Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$ $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. The normal freezing point of mercury is â38.9ºC, and its molar enthalpy of fusion is ÎHfusion = 2.29 kJ/mol. [NCERT] 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. $\therefore$ Internal energy change $(\Delta E)$ during combustion of one mole of Reaction of combustion of octane: If there is trend, use it to predict the molar heat capacity of Fr. Q. $\Delta G_{f}^{o} C a C O_{3}(g)=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}$ Standard vaporization enthalpy of benzene at boiling point is $30.8 \mathrm{kJ} \mathrm{mol}^{-1} ;$ for how long would a $100 \mathrm{W}$ electric heater have to operate in order to vaporize a $100 \mathrm{g}$ sample at the temperature? (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. 5.1 Thermodynamics notes. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$ ... THERMODYNAMICS Interview Questions And Answers <â- CLICK HERE. $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$ is $-92.38 k J$ at $298 K .$ What is 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. $\Delta G^{\circ}=-2.303 R T \log K$ SHOW SOLUTION $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ Q. (ii) $\quad \Delta S=+v e$ because aqueous solution has more disorder than solid. since $\Delta_{r} G^{\circ}$ is negative, the reaction will be spontaneous. (Hint. 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. 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 $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$ Enthalpy is defined as heat content of the system $H=U+P V$ Express the change in internal energy of a system when, (i) No heat is absorbed by the system from the surroundings, but work (, Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2, 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 $24 J m o l^{-1} K^{-1}$. (iii) 1 mol of a liquid X. Q. $\Delta G=120-380=-260 k J$ (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. 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. (ii) Temperature of crytal is increased. When the two bodies one hot and the other cold, are placed in contact with each other, then the hot body loses heat and becomes colder and the cold body gains heat and becomes hotter, and â¦ $C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H=-393.5 k_{0} J m o l^{-1}$ 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 $0^{\circ} C$, $\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$. 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? SHOW SOLUTION We intend them to be used only for the purpose of studying and learning. Predict the sign of entropy change for each of the following changes of state: $=\frac{6.04 \times 2.38}{2.8}=0.5134 k_{0} J$ 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 (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) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. Calculate standard molar entropy change of the formation of $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ SHOW SOLUTION We have transformed classroom in such a way that a student can study anytime anywhere. $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. (i) Liquid to vapours and which is very easy to understand and improve your skill. 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. – oxygen bond in $\mathrm{O}_{2}$ molecules. Place the following systems in order of increasing randomness : The equilibrium constant for a reactions is $10 .$ What will be the value of $\Delta G^{\circ} ?$. Email: help@24houranswers.com $\Delta S_{v a p . Is there any enthalpy change in a cyclic process ? SHOW SOLUTION Given that power$=\frac{\text { Energy }}{\text { time }} \Rightarrow$time$=\frac{\text { energy }}{\text { power }}$You can also Download or view Key Concepts of Thermodynamics & Thermochemistry . Mass of$P=10.32 \mathrm{g}$As heat is taken out, the system must be having thermally conducting walls.$\Delta_{v a p} H^{\ominus}$of$C O=+6.04 \mathrm{kJmol}^{-1}$Therefore$\Delta E=0$under isothermal conditions. This material is made available for the sole purpose of studying and learning - misuse is strictly forbidden. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Power$\left.=\frac{\text { energy }}{\text { time }} \text { and } 1 W=1 \quad J s^{-1}\right)$. What is the sign of$\Delta S$for the forward direction? $c=2.22 \mathrm{mol} \times 24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}=53.28 \mathrm{JK}^{-1}$$\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$2.47 \times 10^{-29}$,$R=8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}, T=298 \mathrm{K}, \mathrm{K}=2.47 \times 10^{-29}$,$\therefore \Delta_{r} G^{\circ}=-2.303 \times\left(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\right) \times(298 \mathrm{K}) \times$,$\log \left(2.47 \times 10^{-29}\right)=163.2 \mathrm{kJ} \mathrm{mol}^{-1}$. No, there is no enthalpy change in a cyclic process because the system returns to the initial state. SHOW SOLUTION (ii)$\quad \Delta n=1-3=-2$. SHOW SOLUTION$g$of$C O_{2}$from carbon and dioxygen gas. (i)$\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { product })-\Sigma S^{\circ}(\text { reactants })C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$(iii) by 2 and add to eqn. Calculate$\Delta S$for the conversion of: One should spend 1 hour daily for 2-3 months to learn and assimilate Thermodynamics â¦$\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$(ii) Calculate the value of$\Delta n$in the following reaction:$\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l) \cdot\left(\Delta \mathrm{U}=-85389 \mathrm{Jmol}^{-1}\right)$. Thus, entropy increases. (ii)$\quad H_{2} O(l)$at$100^{\circ} C \longrightarrow H_{2} O(l)$at$0^{\circ} C$SHOW SOLUTION Compare it with entropy decrease when a liquid sample is converted into a solid. 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. Hence it is non-spontaneous. (iii) Calculation of$\Delta U$SHOW SOLUTION (ii)$\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$What is the value of internal energy for 1 mole of a mono-atomic gas ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i)$H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$, (ii)$H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$. In what way is it different from bond enthalpy of diatomic molecule ? Enthalpy of solution of$N H_{4} N O_{3}=\frac{5.282}{20} \times 80\quad-\left[\Delta G_{f}^{\circ} \mathrm{CaCO}_{3}(\mathrm{s})+2 \Delta G_{f}^{\circ} \mathrm{H}^{+}(a q)\right]A g_{2} O(s) \rightarrow 2 A g(s)+\frac{1}{2} O_{2}(g)$(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.