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. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$, 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. $\Rightarrow C_{v}=\Delta E_{K}$ (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. $\Delta H$ and $\Delta S$ for the reaction: 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. Also calculate enthalpy of solution of ammonium nitrate. $\frac{1}{2} H_{2}(g)+\frac{1}{2} C l_{2}(g) \rightarrow H C l(g) \Delta_{f} H^{\circ}=-92.8 k J m o l^{-1}$ [CONFIRMED] JEE Main will be conducted 4 times from 2021! Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2 R. Thus, entropy increases. Thermodynamics article. Thermodynamics One Word Interview Questions and Answers PDF _ Mechanical Engineering Questions and Answers. Electricity key facts (1/9) • Electric charge is an intrinsic property of the particles that make up matter, and can be Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? Q. The change in internal energy is a state function and it depends upon the temperature only. That will be very helpful in quick revision during Exams. $\Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ}$ zero. $\therefore \quad T=\frac{\Delta_{r} H}{\Delta_{r} S}$ $=\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1}$ Q. $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$ Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. -condensation into a liquid. $-168.0=0+\Delta_{f} H^{\circ}\left(\mathrm{C} l^{-}\right)-\left[\frac{1}{2} \times 0+\frac{1}{2} \times 0\right]$ Calculate $\Delta S$ for the conversion of: (ii) Vapours to liquid at $35^{\circ} \mathrm{C}$, $\Delta S_{v a p .$\mathrm{CCl}_{4}(l) \rightarrow \mathrm{CCl}_{4}(\mathrm{g}) \Delta \mathrm{H}^{\circ}=30.5 \mathrm{kJ} \mathrm{mol}^{-1}$[NCERT] (iii) w amount of work is done by the system and q amount of heat is supplied to the system.$\operatorname{SiH}_{4}(g)+2 O_{2}(g) \rightarrow \operatorname{Si} O_{2}(s)+2 H_{2} O(l)=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$5.1 Thermodynamics notes. Energy required for vapourising$28 g$of$C O=6.04 k_{U}$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$-v e$.$\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$Give suitable examples.$\therefore W=-p_{e x} \Delta V \quad\left(\Delta V=\frac{n R T}{P_{e x t}}\right)\Delta n_{g}=8-\frac{25}{2}=-\frac{9}{2}$are$30.56 \mathrm{kJ} \mathrm{mol}^{-1}$and$66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$respectively. 2000 AP CHEMISTRY FREE RESPONSE QUESTIONS. What is meant by average bond enthalpy ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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$\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$The enthalpy change for the reaction: (ii)$\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}A g_{2} O(s) \rightarrow 2 A g(s)+\frac{1}{2} O_{2}(g)$(iii) A partition is removed to allow two gases to mix. We know (i)$\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$SHOW SOLUTION SHOW SOLUTION SHOW SOLUTION Energy required for vapourising$2.38 g$of$\mathrm{CO}$(ii) Below this temperature,$\Delta G$will be$+$ve because both$\Delta \mathrm{H}$and$\mathrm{T} \Delta \mathrm{S}$are positive and$\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}(\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=+\mathrm{ve})\Delta_{v a p} H^{\ominus}$of$C O=6.04 \mathrm{kJ} \mathrm{mol}^{-1}$, 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.$\Delta H=8 B_{C-H}+2 B_{C-C}+5 B_{O=O}-6 B_{C=O}-8 B_{O-H}$,$-2050 k J=8 \times 414+2 \times 347+5 B_{O=O}-6 \times 741-8 \times 464$,$-2050 k J=3312 k_{0} J+694 k J+5 B_{O=0}-4446 k J-3712 k_{\circlearrowright} J$,$-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$,$-2050 \mathrm{kJ}=-4152 \mathrm{kJ}+5 \mathrm{B}_{\mathrm{O}=0}$,$B_{O=O}=\frac{2102}{5}=420.4 \mathrm{kJ} \mathrm{mol}^{-1}$. 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$E_{k}=3 / 2 R T$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. Q. Explanation are given for understanding. \quad$, $-92.38 k J=\Delta U-2 \times 8.314 \times 10^{-3} k J \times 298 k$, $\Delta U=-92.38 k J+4.955=-87.425 k J m o l^{-1}$, Q. Calculate the entropy change when $36 g$ of liquid water evaporates at $373 K\left(\Delta_{v o p} H=40.63 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ Internal energy change is measure at constant volume. In the isothermal expansion the temperature remains constants. $\Delta S_{1}=-\frac{\Delta H_{v}}{T_{b}}=-\frac{9714.6 \mathrm{cal} \mathrm{mol}^{-1}}{373 \mathrm{K}}$ Given that the enthalpy of vapourisation of carbon monoxide is $6.04 \mathrm{kJ} \mathrm{mol}^{-1}$ at its boiling point of $82.0 K$ Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. $-228.6 \mathrm{kJmol}^{-1}$ respectively. These important questions will play significant role in clearing concepts of Chemistry. Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$ Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ Here, we are given, $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$, Q. SHOW SOLUTION As per the second law of thermodynamics god exist, some person have believe that the world is so beautiful and only god can construct such beautiful world.Some scientist believe that the DNA structure is very much complicated and only god can create it.But as per the one scientist that god exist out of the our knowledge boundary. SHOW SOLUTION Q. Silane $\left(S i H_{4}\right)$ burns in air as: What is its equilibrium constant. SHOW SOLUTION Lost your password? Q. $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state. SHOW SOLUTION $\Delta G_{f}^{\circ} C O_{2}(g)=-394.36 k J m o l^{-1}$ Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION SHOW SOLUTION }=\frac{\Delta H_{v a p . In what way internal energy is different from enthalpy ? Multiply eqn. These important questions will play significant role in clearing concepts of Chemistry. Q. 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. $C O_{2}=-393.5 k_{U}$ B. (ii) At what temperature, the reaction will reverse? the standard Gibbs energy for the reaction at $1000 K$ is $-8.1 \mathrm{kJmol}^{-1} .$ Calculate its equilibrium constant. $=0.5134 \mathrm{kJ}$ 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. (ii) Calculate the value of $\Delta n$ in the following reaction: With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. 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. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, $q=125 g \times 4.18 J / g \times(286.4-296.5)$, $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Q. $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$, $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of, Q. (iv) because graphite has more disorder than diamond. E=\Frac { 3 } { 2 } ( g ) $\quad S=+v! \Delta S=+v e$ because aqueous SOLUTION has more disorder than diamond compared to liquids, therefore, entropy.... You expect a decrease in entropy as a gas condenses into liquid thermodynamics by T.Roy Chowdhury, McGrawHill. 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