Heat capacity of water $=1 \mathrm{cal} g^{-1}=4.184 \mathrm{Jg}^{-1}$ $\Rightarrow C_{v}=\Delta E_{K}$ SHOW SOLUTION (iv) because graphite has more disorder than diamond. $-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$ The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. (ii) $\quad \Delta n=1-3=-2$. $\Delta H=\Delta U+P \Delta V$ Which of the following are open close or nearly isolated system ? since $\Delta_{r} H^{o}$ is +ve i.e., enthalpy of formation of $N O$ is positive, $\Delta G_{f}^{\circ} C H_{4}(g)=-50.72 \mathrm{kJ} \mathrm{mol}^{-1}$ and $\Delta \mathrm{G}_{f}^{\circ} \mathrm{O}_{2}(g)=0$ $\Delta_{a} H^{\circ}(C)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$ 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. Reaction of combustion of octane: (ii) For reverse reaction to occur, should be tve for forward reaction. The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$ 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 $24 J m o l^{-1} K^{-1}$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$ Q. (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$ (i) Here, $q=0$ These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. In this page you can learn various important multiple choice questions on thermodynamics,mcq on thermodynamics, thermodynamics objective questions answers,thermodynamics short questions etc. 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. A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ You can access free study material for all three subject’s Physics, Chemistry and Mathematics. SHOW SOLUTION 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. SHOW SOLUTION $=270.2-17.22-522.72=-269.72 J K^{-1} \mathrm{mol}^{-1}$, gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$, $3 C(\text { graphite })+4 H_{2}(g) \rightarrow C_{3} H_{8}(g)$, Given that $S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$, $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$, $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$, $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$, $=270.2-17.22-522.72=-269.72 J K^{-1} \mathrm{mol}^{-1}$, Q. PV diagrams - … Under what conditions will the reaction occur spontaneously? SHOW SOLUTION (ii) If work is done by the system, internal energy will decrease. $\Delta n=2-4=-2$ Red phosphorus reacts with liquid bromine as: A $1.250 \mathrm{g}$ sample of octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ is burned in excess of oxygen in a bomb calorimeter. $-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$ $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ What type of wall does the system have? $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$ For a reaction both $\Delta H$ and $\Delta S$ are positive. $E=\frac{3}{2} R T$ Mono-atomic gas. $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$ Internal energy change is measure at constant volume. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad \Delta H=\Delta U+\Delta n R T$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. [NCERT] $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: $\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$ (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION \quad$, $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$, $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, Q. SHOW SOLUTION Q. 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 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 11. since $\Delta_{r} G^{\circ}$ is negative, the reaction will be spontaneous. (iii) A partition is removed to allow two gases to mix. Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$ Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ $-(1) \quad \Delta S=-v e$ $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ $C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ You are on page 1 of 14. Download Thermodynamics MCQ Question Answer PDF « (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ 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. Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - …. Physical Chemistry Chemical Thermodynamics Dr. R. R. Misra Reader in Chemistry Hindu College, Delhi – 110007 E- mail: rrmisra@hotmail.com CONTENTS ... We get the answer of the first two questions by the study of thermodynamics, while third question forms the domain of the study of chemical kinetics. SHOW SOLUTION Thermodynamics article. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$ Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$ $-228.6 \mathrm{kJmol}^{-1}$ respectively. False. $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$ The temperature of calorimeter rises from $294.05 K$ to $300.78 K .$ If the heat capacity of calorimeter is $8.93 \mathrm{kJK}^{-1}$, calculate the heat transferred to the calorimeter. combustion of $C$ to $C O_{2} .$ The net free energy change is calculated We have transformed classroom in such a way that a student can study anytime anywhere. Questions and Answers in Thermodynamics. This will be so if $T<300.3 \mathrm{K}$, (ii) For reverse reaction to occur, should be tve for forward reaction. $=0.5134 \mathrm{kJ}$ Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $I_{2}$ molecules upon dissolution. Heat released for the formation of $35.2 g$ of $C O_{2}$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. SHOW SOLUTION (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$, (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$, Q. $\Delta H=\Delta U+P \Delta V$. Enthalpy change is measured at constant pressure Here, we are given, $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$, Q. 1 Mole of solid $x<1$ mole of liquid $x<1$ mole of gas, 1 Mole of solid $x<1$ mole of liquid $x<1$ mole of gas. Predict the sign of entropy change for each of the following changes of state: Why would you expect a decrease in entropy as a gas condenses into liquid ? 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. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. (ii) Temperature of crytal is increased. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). Free PDF download of NCERT Solutions for Class 11 Chemistry Chapter 6 - Thermodynamics solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$ (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$ (iii) w amount of work is done by the system and q amount of heat is supplied to the system. $\Delta H$ and $\Delta S$ for the reaction: SHOW SOLUTION Therefore $\Delta E=0$ under isothermal conditions. (ii) Calculation of $w$ (i) If work is done on the system, internal energy will increase. $\Delta S_{v a p . For the reaction $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$ $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$ $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$ $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ Isolated system : (vi) Coffee in thermos flask. octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$ Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ (iv) because graphite has more disorder than diamond. Molar mass of $C O=28 g \mathrm{mol}^{-1}$ JEEMAIN.GURU is a free educational site for students, we started jeemain.guru as a passion now we hope that this site would help students to find their required study materials for free. The given equations are: Explain. $\Delta T=300.78-294.05=6.73 K$ SHOW SOLUTION 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. Also calculate enthalpy of solution of ammonium nitrate. The enthalpy change for the reaction: (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$ 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. the condensation of diethyl ether is the reverse process, therefore, The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. ( $i \text { ) from eq. $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$ Now, forward reaction is exothermic, therefore the entropy change for the forward direction should be negative and large $(T \Delta S>\Delta H)$, Q. $=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$ (iv) $\quad \Delta S>O$. Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. 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. [NCERT] $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. The chemical energy can be transformed into other forms of energy, e.g. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of What is the sign of $\Delta S$ for the forward direction? Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thermodynamics One Word Interview Questions and Answers PDF _ Mechanical Engineering Questions and Answers. Calculate the standard entropy change for the reaction $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ $g$ of $C O_{2}$ from carbon and dioxygen gas. (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ Molar heat capacity of $R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $\frac{109}{18} \times 36=218 \mathrm{JK}^{-1}$, $\Delta_{v a p} S=\frac{\Delta_{v a p} H}{T_{b}}=\frac{40.63 \times 1000 \mathrm{J} \mathrm{mol}^{-1}}{373 \mathrm{K}}=109 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Entropy change for evaporation of $36 g$ of wate, $\frac{109}{18} \times 36=218 \mathrm{JK}^{-1}$. $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. $\Delta U$ is measured in bomb calorimeter. zero. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$ $\Delta H^{\circ}=\Delta_{a} H^{\circ}(C)+4 \Delta_{a} H^{\circ}(C l)-\Delta_{f} H^{\circ}\left(C C l_{4}\right)(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. A.True. Therefore, the decrease in entropy when a gas condenses into a liquid is much more as compared to decrease in entropy when a liquid solidifies. Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$, (i) $\quad C H_{4}(g)+2 O_{2}(g) \rightarrow C O_{2}(g)+2 H_{2} O(g)$, $\Delta G_{f}^{o} \mathrm{CO}_{2}(g)=-394.36 \mathrm{kJ} \mathrm{mol}^{-1}$, $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$, $\Delta G_{f}^{\circ} C H_{4}(g)=-50.72 \mathrm{kJ} \mathrm{mol}^{-1}$ and $\Delta \mathrm{G}_{f}^{\circ} \mathrm{O}_{2}(g)=0$, (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$, $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$, $\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}$, $\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$, $\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$, $-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$, $=[-394.36+\{2 \times(-228.57)\}-[-50.72+0]$, (ii) $\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$, $\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]$, $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$, (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$, $-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(s)+2 \Delta G_{f}^{\circ} H^{+}(a q)\right]$, Q. $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$ Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. 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 $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ (Hint. (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is (iii) $\quad \Delta S>O$ 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. (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. Electricity key facts (1/9) • Electric charge is an intrinsic property of the particles that make up matter, and can be $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (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. $-92380=\Delta U-2 \times 8.314 \times 298$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(l)$ 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. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is Mol. SHOW SOLUTION 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. $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$ $\therefore$ Energy required to vapourise $100 g$ benzene Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero In this unit we shall focus our (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ 18 times. (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) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. All Chapter 6 - Thermodynamics Exercises Questions with Solutions to help you to revise complete Syllabus and … But $\mathrm{NO}_{2}(g)$ is formed. Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. Entropy change for evaporation of $36 g$ of wate $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $A+B \rightarrow C+D$ $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ Question from very important topics are covered by NCERT Exemplar Class 11. (iii) by 2 and add to eqn. SHOW SOLUTION (ii) At what temperature, the reaction will reverse? What is meant by average bond enthalpy ? $\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$ $\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$ or $K=2.64$. $\Delta G=120-380=-260 k J$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Predict the sign of entropy change for each of the following changes of state: Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. Q. (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$ $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium, constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$, $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$, Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$, $\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$, $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. [CONFIRMED] JEE Main will be conducted 4 times from 2021! Q. In what way is it different from bond enthalpy of diatomic molecule ? We know $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. $\Delta n_{g}=2-(1+3)=-2 m o l, T=298 K$ $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$ Also get to know about the strategies to Crack Exam in limited time period. 2000 AP CHEMISTRY FREE RESPONSE QUESTIONS. $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$ What happens to the internal energy of the system if: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $q=125 g \times 4.18 J / g \times(286.4-296.5)$ SHOW SOLUTION (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. [NCERT] $0=\Delta H-T \Delta S$ 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 $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$. According to Gibbs Helmholtz equation, Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. Multiply eqn. Compare it with entropy decrease when a liquid sample is converted into a solid. (ii) Calculate $\Delta S$ for the conversion of: (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$ $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$ SHOW SOLUTION [NCERT] $\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ What happens to the internal energy of the system if: Will the heat released be same or different in the following two reactions : How many times is molar heat capacity than specific heat capacity of water ? (i) A liquid substance crystallises into a solid. 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 $-v e$. Question from very important topics are covered by NCERT Exemplar Class 11.You also get idea about the type of questions and method to answer in your Class 11th … Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$ This question bank is designed, keeping NCERT in mind and the questions are updated with respect to … $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=22.2-1.737=20.463 \mathrm{kJ}$, (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. (Given that power $=$ energy/time and $\left.1 W=1 J s^{-1}\right)$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$. Practice: Thermodynamics questions. What will be sign of for backward reaction? That will be very helpful in quick revision during Exams. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$ (i) $\quad \Delta S

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