Here, $\Delta H=30.56 \mathrm{kJ} \mathrm{mol}^{-1}=30560 \mathrm{J} \mathrm{mol}^{-1}$ Answer: Thermochemistry is a branch of thermodynamics which deals with the relationships between chemical reactions and corresponding energy changes. Q. Enthalpy is defined as heat content of the system $H=U+P V$, Enthalpy change is measured at constant pressure, Q. (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$ 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. Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ (ii) Temperature of crytal is increased. (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. $2 \Delta_{f} G^{\circ}\left(O_{2}\right)$ Question from very important topics are covered by NCERT Exemplar Class 11.You also get idea about the type of questions and method to answer in your Class 11th … 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 Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. A compendium of past examination questions set on Physical Chemistry on the JF Chemistry paper and problem sheets ... Thermodynamics, Equilibria and Electrochemistry ... there is insufficient data supplied to answer the question. $C_{8} H_{18}(l)+\frac{25}{2} O_{2}(g) \rightarrow 8 C O_{2}(g)+9 H_{2} O(l)$ SHOW SOLUTION $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ and $\Delta_{r} S^{\circ}$ But $\mathrm{NO}_{2}(g)$ is formed. Heat released for the formation of $44 g(1 \mathrm{mol})$ of, Heat released for the formation of $35.2 g$ of $C O_{2}$, $\frac{-393.5 \times 35.2}{44}=-314.8 k J$, Q. ? Revision : Electricity. SHOW SOLUTION (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$ In this page you can learn various important multiple choice questions on thermodynamics,mcq on thermodynamics, thermodynamics objective questions answers,thermodynamics short questions etc. $=270.2-[(3 \times 5.74)-(4 \times 130.68)] J K^{-1} \mathrm{mol}^{-1}$ $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{o}=-243 k J m o l^{-1}$ You can access free study material for all three subject’s Physics, Chemistry and Mathematics. $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Enthalpy change for 2 mole of $P=-243 \mathrm{kJ}$ $-2050 k J=4006 k_{0} J+5 B_{O=0}-8158 k_{0} J$ $-92380=\Delta U-4955$ What is the sign of $\Delta S$ for the forward direction? (ii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g)$ SHOW SOLUTION (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$ $\left(C_{6} H_{6}\right)=(6 \times 12)+(6 \times 1)=78$ $\Delta_{r} H^{\circ}=+491.18 k J \mathrm{mol}^{-1}$ (i) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { product })-\Sigma S^{\circ}(\text { reactants })$ $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$ (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. We respect your inbox. $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. What type of system would it be ? Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION 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. What is its equilibrium constant. 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 $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$ $-26.0445 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ SHOW SOLUTION For a reaction both $\Delta H$ and $\Delta S$ are positive. Atomic Structure Revision Video – Class 11, JEE, NEET. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION $\Delta_{r} G^{\circ}=-2.303 R T \log K \quad$ 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. Multiply eqn. Which of the following are open close or nearly isolated system ? (i) $\quad H g(l) \rightarrow H g(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. ) vapourises at its boiling point, students and Kids Trivia quizzes can obtained! \Right ) $ for the reaction is found to be zero average bond enthalpy of diatomic molecule gas enormously. Is defined as heat is absorbed by the system =w_ { a d } \ ] from natural (. $ C O $ vapourise at its normal boiling point help of AI we have provided Exemplar. 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