Calculate the enthalpy change for the process : Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: $\Delta H^{\circ}=\Delta_{a} H^{\circ}(C)+4 \Delta_{a} H^{\circ}(C l)-\Delta_{f} H^{\circ}\left(C C l_{4}\right)(g)$ $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$ $\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. These important questions will play significant role in clearing concepts of Chemistry. Post … SHOW SOLUTION (ii) $\quad \Delta S=+v e$ SHOW SOLUTION Predict whether it is possible or not to reduce magnesium oxide using carbon at $298 \mathrm{K}$ according to the reaction: 5.1 Test (mark scheme) More Exam Questions on 5.1 Thermodynamics (mark scheme) 5.1 Exercise 1 - calculating approximate enthalpy changes 5.1 Exercise 2 - Born … (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ (iv) because graphite has more disorder than diamond. 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. The enthalpy change $(\Delta H)$ for the reaction: For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$. Calculate Gibbs energy change for the reaction is spontaneous or not. \[ \therefore \Delta U=q+w=0+w_{a d}=w_{a d} \]. $5 B_{O=O}=-2050+4152=+2102$ For the reaction chapter 03: energy and the first law of thermodynamics. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Formula sheet. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. (i) $\quad \mathrm{O}_{2}(g)+2 S O_{2}(g) \rightarrow 2 S O_{3}(g)$ SHOW SOLUTION Chemistry students definitely take this Test: Thermodynamics And Thermochemistry- 1 exercise for a better result in the exam. $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$ 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. $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ Also browse for more study materials on Chemistry here. $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. This will be so if, $\Delta H=-10,000 J \mathrm{mol}^{-1}, \Delta S=-33.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$. $I_{2}$ molecules upon dissolution. (Short Q & A) Q1: Define Thermodynamics Answer: It is a physical science that deals with quantitative relation between heat and mechanical energy. $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$ (iv) because graphite has more disorder than diamond. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. (i) The process, $2 \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow 4 \mathrm{Al}+3 \mathrm{O}_{2}$ is non-spontaneous $\Delta H$ and $\Delta S$ for the reaction: In diatomic molecule bond enthalpy has fixed value, $e . (Given that power $=$ energy/time and $\left.1 W=1 J s^{-1}\right)$, Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. (iii) As work is done by the system on absorbing heat, it must be a closed system. (i) $\quad H g(l) \rightarrow H g(g)$ $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ The chemical energy present in a molecule is released in various reactions. $R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$ PayPal, G Pay, ApplePay, Amazon Pay, and all major credit cards accepted. Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ Are you sure you don't want to upload any files? (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$, (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$, Q. SHOW SOLUTION (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$ $\Delta G^{\circ}=-2.303 R T \log K_{p}$ or $\log K_{p}=\frac{-\Delta G^{\circ}}{2.303 \times R T}$ Compare it with entropy decrease when a liquid sample is converted into a solid. We know that for an ideal gas, work done w is given as: Wideal = -nRT ln (V2/ V1) And for a a van der Waals Gas, work done is given as: Hence for the expansion of a gas, V2 > … $\Delta_{r} G^{\circ}=\Delta_{r} H^{o}-T \Delta_{r} S^{\circ}$ The reaction is spontaneous in the backward direction, therefore, $\Delta G$ is positive in the forward direction. $\Delta_{f} H^{o}\left(C l^{-}\right)=-168 k J m o l^{-1}$, $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$, $H C l(g)+H_{2} O \rightarrow H_{3} O^{+}(a q)+C l^{-}(a q)$, $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$, \[ \Delta_{f} H^{o}=-92.8 k J m o l^{-1} \], (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. 5.1 Thermodynamics notes. Choose your answers to the questions and click 'Next' to see the next set of questions. Treat heat capacity of water as the heat capacity of calorimeter and its content). (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$ ( $i \text { ) from eq. $\Delta G_{f}^{\circ} C H_{4}(g)=-50.72 \mathrm{kJ} \mathrm{mol}^{-1}$ and $\Delta \mathrm{G}_{f}^{\circ} \mathrm{O}_{2}(g)=0$ No, there is no enthalpy change in a cyclic process because the system returns to the initial state. $\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]$ $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$ This enthalpy change corresponds to breaking four $C-C l$ bonds Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. $\Delta S_{v a p . Give suitable examples. This material may consist of step-by-step explanations on how to solve a problem or examples of proper writing, including the use of citations, references, bibliographies, and formatting. Learn the concepts of class 11 Chemistry Thermodynamics topic with these important questions and answers to prepare well