of water vaporised $=\frac{10}{18}=0.56$ $\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$ (iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$ Also get to know about the strategies to Crack Exam in limited time period. taken as zero i.e., $\Delta_{f} H\left[H_{3} O^{+}(a q)\right]=0$ In what way is it different from bond enthalpy of diatomic molecule ? of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of The enthalpy change $(\Delta H)$ for the reaction: Calculate the temperature at which the Gibbs energy change for the reaction will be zero. Register online for Chemistry tuition on Vedantu.com to score more marks in your examination. This is the currently selected item. $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ $\Delta n_{g}=2-(1+3)=-2 m o l, T=298 K$ $=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$ 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. $5 B_{O=O}=-2050+4152=+2102$ $q=c \times \Delta T, \quad c=n \times C_{m}$ since $\Delta_{r} H^{o}$ is +ve i.e., enthalpy of formation of $N O$ is positive, $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $\quad H g(l) \rightarrow H g(g)$ Also calculate the enthalpy of combustion of octane. Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$ 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. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. Which of the following process are accompanied by an increase of entropy: 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 $\left(35^{\circ} \mathrm{C}\right)$, Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$. 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. Therefore, the reaction will not be spontaneous below this temperature. $S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$ (i) Liquid to vapours and SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Isolated system : (vi) Coffee in thermos flask. Calculate standard molar entropy change of the formation of $\Delta H=\Delta E$ during a process which is carried out in a closed vessel $(\Delta v=0)$ or number of moles of gaseous products $=$ number of moles of gaseous reactants or the reaction does not involve any gaseous reactant or product. \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \] (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ It will be greater in reaction, (i) because when water (g) condense to form water (l), heat is released. First law of thermodynamics problem solving. If not at what temperature, the reaction becomes spontaneous. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta U+P \Delta V$. $-92380=\Delta U-4955$ $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. All the commercial liquid fuels are derived from natural petroleum (or crude oil). (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 2000 AP CHEMISTRY FREE RESPONSE QUESTIONS. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. Calculate the standard entropy change for the reaction The process consists of the following reversible steps : 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. We know $20.0 \mathrm{g}$ 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. $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$ Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$ Now, forward reaction is exothermic, therefore the entropy change for the forward direction should be negative and large $(T \Delta S>\Delta H)$, Q. $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)$ SHOW SOLUTION $\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$ Also calculate enthalpy of solution of ammonium nitrate. Welcome to JEEMAIN.GURU, Best educational blog for IIT JEE aspirants. (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ $\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual … SHOW SOLUTION ( $i \text { ) from eq. Treat heat capacity of water as the heat capacity of calorimeter and its content). ( } i v)$ Calculate the standard molar entropy change for the following reactions at $298 K$ (iii) w amount of work is done by the system and q amount of heat is supplied to the system. $=-805.0-457.2-52.3=-1314.5 k J$ $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$ Thermodynamics The normal boiling point of ethanol, C 2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ/mol. -condensation into a liquid. (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ $S_{m C_{3} H_{8}(g)}-\left[3 \times S_{m}^{\circ} C_{(g r a p h i t e)}+4 \times S_{m H_{2}(g)}^{\circ}\right]$ $-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(s)+2 \Delta G_{f}^{\circ} H^{+}(a q)\right]$ Since Gibbs energy change is positive, therefore, at the reaction is not possible. 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. Thus, entropy increases. the condensation of diethyl ether is the reverse process, therefore, Treat heat capacity of water as the heat capacity of calorimeter and its content). You are on page 1 of 14. $-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$ 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 $0^{\circ} \mathrm{C}$. $q=m \times s \times\left(t_{2}-t_{1}\right)$ the condensation of diethyl ether is the reverse process, therefore, $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? (ii) $\quad H_{2} O(l)$ at $100^{\circ} C \longrightarrow H_{2} O(l)$ at $0^{\circ} C$ Lost your password? Questions on Chemistry, Thermodynamics: MCQs test on 'Chemistry, Thermodynamics' with answers, Test: 1, Total Questions: 15 What type of wall does the system have? SHOW SOLUTION This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$ Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ 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. SHOW SOLUTION $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ We know (iii) w amount of work is done by the system and q amount of heat is supplied to the system. $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$ SHOW SOLUTION (ii) 1 mol of a solid X $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. Explain both terms with the help of examples. If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. A-Level Chemistry. 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 $\left(35^{\circ} \mathrm{C}\right)$ SHOW SOLUTION We won’t spam you. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$ $\Delta G_{f}^{\circ} H_{2} O(l)=-237.13 k J m o l^{-1}$ Q. Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero $\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{v a p} S=\frac{\Delta_{v a p} H}{T_{b}}=\frac{40.63 \times 1000 \mathrm{J} \mathrm{mol}^{-1}}{373 \mathrm{K}}=109 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Will the heat released be same or different in the following two reactions : Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. (ii) $\quad \Delta n=1-3=-2$. (iv) $\quad C$ (graphite) $\rightarrow C$ (diamond). $I_{2}$ molecules upon dissolution. What type of system would it be ? $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$ SHOW SOLUTION (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ Thermodynamics Center For Teaching amp Learning. $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ Molar heat capacity of $R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$ Is there any enthalpy change in a cyclic process ? Q. $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION 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. to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. (i) $\quad H_{2} O(l) \rightleftharpoons H_{2} O(g)$ }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$, $\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$, $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$, $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$, $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$. Thermodynamics MCQ Question with Answer Thermodynamics MCQ with detailed explanation for interview, entrance and competitive exams. $\Delta H$ and $\Delta S$ for the reaction: 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. (iii) As work is done by the system on absorbing heat, it must be a closed system. $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$, (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. $C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$ 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. 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 Mol. Calculate $\Delta S$ for the conversion of: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thermodynamics article. $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$ Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$ (i) Dissolution of iodine in a solvent. Express the change in internal energy of a system when 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. SHOW SOLUTION $C_{(G r a p h i t e)}+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ [NCERT] 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. $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ Q. (iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? $-92380=\Delta U-2 \times 8.314 \times 298$ (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $A g_{2} O(s) \rightarrow 2 A g(s)+\frac{1}{2} O_{2}(g)$. (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) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. 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 is its equilibrium constant. SHOW SOLUTION $\Delta_{r} H^{\circ}=-726 \mathrm{kJmol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. This will be so if $T<300.3 \mathrm{K}$, (ii) For reverse reaction to occur, should be tve for forward reaction. [CONFIRMED] JEE Main will be conducted 4 times from 2021! 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) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ Answer : The third … $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ Download Thermodynamics MCQ Question Answer PDF « Q. If not at what temperature, the reaction becomes spontaneous. $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$ SHOW SOLUTION So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. Formula sheet. Last updated on 12.08.2020: Previous question papers are a great way to revise and prepare for Higher secondary exams. $\mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ For, $\Delta G=0$ (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$ What is meant by average bond enthalpy ? (ii) Work is done by the system? which is very easy to understand and improve your skill. $\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]$ The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. At equilibrium $\Delta G=0$ so that $E=\frac{3}{2} R T$ Mono-atomic gas. \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. (Hint. $=(174.8)-(109.12+615.42)$ $C O_{2}=-393.5 k_{U}$ (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. 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. $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (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. 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