Question2: Explain the laws of Thermodynamics.? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. Students can avail this solution at their convenience without hassle. $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$ For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$ Calculate Gibbs energy change for the reaction is spontaneous or not. The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$ $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$, \[ \Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ} \], Thus, calorimeter loses $5.282 \mathrm{kJ}$ of heat during dissolution, of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of, $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$, Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$, Q. [NCERT] Compute the molar heat capacity of these elements and identify any periodic trend. SHOW SOLUTION 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. $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$ (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. .$ in $H_{2}$ molecule $436 k J$ is bond enthalpy. (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ – The reaction will occur spontaneously only when $T \Delta S>\Delta H$ $=22.2-1.737=20.463 \mathrm{kJ}$, (iii) $\Delta U$ Calculate (i) $\Delta H, \quad$ (ii) $w$, since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$, $\therefore \quad \Delta H=+22.2 k_{0} J$, Mol. (i) $\frac{1}{2} N_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow N O(g) ; \Delta_{r} H^{\circ}=90 k J \operatorname{mol}^{-1}$, (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. But $\mathrm{NO}_{2}(g)$ is formed. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$ Treat heat capacity of water as the heat capacity of calorimeter and its content). $\Delta G^{\circ}=-2.303 R T \log K .$ Hence, $\log k=0$ or $K=1$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Fast response time: Used only for emergencies when speed is the single most important factor. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) If work is done on the system, internal energy will increase. SHOW SOLUTION $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. How many times is molar heat capacity than specific heat capacity of water ? $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ $=(174.8)-(109.12+615.42)$ $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$ Chemistry: The Molecular Science (5th Edition) answers to Chapter 18 - Thermodynamics: Directionality of Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 737b 30a including work step by step written by community members like you. 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. 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 $=79.7 \mathrm{cal} g^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. If it's not in your inbox, check your spam folder. Also calculate the enthalpy of combustion of octane. $S^{\circ}(F e(s))=27.28 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ $\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$ Internal energy change is measure at constant volume. In what way is it different from bond enthalpy of diatomic molecule ? 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. These important questions will play significant role in clearing concepts of Chemistry. Q. 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. For, $\Delta G=0$ (ii) $\quad \Delta S>O$ $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, $q=125 g \times 4.18 J / g \times(286.4-296.5)$, $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, Q. Warning: If you try using the HL in an unethical manner, expect to fail your class. Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). $\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}$ (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. What is meant by average bond enthalpy ? as Therefore, the reaction will not be spontaneous below this temperature. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. Automobile radiator system is analyzed as closed system. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$ This is only a preview of the solution. $\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$ Thermodynamics. Since Gibbs energy change is positive, therefore, at the reaction is not possible. (ii) Below this temperature, $\Delta G$ will be $+$ve because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$ (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$ eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. SHOW SOLUTION Q. 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. The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$ Red phosphorus reacts with liquid bromine as: Thus $A l_{2} O_{3}$ cannot be reduced by $C$, (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$, $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$, $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$, On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$, Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$. $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$ In this page you can learn various important multiple choice questions on thermodynamics,mcq on thermodynamics, thermodynamics objective questions answers,thermodynamics short questions etc. Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$ Under what conditions will the reaction occur spontaneously? 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 $=79.7 \mathrm{cal} g^{-1}$ $416 \mathrm{kJ} \mathrm{mol}^{-1}$ $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$ (Last Updated On: February 22, 2020) This is the complete list of Multiples Choice Questions Series in Thermodynamics as one of the General Engineering and Applied Sciences (GEAS) topic. [NCERT] SHOW SOLUTION $\Rightarrow K_{p}=$ antilog $0.4230=2.649$, $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$. combustion of $C$ to $C O_{2} .$ The net free energy change is calculated $\Delta H^{\circ}=\Delta_{a} H^{\circ}(C)+4 \Delta_{a} H^{\circ}(C l)-\Delta_{f} H^{\circ}\left(C C l_{4}\right)(g)$ If not at what temperature, the reaction becomes spontaneous. As there is little order in gases are compared to liquids, therefore, entropy of gas decreases enormously on $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ Also calculate enthalpy of solution of ammonium nitrate. SHOW SOLUTION SHOW SOLUTION Q. Upload a file Which formula forms a link between the Thermodynamics and Electro chemistry? SHOW SOLUTION $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$ (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$ Compare it with entropy decrease when a liquid sample is converted into a solid. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). 5.1 Test (mark scheme) More Exam Questions on 5.1 Thermodynamics (mark scheme) 5.1 Exercise 1 - calculating approximate enthalpy changes 5.1 Exercise 2 - Born … $\therefore \quad \Delta H=+22.2 k_{0} J$ Isolated system : (vi) Coffee in thermos flask. Calculate the standard molar entropy change for the following reactions at $298 K$. $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$ Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$ Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: What type of wall does the system have ? 