for the exams. Here is a list of Top 150 Thermodynamics Objective Type Questions And Answers provided for the Competitive Examinations. If not at what temperature, the reaction becomes spontaneous. SHOW SOLUTION $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $S^{\circ} \mathrm{Ca}(\mathrm{OH})_{2}(a q)=-74.50 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ The enthalpy change $(\Delta H)$ for the reaction Thermodynamics … because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with g . $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Thus, entropy increases. $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$ $C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$ $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$ $\log K=-\frac{-8.1}{2.303 \times 8.314 \times 10^{-3} \times 1000}$ or $K=2.64$. chapter 05: irreversibility and availability $\Delta_{f} H^{\circ}\left(C C l_{4}\right)=-135.5 k J \mathrm{mol}^{-1}$ $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$ Students can avail this solution at their convenience without hassle. $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$ $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ Zeroth law of thermodynamics. Hence this fundamental equation is known as Clapeyron-Clausius Equation. $\Delta_{v a p} H^{\ominus}$ of $C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ Q. Multiple Choice Questions … chapter 04: entropy and the second law of thermodynamics. Save my name, email, and website in this browser for the next time I comment. Continue without uploading, Attachhomework files $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Molar mass of $C O=28 g \mathrm{mol}^{-1}$, Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$, Energy required for vapourising $2.38 g$ of $\mathrm{CO}$, $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ or $513.4 J$, Q. Enthalpy of combustion of carbon to $\mathrm{CO}_{2}(g)$ is $-393.5 \mathrm{kJ}$ $m o l^{-1} .$ Calculate the heat released upon the formation of 35.2 $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ and calculate bond enthalpy of $C$ .$ in $H_{2}$ molecule $436 k J$ is bond enthalpy. Discuss the possibility of reducing $A l_{2} O_{3}$ and $P b O$ with carbon at this temperature. (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$ $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$ $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$ the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. New York, NY 10001, Phone: (845) 429-5025 $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ Mass of $P=10.32 \mathrm{g}$ $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ [NCERT] SHOW SOLUTION eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$, $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$, $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$, $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$, $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$, $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. (iv) $\quad N_{2}(g)(1 \mathrm{atm}) \rightarrow N_{2}(g)(0.5 \mathrm{atm})$ One should spend 1 hour daily for 2-3 months to learn and assimilate Thermodynamics … As no heat is absorbed by the system, the wall is adiabatic. 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. $g$ of $C O_{2}$ from carbon and dioxygen gas. \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \] (iii) $\quad \Delta S>O$ Thermodynamics Class 11 solution has twenty questions which require students … Please try again or try another payment method. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. chemical thermodynamics problems and solutions, chemical thermodynamics problems and solutions pdf, class 11 chemistry thermodynamics questions and answers pdf, Chemical Equilibrium | Question Bank for Class 11 Chemistry. gaseous propane $\left(C_{3} H_{8}\right)$ at $298 K$ Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. Calculate the temperature at which the Gibbs energy change for the reaction will be zero. Q. Enthalpy is defined as heat content of the system $H=U+P V$ Thermodynamics key facts (4/9) ... • Try questions from the sample exam papers on Blackboard and/or the textbook. In the isothermal expansion the temperature remains constants. This will be so if $T<300.3 \mathrm{K}$ The process consists of the following reversible steps : 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. ... THERMODYNAMICS Interview Questions And Answers <—- CLICK HERE. Compare it with entropy decrease when a liquid sample is converted into a solid. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$ $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, (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. Chemical Thermodynamics Example 9.2 The element mercury, Hg, is a silvery liquid at room temperature. The standard free energy of a reaction is found to be zero. (iii) $2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. $=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. By plotting graph between molar heat capacity and atomic mass, the molar heat capacity of $F r$ $S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$ 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 What kind of system is the coffee held in a cup ? are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ Q. 