$\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)\Delta H=\Delta E$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. [NCERT] 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$=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION (iv)$\quad C$(graphite)$\rightarrow C$(diamond). If it's not in your inbox, check your spam folder. since, the value of$\Delta G_{r}^{\circ}$is negative, therefore, the reaction is You will get here all the important questions with answers for class 11 Chemistry Therodynamics and chapters. In the isothermal expansion the temperature remains constants. Visit eSaral Website to download or view free study material for JEE & NEET. Continue without uploading, Attachhomework files$\Rightarrow C_{v}=\Delta E_{K}$Question 1. Place the following systems in order of increasing randomness : (ii)$\quad H C l$is added to$A g N O_{3}$solution and precipitate of$A g C l$is obtained.$\Delta S_{3}=-\frac{\Delta H_{f}}{T_{f}}=-\frac{1435 \mathrm{cal} \mathrm{mol}^{-1}}{273 \mathrm{K}}=-5.26 \mathrm{cal} K^{-1} \mathrm{mol}^{-1}$,$\Delta H_{v}=\left(5397 \operatorname{cal} g^{-1}\right)\left(180 g m o l^{1}\right)=97146 \mathrm{calmol}^{1}$,$\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$. (iii)$2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$Question3: Explain the following terms: Isolated system, Open system and closed system and give example where â¦ 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. so$\mathrm{NO}(g)$is unstable. 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$24 J m o l^{-1} K^{-1}\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$SHOW SOLUTION$-228.6 \mathrm{kJmol}^{-1}$respectively. These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Thus$A l_{2} O_{3}$cannot be reduced by$C$[NCERT] (ii) At what temperature, the reaction will reverse? Let us calculate$T$at which$\Delta_{r} G^{\circ}$becomes zero (2) Answer: Heat absorbed by the system, (q) = + 701 J Work done by the system â¦ SHOW SOLUTION Warning: If you try using the HL in an unethical manner, expect to fail your class. What type of wall does the system have ?$\Rightarrow \Delta H$during vapourisation of$28 g=6.04 \mathrm{kJ}$Explain both terms with the help of examples. Q.$R=8.314 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$How many times is molar heat capacity than specific heat capacity of water ? SHOW SOLUTION$\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. because$\Delta G_{r}$is$+22500 \mathrm{kJ} .$When this process is coupled with Formula sheet. (ii)$\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })\Delta G=\Delta H-T \Delta S-v e=\Delta H-(+v e)(-v e)$. Therefore, the reaction will not be spontaneous below this temperature.$S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$Thus$A l_{2} O_{3}$cannot be reduced by$C$, (ii) Let us now calculate the$\Delta G$value for reduction of$P b O$,$2 P b O \longrightarrow 2 P b+O_{2} ; \Delta+120 k J$,$C+O_{2} \longrightarrow C O_{2} ; \Delta G=-380 k J$, On adding,$2 P b O+C \longrightarrow 2 P b+C O_{2}$, Hence$P b O$can bereduced by carbon because$\Delta \mathrm{G}$of couple reaction is$-v e$. (i)$\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]\Rightarrow$Enthalpy change,$\Delta H=\Delta E+\Delta n_{g} R T$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. SHOW SOLUTION SHOW SOLUTION We have provided Thermodynamics Class 11 Chemistry MCQs Questions with Answers â¦ By plotting graph between molar heat capacity and atomic mass, the molar heat capacity of$F r$, (atomic mass$=223$) would be$33.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Q. 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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Standard vaporization enthalpy of benzene at boiling point is$30.8 \mathrm{kJ} \mathrm{mol}^{-1} ;$for how long would a$100 \mathrm{W}$electric heater have to operate in order to vaporize a$100 \mathrm{g}$sample at the temperature? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. These type of Thermodynamics Objective Questions are asked in various competitive examination, and Technical section of Railway Recruitment Examinations, BHEL, SAIL, HAL, BEL and Other Mechanical Engineering Examination. 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$E_{k}=3 / 2 R T$Q. SHOW SOLUTION SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol.$\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$g .$0=\Delta H-T \Delta S$or$\Delta H=T \Delta S$or$T=\frac{\Delta H}{\Delta S}$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 ? 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.,$\Delta U=\Delta E_{K}$So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. ($i \text { ) from eq. (ii) $\quad \mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s}) \rightarrow \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ Also, the order of entropy for the three phases of the matter is $S(g)>>S(l)>S(s)$ Now, forward reaction is exothermic, therefore the entropy change for the forward direction should be negative and large $(T \Delta S>\Delta H)$, Q. (ii) At what temperature, the reaction will reverse? Q. The temperature falls from $296.5 \mathrm{Kto} 286.4 \mathrm{K.Find}$ out the value of $q$ for calorimeter and its contents. 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 $0^{\circ} \mathrm{C}$. Predict the sign of entropy change in the following reactions: $=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. octane $=\frac{60.0989}{1.250} \times 114=5481.02 k J \mathrm{mol}^{-1}$ We know Material may not be reproduced in part or whole without written consent of the. Professionals, Teachers, Students and Kids Trivia Quizzes to test your knowledge on the subject. At equilibrium $\Delta G=0$ so that Under what condition $\Delta H$ becomes equal to $\Delta E ?$ Save my name, email, and website in this browser for the next time I comment. $=-800.78 \mathrm{kJ} \mathrm{mol}^{-1}$ 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. 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