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 $\left(35^{\circ} \mathrm{C}\right)$ Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. (iv) $\quad \Delta S>O$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION (ii) At what temperature, the reaction will reverse? Q. $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. $C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$ Multiply eqn. Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ – Molar heat capacity $=$ specific heat capacity $\times$ molar mass Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$ Thus, entropy increases. 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. g . 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. A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? Visit eSaral Website to download or view free study material for JEE & NEET. What is the change in internal energy for the process? 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 … 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. At equilibrium $\Delta G=0$ so that Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta U$ is measured in bomb calorimeter. (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. No, there is no enthalpy change in a cyclic process because the system returns to the initial state. We also acknowledge previous National Science … $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$ $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$ Chemical Thermodynamics Example 9.2 The element mercury, Hg, is a silvery liquid at room temperature. SHOW SOLUTION Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ 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. Give suitable examples. to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. $\Delta G=\Delta H-T \Delta S$ (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) Due to settling of solid $A g C l$ from solution, entropy decreases. $\Delta G_{f}^{o} H^{+}(a q)=0$ and In this no mass (water) cross the boundary. – The required equation for the formation of $C H_{3} O H(l)$ is : MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. For the water gas reaction : Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$, Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$, Enthalpy change for 2 mole of $P=-243 \mathrm{kJ}$, Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$, Q. Thermodynamics key facts (4/9) ... • Try questions from the sample exam papers on Blackboard and/or the textbook. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. Predict the sign of entropy change for each of the following changes of state: In what way is it different from bond enthalpy of diatomic molecule ? $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ 5 Penn Plaza, 23rd Floor SHOW SOLUTION (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. – For a spontaneous process, $\Delta G<0 .$ Also entropy change (\DeltaS) during polymerization is negative. SHOW SOLUTION $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ We have transformed classroom in such a way that a student can study anytime anywhere. $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . $g$ of $C O_{2}$ from carbon and dioxygen gas. Explain both terms with the help of examples. Q2: Which law of thermodynamics evaluate thermodynamic parameters? 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. Question3: Explain the following terms: Isolated system, Open system and closed system and give example where … $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ chapter 02: work and heat. Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Will the heat released be same or different in the following two reactions : Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. zero. [NCERT] $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ You will get a negotiable price quote with no obligation. What type of wall does the system have? Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. SHOW SOLUTION $\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ 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. Given that $S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$ $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$, (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$ Q. – oxygen bond in $\mathrm{O}_{2}$ molecules. $\Delta G^{\circ}=-2.303 R T \log K$ $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. SHOW SOLUTION $-228.6 \mathrm{kJmol}^{-1}$ respectively. ? $20.0 \mathrm{g}$ 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Welcome to 5.1 THERMODYNAMICS. (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. (Hint. Our 1000+ Thermodynamics questions and answers focuses on all areas of Thermodynamics covering 100+ topics. (iv) because graphite has more disorder than diamond. According to Gibbs Helmholtz equation, $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$ $q=m \times s \times\left(t_{2}-t_{1}\right)$ The standard free energy of a reaction is found to be zero. $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ and calculate bond enthalpy of $C$ (ii) Calculation of $w$ (i) A liquid substance crystallises into a solid. SHOW SOLUTION $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$ In the isothermal expansion the temperature remains constants. $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$ On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$ 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. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\therefore$ Energy required to vapourise $100 g$ benzene They will be ignored! Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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 $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$ (ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$ $\Delta H=\Delta U+P \Delta V$. Normal response time: Our most experienced, most successful tutors are provided for maximum expertise and reliability. The process consists of the following reversible steps : Formula sheet. $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ 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. -condensation into a liquid. are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. \[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \] (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. [NCERT] $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. For a reaction both $\Delta H$ and $\Delta S$ are positive. SHOW SOLUTION You can access free study material for all three subject’s Physics, Chemistry and Mathematics. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$ $-\Delta H_{\text {reaction }}=-2.05 \times 10^{3} \mathrm{kJmol}$ $\Delta_{v a p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}, T_{b}=373 \mathrm{K}$ $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$ 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 $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$, 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. 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