Q. as Subtract eq. $=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ As no heat is absorbed by the system, the wall is adiabatic. $\Delta G=120-380=-260 k J$ This will be so if $T<300.3 \mathrm{K}$ chemical thermodynamics problems and solutions, chemical thermodynamics problems and solutions pdf, class 11 chemistry thermodynamics questions and answers pdf, JEE Main Previous Year Questions Topicwise. Please enter your email address. (ii) At what temperature, the reaction will reverse? The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. change. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$ $T=\frac{\Delta H}{\Delta S}=\frac{-10000 J m o l^{-1}}{-33.3 J K^{-1} m o l^{-1}}=300.3 \mathrm{K}$ SHOW SOLUTION SHOW SOLUTION (i) $\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(l)$ What is its equilibrium constant. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ 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. 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 $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$. The standard Gibbs energy of reaction (at $1000 K)$ is $-8.1$ $k J m o l^{-1} .$ Calculate its equilibrium constant. (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$ $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Therefore $\Delta E=0$ under isothermal conditions. The coffee held in a cup is an open system because it can exchange matter (water vapour) and energy (heat) with the surroundings. By plotting graph between molar heat capacity and atomic mass, the molar heat capacity of $F r$ Questions and Answers in Thermodynamics. Predict the entropy change (positive/negative) in the following : The heat released in the above two reactions will be different. (i) Liquid to vapours and $\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$ (ii) Work is done by the system? $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$ $=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$ $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$, $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$, $K_{p}$ for this conversion is $2.47 \times 10^{-29}$, $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$, $\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$, $\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. Q. $-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(s)+2 \Delta G_{f}^{\circ} H^{+}(a q)\right]$ The reaction is spontaneous in the backward direction, therefore, $\Delta G$ is positive in the forward direction. Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$ $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$ SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$ (ii) At what temperature, the reaction will reverse? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. g . which is very easy to understand and improve your skill. are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Predict the sign of entropy change in the following reactions: Download Thermodynamics MCQ Question Answer PDF « Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ (i) The process, $2 \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow 4 \mathrm{Al}+3 \mathrm{O}_{2}$ is non-spontaneous Q. Molar mass of $C O=28 g \mathrm{mol}^{-1}$ SHOW SOLUTION 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 $0^{\circ} C$, $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$. so $\mathrm{NO}(g)$ is unstable. Click Here for Detailed Notes of any chapter. $\Delta S_{v a p . 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. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. (i) At what temperature the reaction will occur spontaneously from left to right? SHOW SOLUTION Compute the molar heat capacity of these elements and identify any periodic trend. $=53.28 \mathrm{JK}^{-1} \times 20 \mathrm{K}=1065.6 \mathrm{V}$ 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. $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$ Q. 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. constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Internal energy change is measure at constant volume. Thermodynamics article. Q. Reaction of combustion of octane: (iii) A partition is removed to allow two gases to mix. Formula sheet. ( } i v)$ SHOW SOLUTION A comprehensive database of more than 19 thermodynamics quizzes online, test your knowledge with thermodynamics quiz questions. 2000 AP CHEMISTRY FREE RESPONSE QUESTIONS. (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$ Welcome to 5.1 THERMODYNAMICS. Predict the sign of entropy change for each of the following changes of state: Why would you expect a decrease in entropy as a gas condenses into liquid ? Heat released for the formation of $35.2 g$ of $C O_{2}$ [NCERT] the condensation of diethyl ether is the reverse process, therefore, $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. Only coupons for themes and useful news bulletins. 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. $\Delta_{f} H^{\circ}\left(C C l_{4}\right)=-135.5 k J \mathrm{mol}^{-1}$ $=+21.83 \mathrm{kJ} \mathrm{mol}^{-1}$, (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. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET. $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$ $\frac{-393.5 \times 35.2}{44}=-314.8 k J$. Mass of $P=10.32 \mathrm{g}$ (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $C_{p}=\left(1.0 \mathrm{cal} K^{-1} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=18.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \]. Q. $\Delta S_{\text {Reaction }}=\Sigma S_{m(\text { products })}^{\circ}-\Sigma S_{m(\text { reactants })}^{\circ}$ A-Level Chemistry. The enthalpy change $(\Delta H)$ for the reaction: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION $\Delta G=\Delta H-T \Delta S$ $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$ Download free printable worksheets Chemistry pdf of CBSE and kendriya vidyalaya Schools as per latest syllabus in pdf, CBSE Class 11 Chemistry Worksheet - Thermodynamics (1) CBSE,CCE and NCERT students can refer to the attached file. Practice: Thermodynamics questions. Q. What will be sign of for backward reaction? Jump to Page . SHOW SOLUTION (iii) As work is done by the system on absorbing heat, it must be a closed system. Also calculate the enthalpy of combustion of octane. All the commercial liquid fuels are derived from natural petroleum (or crude oil). $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$ 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. $\Delta_{a} H^{o}(C l)=715.0 \mathrm{kJ} \mathrm{mol}^{-1}$ All educational material on the website has been prepared by the best teachers having more than 20 years of teaching experience in various Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: According to Gibbs Helmholtz equation, $q=m \times s \times\left(t_{2}-t_{1}\right)$ In the reaction $C_{3} H_{8}(g)+5 O_{2} \rightarrow 3 C O_{2}+4 H_{2} O(g) \quad$ if SHOW SOLUTION $\Delta_{r} G^{\circ}=\Delta_{f} G^{\circ}\left(S i O_{2}\right)+2 \Delta_{f} G^{\circ}\left(H_{2} O\right)-\left[\Delta_{f} G^{\circ}\left(S i H_{4}\right)\right]+$ (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ Thermodynamics Questions and Answers pdf free download 1. What is the sign of $\Delta S$ for the forward direction? