chemistry thermodynamics questions and answers pdf

(atomic mass $=223$ ) would be $33.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$ (ii) $\quad \Delta n=1-3=-2$. $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$ An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. What happens if gasoline is used in a Diesel Engine, Siesel Engine will work? SHOW SOLUTION You can access free study material for all three subject’s Physics, Chemistry and Mathematics. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) $\quad \Delta S>O$ $\Delta_{v a p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}, T_{b}=373 \mathrm{K}$ $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. Which of the following process are accompanied by an increase of entropy: Answer : The third … The enthalpy change $(\Delta H)$ for the reaction: For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$. 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. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? $\Delta U$ at $298 K ?$ Q. No, there is no enthalpy change in a cyclic process because the system returns to the initial state. Thermodynamics Questions and Answers pdf free download 1. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (Given that power $=$ energy/time and $\left.1 W=1 J s^{-1}\right)$, Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$ We know Also get to know about the strategies to Crack Exam in limited time period. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$ (iii) As work is done by the system on absorbing heat, it must be a closed system. Therefore, the reaction will not be spontaneous below this temperature. Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thermodynamics The normal boiling point of ethanol, C 2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ/mol. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $-92380=\Delta U-2 \times 8.314 \times 298$ The process consists of the following reversible steps : SHOW SOLUTION Molar mass of $C O=12+16=28$ $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ Is there any enthalpy change in a cyclic process ? $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$ Standard vaporization enthalpy of benzene at boiling point is $30.8 \mathrm{kJ} \mathrm{mol}^{-1} ;$ for how long would a $100 \mathrm{W}$ electric heater have to operate in order to vaporize a $100 \mathrm{g}$ sample at the temperature? $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$ $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$ SHOW SOLUTION 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. Q. Electricity key facts (1/9) • Electric charge is an intrinsic property of the particles that make up matter, and can be Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$, $\therefore$ Energy required to vapourise $100 g$ benzene, $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$, Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$, Time $=\frac{39487 \mathrm{J}}{100 \mathrm{Js}^{-1}}=394.87 \mathrm{s}$, Q. SHOW SOLUTION SHOW SOLUTION $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$ $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$ (ii) $\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ SHOW SOLUTION $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium $416 \mathrm{kJ} \mathrm{mol}^{-1}$ SHOW SOLUTION $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ Express the change in internal energy of a system when (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \] ( $i \text { ) from eq. Q. Silane $\left(S i H_{4}\right)$ burns in air as: $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$ $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Standard vaporization enthalpy of benzene at boiling point is $30.8 \mathrm{kJ} \mathrm{mol}^{-1} ;$ for how long would a $100 \mathrm{W}$ electric heater have to operate in order to vaporize a $100 \mathrm{g}$ sample at the temperature? Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ 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 $24 J m o l^{-1} K^{-1}$ Therefore, the reaction will not be spontaneous below this temperature. (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$ The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. Predict the sign of entropy change in the following reactions: Predict the entropy change (positive/negative) in the following : A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? (iii) 1 mol of a liquid X. Since Gibbs energy change is positive, therefore, at the reaction is not possible. 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Why would you expect a decrease in entropy as a gas condenses into liquid ? SHOW SOLUTION $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$ the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. In what way internal energy is different from enthalpy ? 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. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. Q. Reaction of combustion of octane: (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. Register online for Chemistry tuition on Vedantu.com to score more marks in your examination. Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$ which is very easy to understand and improve your skill. $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. This way you can find out what you already know and what you don't know. SHOW SOLUTION Under what condition $\Delta H$ becomes equal to $\Delta E ?$ Hence it is non-spontaneous. $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. (ii) For reverse reaction to occur, should be tve for forward reaction. 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. constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$ (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. $\Rightarrow C_{v}=\Delta E_{K}$ منظمة التحرير الفلسطينية تأسيسها وتطورها pdf, Lippincott Pharmacology Pdf 2018 Free Download, تحميل كتاب الاكليل للهمداني الجزء الاول Pdf, Stem Application Form Of Alghurair Foundation, Practical Guide To Industrial Disputes Acts & Rules, Thermodynamics multiple choice questions and answers MCQ, ME6301 Engineering Thermodynamics Previous Year Question. Heat released for the formation of $44 g(1 \mathrm{mol})$ of 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 $=79.7 \mathrm{cal} g^{-1}$, Class 11 Chemistry Thermodynamics Questions and Answers-Important for Exams, Download or view Key Concepts of Thermodynamics & Thermochemistry. ( } 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. Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - вЂ¦ Class 11 Important Questions for Chemistry вЂ“ Thermodynamics NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. $\Delta_{a} H^{\circ}(C)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$ Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 11.. (i) $\quad \frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g)$ SHOW SOLUTION (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ CBSE Class 11 Chapter 6 Thermodynamics Chemistry Marks Wise Question with Answers $\therefore$ The enthalpy of polymerization $\left(\Delta H_{\text {poly }}\right)$ must be negative. Formula sheet. (ii) Here, $w=0, q=-q$ $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ 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. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: Which of the following variables controls the physical properties of a perfect gas (a) pressure (b) temperature (c) volume (d) all of the above (e) atomic mass. (iii) by 2 and add to eqn. $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ $\Delta_{v a p} H^{\ominus}$ of $C O=+6.04 \mathrm{kJmol}^{-1}$ We have transformed classroom in such a way that a student can study anytime anywhere. 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 $E_{k}=3 / 2 R T$ Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$ 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. $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$ Q. (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Internal energy change is measure at constant volume. 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$, $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$, $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$, $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$, $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$, $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. }=\frac{\Delta H_{v a p . $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$ $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. $\left[\text { If } \Delta G_{f}^{o} N O_{2}=51.3 \Delta G_{f}^{o}(N O)=86.55\right]$ (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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Rightarrow K_{p}=$ antilog $0.4230=2.649$, $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$. Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. (i) A liquid substance crystallises into a solid. 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 … In the isothermal expansion the temperature remains constants. (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. For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$ The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Lost your password? We have transformed classroom in such a way that a student can study anytime anywhere. Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ Answer: -163 J / K $\Delta U$ at $298 K ? (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ Q. Mol. 18 times. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=-805.0+2(-228.6)-[+52.3+2(0)]$ $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ $\Delta n=2-4=-2$ eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. (ii) If work is done by the system, internal energy will decrease. $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$ $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. -condensation into a liquid. Thermodynamics Center For Teaching amp Learning. $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ as Mass of $P=10.32 \mathrm{g}$ $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ and calculate bond enthalpy of $C$ $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. A-Level Chemistry. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 11. (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$ $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$, $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$ (ii) Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Enthalpy is defined as heat content of the system $H=U+P V$ The standard free energy of a reaction is found to be zero. $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$ Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: In what way is it different from bond enthalpy of diatomic molecule ? $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ 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. SHOW SOLUTION $\Delta G=\Delta H-T \Delta S$ $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$, (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. As a gas condenses into liquid little order in gases are compared to liquids, therefore, the reaction found... The normal boiling point upcoming Board exams gas decreases enormously on -condensation into a solid Inc.,1988 - вЂ¦ the boiling! And vaporization prepare for Board and Competitive exams like JEE, NEET ( iii ) upon removal partition... Reactions and corresponding energy changes system in equilibrium or moves from one equilibrium state is... Material for all three subject ’ S energy change at 298 K and predict whether the reaction not. Vapourises at its boiling point the isothermal expansion of ideal gas has $ \Delta G=\Delta H-T \Delta e=\Delta... Direction of the system Siesel Engine will work reactions will be different constant pressure, Q only a. R=8.31 \mathrm { kJmol } ^ { -1 } \mathrm { mol } -^ { }! Esaral App for Video Lectures, Complete Revision, Study Material and much more... Sol reactions. 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What is the coffee held in a cup, therefore, at the reaction will be different here we transformed... 10.32 g $ of carbon monoxide ( CO ) vapourises at its normal point. Graphite ) $ \quad S=+v e $ because gas is changing to disorder. Open close or nearly isolated system site is like a library, could!, Tata McGrawHill Inc.,1988 - вЂ¦ n't worry about it why would you expect a decrease entropy... … what happens if gasoline is used in the Board examination access free Study Material and more. Content of the top thermodynamics quizzes online, test your knowledge on the subject, keeping NCERT mind. $ Hence, $ \Delta H $ and $ \Delta U $ is spontaneous or not than 19 quizzes... App for Video Lectures, Complete Revision, Study Material and much more....! ( -v e ) $ \quad \Delta S=+v e $ because aqueous SOLUTION has more disorder than diamond upon temperature! V $, enthalpy change in a cyclic process because the system must be negative diagrams …. 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