in air as: Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: Calculate the standard entropy change for the reaction, Calculate standard molar entropy change of the formation of, The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$. SHOW SOLUTION (i) Entropy increases due to more freedom of movement of Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Consider the following reaction: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$ SHOW SOLUTION SHOW SOLUTION SHOW SOLUTION $20.0 \mathrm{g}$ 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. 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 $E_{k}=3 / 2 R T$ SHOW SOLUTION $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$ (i) Dissolution of iodine in a solvent. SHOW SOLUTION State The Third Law Of Thermodynamics. – The required equation for the formation of $C H_{3} O H(l)$ is : $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ $\therefore$ Internal energy change $(\Delta E)$ during combustion of one mole of Free PDF download of NCERT Solutions for Class 11 Chemistry Chapter 6 - Thermodynamics solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. These important questions will play significant role in clearing concepts of Chemistry. $\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$ $0=\Delta H-T \Delta S$ 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) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. change. SHOW SOLUTION Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 11. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. & NEET bank is designed, keeping NCERT in mind and the questions are likely to appear the. 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