$\Delta_{a} H^{\circ}(C)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$ $-92380=\Delta U-4955$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. SHOW SOLUTION (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. to do mechanical work as burning of fuel in an engine, provide electrical energy as in dry cell, etc. $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$ Q. Reaction of combustion of octane: $q=125 g \times 4.18 J / g \times(286.4-296.5)$ Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ Given that $S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$ (i) $\quad \Delta S\Delta H)$, Q. Here is a list of Thermodynamics MCQs with Answers (Multiple Choice Questions) is given below. Q. Silane $\left(S i H_{4}\right)$ burns in air as: $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ $=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$ The students those are preparing for the Govt. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ SHOW SOLUTION $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ 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. ( } 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. This enthalpy change corresponds to breaking four $C-C l$ bonds $E=\frac{3}{2} R T$ Mono-atomic gas. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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 $E_{k}=3 / 2 R T$ Also calculate enthalpy of solution of ammonium nitrate. Q. SHOW SOLUTION $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ For the water gas reaction : (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iv) $\quad \Delta S>O$. $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state. 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. Formula sheet. The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. In what way internal energy is different from enthalpy ? ©2020 24houranswers.com. SHOW SOLUTION $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ (iii) A partition is removed to allow two gases to mix. $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ The equilibrium constant for a reactions is $10 .$ What will be the value of $\Delta G^{\circ} ?$ $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ (i) At what temperature the reaction will occur spontaneously from left to right? We'll send you an email right away. SHOW SOLUTION Therefore, $C_{v}=3 / 2 R$, At $(T+1) K,$ the kinetic energy per mole $\left(E_{k}\right)=3 / 2 R(T+1)$ Therefore, increase in the average kinetic energy of the gas for $1^{\circ} \mathrm{C}(\text { or } 1 \mathrm{K})$ rise in temperature $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$, 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., $\Delta U=\Delta E_{K}$, Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$, Q. Normal response time: Our most experienced, most successful tutors are provided for maximum expertise and reliability. $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? One should spend 1 hour daily for 2-3 months to learn and assimilate Thermodynamics â¦ (ii), (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$, (iii) $\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$, (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$, $\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$, Subtract eq. $\therefore \quad \Delta H=+22.2 k_{0} J$ 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. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden. \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \]. 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. Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$ (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$ In what way is it different from bond enthalpy of diatomic molecule ? $\Delta n_{g}=2-(1+3)=-2 m o l, T=298 K$ Here you can get Class 11 Important Questions Chemistry based on NCERT Text book for Class XI.Chemistry Class 11 Important Questions are very helpful to score high marks in board exams. We respect your privacy. – oxygen bond in $\mathrm{O}_{2}$ molecules. (ii) If work is done by the system, internal energy will decrease. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, $q=125 g \times 4.18 J / g \times(286.4-296.5)$, $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, Q. $\Delta S_{v a p . $=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$ Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. Q. $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$ Our 1000+ Thermodynamics questions and answers focuses on all areas of Thermodynamics covering 100+ topics. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=(174.8)-(109.12+615.42)$ What type of wall does the system have? Q. $\Delta H=\Delta U+P \Delta V$. 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. If there is trend, use it to predict the molar heat capacity of Fr. Which formula forms a link between the Thermodynamics and Electro chemistry? (Hint. (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$ What is the sign of $\Delta S$ for the forward direction? But $\mathrm{NO}_{2}(g)$ is formed. $\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$ Calculate standard molar entropy change of the formation of THERMODYNAMICS Mechanical Interview Questions And Answers pdf free download for gate,objective questions,mcqs,online test quiz bits,lab viva manual. $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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 $24 J m o l^{-1} K^{-1}$ $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. 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. (Files = Faster Response). $-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$ $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ Entropy change for evaporation of $36 g$ of wate Welcome to 5.1 THERMODYNAMICS. $S^{\circ} \mathrm{Ca}(\mathrm{OH})_{2}(a q)=-74.50 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Please use the purchase button to see the entire solution. $=-805.0+2(-228.6)-[+52.3+2(0)]$ SHOW SOLUTION $-92380=\Delta U-2 \times 8.314 \times 298$ Click Here for Detailed Notes of any chapter. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and chapters. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. Calculate $\Delta S$ for the conversion of: Professionals, Teachers, Students and Kids Trivia Quizzes to test your knowledge on the subject. }=\frac{\Delta H_{v a p . Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (iii) $\quad H_{2} O(l)$ at $0^{\circ} C \rightarrow H_{2} O(s)$ at $0^{\circ} C$ Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. We know (ii) Work is done by the system? SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers Pdf free download. SHOW SOLUTION Also get to know about the strategies to Crack Exam in limited time period. $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$ $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ 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. Here, we are given, $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$, Q. Is there any enthalpy change in a cyclic process ? $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$ $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (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. Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. Enthalpy of combustion of octane, Heat transferred $=$ Heat capacity $\times \Delta T$, $=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$, Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$, $\therefore$ Internal energy change $(\Delta E)$ during combustion of one mole of, octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$, $C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$, $\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$, $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$, $=\left(5481.02 \times 10^{3}\right)+(-4.5 \times 8.314 \times 300.78)$, $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$, $=5492273.082 \mathrm{Jmol}^{-1}=5492.27 \mathrm{kJ} \mathrm{mol}^{-1}$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Molar mass of $C O=28 g \mathrm{mol}^{-1}$ Automobile radiator system is analyzed as closed system. Heat released for the formation of $35.2 g$ of $C O_{2}$ Q. You may read our privacy policy for more info. $\Delta H=\Delta E$ 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. $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ Calculate enthalpy change when $2.38 g$ of carbon monoxide (CO) vapourises at its normal boiling point. -condensation into a liquid. $=\frac{30.8}{78} \times 100=39.487 k_{U}=39487 J$ $\Delta G=120-380=-260 k J$ $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$, $\Delta_{r} S^{o}=197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}$, $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$, $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$, $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$. $-(1) \quad \Delta S=-v e$ $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$ 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. These important questions will play significant role in clearing concepts of Chemistry. $\frac{-393.5 \times 35.2}{44}=-314.8 k J$. $\Delta G_{f}^{\circ} C H_{4}(g)=-50.72 \mathrm{kJ} \mathrm{mol}^{-1}$ and $\Delta \mathrm{G}_{f}^{\circ} \mathrm{O}_{2}(g)=0$ [CBSE Sample Paper for 2006] [1] ... THERMODYNAMICS Interview Questions And Answers <â- CLICK HERE. Email: help@24houranswers.com
$O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$ $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$ Thus, calorimeter loses $5.282 \mathrm{kJ}$ of heat during dissolution $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ 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. 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