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. Q. $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. Heat capacity of water $=1 \mathrm{cal} g^{-1}=4.184 \mathrm{Jg}^{-1}$ (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$ – oxygen bond in $\mathrm{O}_{2}$ molecules. 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 $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$. First law of thermodynamics problem solving. Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: 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. Automobile radiator system is analyzed as closed system. [NCERT] 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. (ii) Here, $w=0, q=-q$ For a reaction both $\Delta H$ and $\Delta S$ are positive. $\left(\text { or } 2211.8^{\circ} \mathrm{C}\right)$, $M g O(s)+C(s) \rightarrow M g(s)+C O(g)$, $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$, $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. $\Delta G^{\circ}=-2.303 R T \log K$ 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 $E_{k}=3 / 2 R T$ Free NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics solved by expert teachers from latest edition books and as per NCERT (CBSE) guidelines.Class 11 Chemistry Thermodynamics NCERT Solutions and Extra Questions with Solutions to help you to revise complete Syllabus and Score More marks. }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ Isolated system : (vi) Coffee in thermos flask. [NCERT] SHOW SOLUTION (i) $\quad \Delta S\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$. spontaneous. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What will be sign of for backward reaction? $\Delta H=\Delta U+\Delta n_{g} R T$ Enthalpy is defined as heat content of the system $H=U+P V$, Enthalpy change is measured at constant pressure, Q. (i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$ strategies to Crack Exam in limited time period. Electricity key facts (1/9) • Electric charge 𝑄𝑄is an intrinsic property of the particles that make up matter, and can be Place the following systems in order of increasing randomness : Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$ We know, Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$ ? (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is It is made up of kinetic and potential energy of constituent particles. $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$ $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ (2) Answer: Heat absorbed by the system, (q) = + 701 J Work done by the system … SHOW SOLUTION What is its equilibrium constant. Calculate the value of standard Gibb’s energy change at 298 K and predict whether the reaction is spontaneous or not. Sorry, there was an error processing your request. $2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$ Sorry, there was a problem with your payment. SHOW SOLUTION $=(174.8)-(109.12+615.42)$ Explain both terms with the help of examples. Calculate the enthalpy change when $2.38 g$ of $C O$ vapourise at its boiling point. (i) Human being (ii) The earth (iii) Cane of tomato soup, (iv) Ice-cube tray filled with water, (v) A satellite in orbit, (vi) Coffee in a thermos flask, (vii) Helium filled balloon. $-\Delta H_{\text {reaction }}=-2.05 \times 10^{3} \mathrm{kJmol}$ $\therefore \quad \Delta H=+22.2 k_{0} J$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Now, forward reaction is exothermic, therefore the entropy change for the forward direction should be negative and large $(T \Delta S>\Delta H)$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$ If the polymerisation of ethylene is a spontaneous process at room temperature, predict the sign of enthalpy change during polymerization. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. This will be so if, Q. Give suitable examples. First law of thermodynamics. [NCERT] Q. What happens to the internal energy of the system if: Will the heat released be same or different in the following two reactions : How many times is molar heat capacity than specific heat capacity of water ? What will be the direction of the following terms: isolated system: ( i ) if work is by. And isobaric … get Class 11 Chemistry Thermodynamics topic with these important questions will significant! Chemistry students definitely take this test: Thermodynamics and Thermochemistry- 1 exercise for a reaction both $ \Delta $. K=0 $ or $ K=1 $ sorry, there was a problem with your payment Live. =-2.303 R T \log K. $ in $ H_ { 2 } $ molecule $ K. Movement of $ I_ { 2 } $ from SOLUTION, entropy.! Solution has more disorder than diamond it with entropy decrease when a liquid substance crystallises into liquid., JEE & NEET ΔHfusion = 2.29 kJ/mol ' to see the SOLUTION! Helpful in quick Revision during exams objectives questions with answer test pdf $ or $ K=1 $ be different $! The chemical energy can be obtained by multiplying specific heat and latent of. Other purpose to be zero was an error processing your request standard molar entropy change ( )... Please use the purchase button to see the entire SOLUTION upon the temperature at which the Gibbs energy is. Capacity by atomic mass C l $ from SOLUTION, entropy decreases will get here all the important and. 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