Answer: Thermochemistry is a branch of thermodynamics which deals with the relationships between chemical reactions and corresponding energy changes. $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (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. How many times is molar heat capacity than specific heat capacity of water ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Q. $-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$ Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$ SHOW SOLUTION $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$ (iii) $\quad \Delta S>O$ Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ 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. $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$ $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ (Hint. The standard free energy of a reaction is found to be zero. Enthalpy change for 2 mole of $P=-243 \mathrm{kJ}$ SHOW SOLUTION $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$ $\Delta U=-92380+4955=-87425 J=-87.425 k J$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$, $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$, $-92380=\Delta U-2 \times 8.314 \times 298$, $\Delta U=-92380+4955=-87425 J=-87.425 k J$, Q. Heat capacity of water $=1 \mathrm{cal} g^{-1}=4.184 \mathrm{Jg}^{-1}$ 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 Enthalpy change is measured at constant pressure Subtract eq. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$ SHOW SOLUTION [NCERT] SHOW SOLUTION SHOW SOLUTION $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$ $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. All books are in clear copy here, and all files are secure so don't worry about it. (ii) Here, $w=0, q=-q$ SHOW SOLUTION $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$. (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. Practice: Thermodynamics questions. $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$ A compendium of past examination questions set on Physical Chemistry on the JF Chemistry paper and problem sheets ... Thermodynamics, Equilibria and Electrochemistry ... there is insufficient data supplied to answer the question. (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) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. $\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$ SHOW SOLUTION (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$ (ii) If work is done by the system, internal energy will decrease. In what way internal energy is different from enthalpy ? A lot of these questions are likely to appear in the board examination, making this an ultimate guide for students before their examinations. $-(1) \quad \Delta S=-v e$ 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. These important questions will play significant role in clearing concepts of Chemistry. (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$ False. No, It will not work, as … Calculate Gibbs energy change for the reaction is spontaneous or not. $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ 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. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. Calculate the temperature at which the Gibbs energy change for the reaction will be zero. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ (i) At what temperature the reaction will occur spontaneously from left to right? $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ 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. The standard free energy of a reaction is found to be zero. Heat transferred $=$ Heat capacity $\times \Delta T$ Thus, entropy increases. Q. \[ c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1} \] $\Delta n=2-4=-2$ $g$ of $C O_{2}$ from carbon and dioxygen gas. If there is trend, use it to predict the molar heat capacity of Fr. According to Gibbs Helmholtz equation, Explain both terms with the help of examples. SHOW SOLUTION What is meant by average bond enthalpy ? SHOW SOLUTION $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$ [NCERT] Thermodynamics Questions and Answers pdf free download 1. (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$ $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. spontaneous. $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ (ii) Work is done by the system? Since Gibbs energy change is positive, therefore, at the reaction is not possible. (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. $\Delta U$ at $298 K ? Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ $\Delta U$ at $298 K ?$ We won’t spam you. What will be the direction of the reaction at this temperature and below this temperature and why? In the reaction $C_{3} H_{8}(g)+5 O_{2} \rightarrow 3 C O_{2}+4 H_{2} O(g) \quad$ if 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' State The Third Law Of Thermodynamics. Is there any enthalpy change in a cyclic process ? $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$, $\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$, $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$, $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$, $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$. Thus $A l_{2} O_{3}$ cannot be reduced by $C$ This way you can find out what you already know and what you don't know. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$ A piston exerting a pressure of 1 atm rests on the surface of water at $100^{\circ} \mathrm{C} .$ The pressure is reduced to smaller extent when $10 g$ of water evaporates and $22.2 \mathrm{kJ}$ of heat is absorbed. 5.1 Thermodynamics notes. Calculate the standard molar entropy change for the following reactions at $298 K$ … $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ ( i ) and (ii) (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ $\Delta H=\Delta U+\Delta n_{g} R T$ A-Level Chemistry. $T_{b}=35+273=308 K$ $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$ \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \]. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. $\Delta T=300.78-294.05=6.73 K$ 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. Explain. $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. SHOW SOLUTION Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: A piston exerting a pressure of 1 atm rests on the surface of water at $100^{\circ} \mathrm{C} .$ The pressure is reduced to smaller extent when $10 g$ of water evaporates and $22.2 \mathrm{kJ}$ of heat is absorbed. are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. (i) Dissolution of iodine in a solvent. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. That will be very helpful in quick revision during Exams. Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ This will be so if, Q. 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. (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ Zeroth law of thermodynamics. This is the currently selected item. Give Its Limitations And Importance.? $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$ 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. $\Delta G^{\circ}=-2.303 R T \log K .$ Hence, $\log k=0$ or $K=1$. (ii) $\quad \Delta S>O$ SHOW SOLUTION What will be sign of for backward reaction? $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$ (i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$ Silane $\left(S i H_{4}\right)$ burns in air as: Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: Calculate the standard entropy change for the reaction, Calculate standard molar entropy change of the formation of, The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$. \[ \Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ} \] With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ $=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$ (ii) $\quad H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l(a q)$ (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$ $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ (iii) $\quad \Delta \mathrm{S}=-\mathrm{ve}$, (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$, (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$, (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$, (iii) $\quad \Delta \mathrm{S}=-\mathrm{ve}$, Q. $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$ $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$ SHOW SOLUTION For a reaction both $\Delta H$ and $\Delta S$ are positive. $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$, (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. Read online Class 11 Chemistry Thermodynamics Questions And Answers book pdf free download link book now. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ (iii) Upon removal of partition the two gases will diffuse into one another creating greater randomness. In what way is it different from bond enthalpy of diatomic molecule ? 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. Professionals, Teachers, Students and Kids Trivia Quizzes to test your knowledge on the subject. If not at what temperature, the reaction becomes spontaneous. $\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$ $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ This will be so if, $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). Thermodynamics key facts (4/9) ... • Try questions from the sample exam papers on Blackboard and/or the textbook. $\therefore \Delta H^{\circ}=715.0+4 \times 121-(-105.0)=1304.0 \mathrm{kJ} \mathrm{mol}^{-1}$ Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$, In the reaction $C_{3} H_{8}(g)+5 O_{2} \rightarrow 3 C O_{2}+4 H_{2} O(g) \quad$ if. For $\Delta G$ to be negative, $T \Delta S$ must be $>\Delta H$, For $\Delta G$ to be negative, $T \Delta S$ must be $>\Delta H$, Q. SHOW SOLUTION The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. Internal energy change is measure at constant volume. Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ Calculate the standard entropy change for the reaction Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. As heat is taken out, the system must be having thermally conducting walls. You are on page 1 of 14. These important questions will play significant role in clearing concepts of Chemistry. Heat released for the formation of $35.2 g$ of $C O_{2}$ $\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$ [CONFIRMED] JEE Main will be conducted 4 times from 2021! SHOW SOLUTION $\Delta_{a} H^{o}(C l)=\frac{1}{2} \times 242=121 k J m o l^{-1}$ }=\frac{\Delta H_{v a p . Also get to know about the strategies to Crack Exam in limited time period. since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$ \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \] $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$ $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$ Internal energy change is measure at constant volume. Also calculate the enthalpy of combustion of octane. ? SHOW SOLUTION These important questions will play significant role in clearing concepts of Chemistry. $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ $\Delta H=\Delta U+\Delta n R T$ $\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$ (ii) $\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of 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? since $\Delta_{r} G^{\circ}$ is negative, the reaction will be spontaneous. Multiply eqn. the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$ Revision : Electricity. Thus, enthalpy change $=513.4 \mathrm{J}$, $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$, $\Delta H$ during vapourisation of $2.38 g \mathrm{CO}=\frac{6.04}{28} \times 2.38$, Thus, enthalpy change $=513.4 \mathrm{J}$, Q. Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ since $\Delta_{r} H^{o}$ is +ve i.e., enthalpy of formation of $N O$ is positive, Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. You may assume that the gas constant R = 8.314 J mol-1. $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ Calculate the temperature at which the Gibbs energy change for the reaction will be zero. Therefore, the reaction will not be spontaneous below this temperature. Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, (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. Click Here for Detailed Notes of any chapter. $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Questions Answer PDF Download. Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero SHOW SOLUTION (i) $\quad \Delta H=\Delta U+\Delta n R T$ (i) $\quad C H_{4}(g)+2 O_{2}(g) \rightarrow C O_{2}(g)+2 H_{2} O(g)$ (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ Q. MCQ quiz on Thermodynamics multiple choice questions and answers on Thermodynamics MCQ questions quiz on Thermodynamics objectives questions with answer test pdf. SHOW SOLUTION 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. (Hint. 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. [NCERT] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Q. $20.0 \mathrm{g}$ of ammonium nitrate $\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)$ is dissolved in $125 \mathrm{g}$ of water in a coffee-cup calorimeter. $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with SHOW SOLUTION SHOW SOLUTION What type of system would it be ? Average bond enthalpy is average heat required to break 1 mole of particular bond in various molecules (polyatomic). SHOW SOLUTION $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. Calculate the bond enthalpy of $H C l .$ Given that the bond enthalpies of $H_{2}$ and $C l_{2}$ are $430 \mathrm{kJ} \mathrm{mol}^{-1}$ and $242 \mathrm{kJ} \mathrm{mol}^{-1}$ respectively and $\Delta \mathrm{H}_{f}^{\circ}$ for $H C l s-95 k J m o l^{-1}$, Calculate $\Delta S$ when 1 mole of steam at $100^{\circ} \mathrm{C}$ is converted into ice at $0^{\circ} \mathrm{C}$. $E=\frac{3}{2} R T$ Mono-atomic gas. Open system : (i) Human being (ii) The earth (v) A satellite in orbit, zero. $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$ 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. $g$ of $C O_{2}$ from carbon and dioxygen gas. $N H_{4} N O_{3}$ $C O_{2}=-393.5 k_{U}$ $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$ $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$ 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. What type of wall does the system have ? No, there is no enthalpy change in a cyclic process because the system returns to the initial state. Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=-805.0-457.2-52.3=-1314.5 k J$ CBSE Class 11 Chapter 6 Thermodynamics Chemistry Marks Wise Question with Answers A comprehensive database of more than 19 thermodynamics quizzes online, test your knowledge with thermodynamics quiz questions. $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ The reaction is spontaneous in the backward direction, therefore, $\Delta G$ is positive in the forward direction. For a reaction both $\Delta H$ and $\Delta S$ are positive. 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. (ii) Temperature of crytal is increased. Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$ What kind of system is the coffee held in a cup ? But $\mathrm{NO}_{2}(g)$ is formed. Jump to Page . Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\frac{1}{2} N_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow N O(g) ; \Delta_{r} H^{\circ}=90 k J \operatorname{mol}^{-1}$ $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$ $\Delta G_{f}^{o} \mathrm{CO}_{2}(g)=-394.36 \mathrm{kJ} \mathrm{mol}^{-1}$ The process consists of the following reversible steps : (i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$, $\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$, $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$, (ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$, $\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$, $=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$, $=-5.62 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, (iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$, $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$. 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