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. This way you can find out what you already know and what you don't know. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ False. You may assume that the gas constant R = 8.314 J mol-1. Comment on the thermodynamic stability of $N O(g),$ given. These important questions will play significant role in clearing concepts of Chemistry. SHOW SOLUTION $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$ As heat is taken out, the system must be having thermally conducting walls. $\Delta G=\Delta H-T \Delta S=(+)-T(+)$ The enthalpy change $(\Delta H)$ for the reaction: For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$. Thermodynamics is the study of energy transformations. SHOW SOLUTION Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $\Delta G^{\circ}=-2.303 R T \log K$ The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. $S^{\circ}(F e(s))=27.28 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Molar mass of $C O=12+16=28$ $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. $20.0 \mathrm{g}$ 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. Heat transferred $=$ Heat capacity $\times \Delta T$ (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$ (atomic mass $=223$ ) would be $33.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. Therefore, the reaction will be spontaneous above $2484.8 \mathrm{K}$ [NCERT] (ii) $\quad \Delta S>O$ $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. SHOW SOLUTION $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ $20.0 \mathrm{g}$ 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. Under what condition $\Delta H$ becomes equal to $\Delta E ?$ (i) No heat is absorbed by the system from the surroundings, but work (w) is done on the system. Calculate the enthalpy change when $10.32 g$ of phosphorus reacts with an excess of bromine. $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$, 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. Multiply eqn. spontaneous. Explain both terms with the help of examples. ? so $\mathrm{NO}(g)$ is unstable. The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. First law of thermodynamics problem solving. SHOW SOLUTION $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (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. $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$ $\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$ (iii) As work is done by the system on absorbing heat, it must be a closed system. $R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $\Rightarrow$ Enthalpy change, $\Delta H=\Delta E+\Delta n_{g} R T$ $\therefore \quad \Delta S_{\text {total}}=-26.0445-5.62-5.26$, $\Rightarrow \quad-36.9245 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$, Download India's Leading JEE | NEET | Class 9,10 Exam preparation app. $\therefore$ Internal energy change $(\Delta E)$ during combustion of one mole of In what way internal energy is different from enthalpy ? (i) For spontaneity from left to right, $\Delta G$ should be $-v e$ for the given reaction. Express the change in internal energy of a system when SHOW SOLUTION All books are in clear copy here, and all files are secure so don't worry about it. Q. – The required equation for the formation of $C H_{3} O H(l)$ is : [NCERT] Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state. Thus, entropy increases. (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$ R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $-228.6 \mathrm{kJmol}^{-1}$ respectively. $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium, constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$, $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$, Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$, $\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$, $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. What will be the direction of the reaction at this temperature and below this temperature and why? What type of wall does the system have? SHOW SOLUTION Q. Silane $\left(S i H_{4}\right)$ burns in air as: CBSE Chemistry Chapter 6 Thermodynamics class 11 Notes Chemistry in PDF are available for free download in myCBSEguide mobile app. We respect your inbox. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. No, It will not work, as … $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$, $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. 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. Thermodynamics Center For Teaching amp Learning. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $=22.2-1.737=20.463 \mathrm{kJ}$, (iii) $\Delta U$ Calculate (i) $\Delta H, \quad$ (ii) $w$, since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$, $\therefore \quad \Delta H=+22.2 k_{0} J$, Mol. Electricity key facts (1/9) • Electric charge is an intrinsic property of the particles that make up matter, and can be Calculate the enthalpy change for the process : (iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION $=0.5134 \mathrm{kJ}$ If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ You can access free study material for all three subject’s Physics, Chemistry and Mathematics. Questions Answer PDF Download. SHOW SOLUTION First law of thermodynamics. Predict the sign of entropy change in the following reactions: Predict the entropy change (positive/negative) in the following : A reversible reaction has $\Delta G^{\circ}$ negative for forward reaction? SHOW SOLUTION $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$ In what way is it different from bond enthalpy of diatomic molecule ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Thus $A l_{2} O_{3}$ cannot be reduced by $C$ $\Delta G=\Delta H-T \Delta S$ $\Delta U=-92380+4955=-87425 J=-87.425 k J$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$, $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$, $-92380=\Delta U-2 \times 8.314 \times 298$, $\Delta U=-92380+4955=-87425 J=-87.425 k J$, Q. Adding eq. $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. 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 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta G_{f}^{o} H^{+}(a q)=0$ and 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. $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ 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. -condensation into a liquid. 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