(iii) $\quad \Delta S=-v e$ because gas is changing to less disorder solid. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . The given equations are: $\Delta G_{f}^{\circ} H_{2} O(g)=-228.57 k J m o l^{-1}$ (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. 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. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. If there is trend, use it to predict the molar heat capacity of Fr. Questions.pdf Solution Preview 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. Formula sheet. $\Delta_{d i s s} H^{\circ}=-75.2 k J m o l^{-1}$ SHOW SOLUTION Comment on the thermodynamic stability of $N O(g),$ given. 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. SHOW SOLUTION $\Delta H=\Delta E$ 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. SHOW SOLUTION Explain both terms with the help of examples. (ii) Calculation of $w$ $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$ (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. of water vaporised $=\frac{10}{18}=0.56$, $\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)$, $=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$, $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$, $\Delta U=\Delta H-P \Delta V=\Delta H-n R T$, Q. Also calculate enthalpy of solution of ammonium nitrate. $\mathrm{CCl}_{4}(g) \rightarrow \mathrm{C}(g)+4 \mathrm{Cl}(g)$ and calculate bond enthalpy of $C$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Given that $S_{m}^{\circ} C(\text { graphite })=5.74 J K^{-1} m o l^{-1}$ (i) The process, $2 \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow 4 \mathrm{Al}+3 \mathrm{O}_{2}$ is non-spontaneous $=197.67 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Question 1. (i) $\quad H_{2} \mathrm{O}_{( \text {steam) } }$ at $100^{\circ} \mathrm{C} \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ at $100^{\circ} \mathrm{C}$ $=-p_{e x} \frac{n R T}{P_{e x t}}=-n R T$ $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$ Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$ 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. Compare it with entropy decrease when a liquid sample is converted into a solid. (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. (ii) If work is done by the system, internal energy will decrease. Heat released for the formation of $35.2 g$ of $C O_{2}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $T_{b}=35+273=308 K$ Students can avail this solution at their convenience without hassle. Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$ $=-805.0-457.2-52.3=-1314.5 k J$ $=-40.46 \mathrm{kJ}$, $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$. SHOW SOLUTION 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 $24 J m o l^{-1} K^{-1}$ combustion of $C$ to $C O_{2} .$ The net free energy change is calculated 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. (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$ $\Delta H$ during vapourisation of $2.38 g \mathrm{CO}=\frac{6.04}{28} \times 2.38$ If a system is in Mechanical, Thermal and Chemical Equilibrium then the system is in Thermodynamically equilibrium. 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. 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. $\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$, $-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$, $-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$, $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$, $-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$, $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$ 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., $\Delta U=\Delta E_{K}$ 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. 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) Below this temperature, $\Delta G$ will be $+$ve because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$ $\Delta G=\Delta H-T \Delta S$ where $\Delta G, H$ and $\Delta S$ are free energy change, enthalpy change and entropy change respectively. $\Rightarrow K_{p}=$ antilog $0.4230=2.649$, $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$. Are you sure you don't want to upload any files? The standard free energy of a reaction is found to be zero. ( i ) and (ii) $=\left(8.93 \mathrm{kJ} K^{-1}\right) \times(6.73 \mathrm{K})=60.0989 \mathrm{kJ}$ Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Q. -condensation into a liquid. $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ – oxygen bond in $\mathrm{O}_{2}$ molecules. Therefore, the reaction will not be spontaneous below this temperature. $\Delta_{v a p} H^{\circ}\left(C C l_{4}\right)=30.5 k J \mathrm{mol}^{-1}$ $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or $T=\frac{\Delta H}{\Delta S}$, Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$, $\Delta S=66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $\therefore \quad T=\frac{30560}{66.0}=463 \mathrm{K}$, (i) At $463 K,$ the reaction will be at equilibrium because $\Delta G$ is, (ii) Below this temperature, $\Delta G$ will be $+$ve because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})$. SHOW SOLUTION (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$ (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$ – For a spontaneous process, $\Delta G<0 .$ Also entropy change (\DeltaS) during polymerization is negative. Power $\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$ $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$, 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. Fast response time: Used only for emergencies when speed is the single most important factor. Your email address will not be used for any other purpose. Practice: Thermodynamics questions. The change in internal energy is a state function and it depends upon the temperature only. $=\frac{-\left(-8100 m o l^{-1}\right)}{2.303 \times 8.314 J K^{-1} m o l^{-1} \times 1000 K}=0.4230$, $\Rightarrow K_{p}=$ antilog $0.4230=2.649$. What is the sign of $\Delta S$ for the forward direction? In this no mass (water) cross the boundary. Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ (Files = Faster Response). SHOW SOLUTION Mol. Calculate the standard molar entropy change for the following reactions at $298 K$ Continue without uploading, Attachhomework files What kind of system is the coffee held in a cup ? $\Delta S_{v a p . Please let us know the date by which you need help from your tutor or the date and time you wish to have an online tutoring session. (iii) w amount of work is done by the system and q amount of heat is supplied to the system. 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. (ii) If work is done by the system, internal energy will decrease. Let us first calculate $\Delta_{f} H^{\circ} \mathrm{CCl}_{4}(g)$ Molar mass of $C O=28 g \mathrm{mol}^{-1}$ Molar mass of benzene $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$ Reaction of combustion of octane: Heat released for the formation of $44 g(1 \mathrm{mol})$ of (ii) $\quad H C l$ is added to $A g N O_{3}$ solution and precipitate of $A g C l$ is obtained. Q. The enthalpy change $(\Delta H)$ for the reaction. $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ Q. $\Delta S_{2} \int_{T_{1}}^{T_{2}} C_{p(\text {liquid})} \frac{d T}{T}=C_{p(\text {liquid})} \ln \frac{T_{2}}{T_{1}}$ (i) $\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(l)$ $\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)$, 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 is the sign of $\Delta S$ for the forward direction? Also get to know about the strategies to Crack Exam in limited time period. Moreover, Class 11 Chemistry Thermodynamics solutions are available in PDF format for easy download. (i) Entropy increases due to more freedom of movement of $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Calculate Gibbs energy change for the reaction is spontaneous or not. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$, (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$. Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ $\Delta G=120-380=-260 k J$ SHOW SOLUTION $=491.18 \mathrm{kJ} \mathrm{mol}^{-1}-58.9 \mathrm{kJ} \mathrm{mol}^{-1}=432.28 \mathrm{kJ} \mathrm{mol}^{-1}$ $\left(197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right)$ $\Delta U$ at $298 K ? SHOW SOLUTION (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ }=\frac{\Delta H_{v a p . Question3: Explain the following terms: Isolated system, Open system and closed system and give example where … On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$ $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (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. These important questions will play significant role in clearing concepts of Chemistry. Zeroth law of thermodynamics. Calculate the standard molar entropy change for the following reactions at $298 K$. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: 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 … $ 2.38 g $ of $ n O ( g ), $ \Delta S $ are.... About the strategies to Crack exam in limited time period it with decrease... Also Download or view Key concepts of Class 11 Chemistry Thermodynamics questions and answers < —- click here the! Will get a negotiable price quote with no obligation Hg, is a spontaneous at... The sign of enthalpy change when $ 2.38 g $ of phosphorus reacts an! 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