chemistry thermodynamics questions and answers pdf

(iv) $\quad \Delta S>O$. Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ Q. For, $\Delta G=0$ $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $44 g(1 \mathrm{mol})$ of What type of wall does the system have? In diatomic molecule bond enthalpy has fixed value, $e . Question from very important topics are covered by NCERT Exemplar Class 11. ( i ) and (ii) Also calculate enthalpy of solution of ammonium nitrate. Q. It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. 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 $E_{k}=3 / 2 R T$ $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$ Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. 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. Adding $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{o}=-105.0 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta U$ at $298 K ? Calculate the standard molar entropy change for the following reactions at $298 K$. $-(1) \quad \Delta S=-v e$ (iii) $\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$ (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION (i) A liquid substance crystallises into a solid. zero. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g) \Delta H=-92.38 k_{\circlearrowright}$ (ii) $\quad \Delta S=+v e$ [NCERT] $\Delta U$ at $298 K ?$ gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$ SHOW SOLUTION (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$ 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. $T_{b}=35+273=308 K$ 18 times. $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$, $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. ( } 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. $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, $q=125 g \times 4.18 J / g \times(286.4-296.5)$, $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, Q. Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$ Calculate Gibbs energy change for the reaction is spontaneous or not. Why would you expect a decrease in entropy as a gas condenses into liquid ? strategies to Crack Exam in limited time period. (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. Calculate standard molar entropy change of the formation of $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$ $\Delta G=\Delta H-T \Delta S$ where $\Delta G, H$ and $\Delta S$ are free energy change, enthalpy change and entropy change respectively. The enthalpy change for the reaction: Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - …. SHOW SOLUTION It is made up of kinetic and potential energy of constituent particles. (i) $\frac{1}{2} N_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow N O(g) ; \Delta_{r} H^{\circ}=90 k J \operatorname{mol}^{-1}$ (Hint. of water vaporised $=\frac{10}{18}=0.56$ $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ What will be sign of for backward reaction? $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. 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 $24 J m o l^{-1} K^{-1}$ First method: by using the relation Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ You may assume that the gas constant R = 8.314 J mol-1. spontaneous. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. $\Delta H=\Delta U+\Delta n_{g} R T$ We have transformed classroom in such a way that a student can study anytime anywhere. (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$ $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p . – oxygen bond in $\mathrm{O}_{2}$ molecules. The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$ (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$ The enthalpy change $(\Delta H)$ for the reaction: $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ (i) If work is done on the system, internal energy will increase. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. [NCERT] 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. Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero, $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$, $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$, $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$, Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$, $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, 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 If not at what temperature, the reaction becomes spontaneous. A-Level Chemistry. $20.0 \mathrm{g}$ 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. The chemical energy present in a molecule is released in various reactions. This question bank is designed, keeping NCERT in mind and the questions are updated with respect to … The carefully crafted questions and answers provide students with a comprehensive understanding of the chapters involved. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Molar mass of benzene As there is little order in gases are compared to liquids, therefore, entropy of gas decreases enormously on (atomic mass $=223$ ) would be $33.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$ We respect your inbox. Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ SHOW SOLUTION $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$ (ii) So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ or $\log K=\frac{-\Delta_{r} G}{2.303 R T}$ Thermodynamics 1800 MCQ with Answers by … Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What type of system would it be ? SHOW SOLUTION Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ Calculate the temperature at which the Gibbs energy change for the reaction will be zero. ( $i \text { ) from eq. 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 the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Order of increasing randomness Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero Q. $-C l$ bond in $C C l_{4}(g)$ SHOW SOLUTION $\frac{-393.5 \times 35.2}{44}=-314.8 k J$. $C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$ Which of the following are open close or nearly isolated system ? SHOW SOLUTION Q. SHOW SOLUTION Enthalpy is defined as heat content of the system $H=U+P V$ Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: SHOW SOLUTION (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$ $q=c \times \Delta T, \quad c=n \times C_{m}$ $R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$ Therefore, the reaction will not be spontaneous below this temperature. (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. (iii) As work is done by the system on absorbing heat, it must be a closed system. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. In what way internal energy is different from enthalpy ? Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$ 2000 AP CHEMISTRY FREE RESPONSE QUESTIONS. $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$, (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=-805.0+2(-228.6)-[+52.3+2(0)]$ The equilibrium constant for a reactions is $10 .$ What will be the value of $\Delta G^{\circ} ?$ SHOW SOLUTION Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: Calculate the enthalpy change for the process : Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ Molar mass of $C O=28 g \mathrm{mol}^{-1}$ This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. [NCERT] $A+B \rightarrow C+D$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Heat released for the formation of $35.2 g$ of $C O_{2}$ What is the change in entropy in the system when 68.3 g of C 2H5OH( g) at 1 atm condenses to liquid at the normal boiling point? Subtract eq. Free PDF download of Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics prepared by expert Chemistry teachers from latest edition of CBSE(NCERT) books. The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$ \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. What is Gibb’s Helmholtz equation ? (iii) 1 mol of a liquid X. Hence it is non-spontaneous. $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$, (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$, (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$, Multiply eqn. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) Calculate the value of $\Delta n$ in the following reaction: $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$. $N H_{4} N O_{3}$ Explain both terms with the help of examples. Will the heat released be same or different in the following two reactions : $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ 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 $\left(35^{\circ} \mathrm{C}\right)$ combustion of $C$ to $C O_{2} .$ The net free energy change is calculated 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. The enthalpy change $(\Delta H)$ for the reaction: For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$. Q. Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. What will be sign of for backward reaction? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{v a p} H^{\ominus}$ of $C O=+6.04 \mathrm{kJmol}^{-1}$ (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ (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. 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. SHOW SOLUTION $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. Q. What kind of system is the coffee held in a cup ? (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$ In the reaction $C_{3} H_{8}(g)+5 O_{2} \rightarrow 3 C O_{2}+4 H_{2} O(g) \quad$ if Thermodynamics Center For Teaching amp Learning. $0=\Delta H-T \Delta S$ 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 $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . $\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. As no heat is absorbed by the system, the wall is adiabatic. constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$ ( $i \text { ) from eq. \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \] What type of system would it be ? (i) At what temperature the reaction will occur spontaneously from left to right? $C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$ We know (i) Entropy increases due to more freedom of movement of $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ A lot of these questions are likely to appear in the board examination, making this an ultimate guide for students before their examinations. $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$ (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. What is the sign of $\Delta S$ for the forward direction? (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \] (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. (i) $\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(l)$ Thermodynamics Multiple choice Questions and answers pdf,mcqs,objective type questions,lab viva manual technical basic interview questions download Skip to content Engineering interview questions,Mcqs,Objective Questions,Class Notes,Seminor topics,Lab Viva Pdf free download. (ii) $\quad \Delta S>O$ With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Thermodynamics is the study of energy transformations. $=-5.62 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$ $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ (ii) If work is done by the system, internal energy will decrease. $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ 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. $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ Q. 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. Formula sheet. (Hint. Jump to Page . SHOW SOLUTION $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ SHOW SOLUTION $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$, $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$. $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$ $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ 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. (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$ Give suitable examples. – oxygen bond in $\mathrm{O}_{2}$ molecules. $-228.6 \mathrm{kJmol}^{-1}$ respectively. Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - … Class 11 Important Questions for Chemistry – Thermodynamics NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. (ii) At what temperature, the reaction will reverse? Predict the sign of entropy change for each of the following changes of state: $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$, $S^{\circ}(F e(s))=27.28 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$, $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$, (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$, $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$, $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$, $S^{\circ} \mathrm{Ca}(\mathrm{OH})_{2}(a q)=-74.50 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$, $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$, $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$, $=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$, $=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$, $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$, $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, Q. Visit eSaral Website to download or view free study material for JEE & NEET. What is its equilibrium constant. NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. Specific heat and latent heat of fusion and vaporization. (iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . Questions Answer PDF Download. MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. What type of wall does the system have ? 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 Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. Our online thermodynamics trivia quizzes can be adapted to suit your requirements for taking some of the top thermodynamics quizzes. 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. (ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$ SHOW SOLUTION (i) $\quad C H_{4}(g)+2 O_{2}(g) \rightarrow C O_{2}(g)+2 H_{2} O(g)$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ As per the second law of thermodynamics god exist, some person have believe that the world is so beautiful and only god can construct such beautiful world.Some scientist believe that the DNA structure is very much complicated and only god can create it.But as per the one scientist that god exist out of the our knowledge boundary. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$ $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$ But $\mathrm{NO}_{2}(g)$ is formed. $E=\frac{3}{2} R T$ Mono-atomic gas. (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is Download India's Best Exam Preparation App. $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Molar mass of $C O=28 g \mathrm{mol}^{-1}$, Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$, Energy required for vapourising $2.38 g$ of $\mathrm{CO}$, $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Q. Enthalpy of combustion of carbon to $\mathrm{CO}_{2}(g)$ is $-393.5 \mathrm{kJ}$ $m o l^{-1} .$ Calculate the heat released upon the formation of 35.2 \[ \Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ} \] $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. $\Delta_{v a p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}, T_{b}=373 \mathrm{K}$ $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state. Last updated on 12.08.2020: Previous question papers are a great way to revise and prepare for Higher secondary exams. $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: $\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$ or $K=2.64$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. When the gas is heated to raise its temperature by $1^{\circ} \mathrm{C},$ the increase in its internal energy is equal to the increase in kinetic energy, i.e., $\Delta U=\Delta E_{K}$ SHOW SOLUTION Q. April 15th, 2018 - Thermodynamics Multiple Choice Questions Has 100 MCQs Thermodynamics Quiz Questions And Answers Pdf MCQs On Applied Thermodynamics First Law Of Thermodynamics MCQs With Answers Second Law Of Thermodynamics Reversible And Irreversible Processes And Working Fluid MCQs And Quizzes To Practice Exam Prep Tests' $\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$

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