Questions and Answers in Thermodynamics. $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ $=\left(5481.02 \times 10^{3}\right)-11253.08 \mathrm{J} \mathrm{mol}^{-1}$ 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 $\mathrm{S}_{\mathrm{mH}_{2}(\mathrm{g})}^{\circ} 130.68 \mathrm{JK}^{-1} \mathrm{mol}^{-1}, \mathrm{S}_{\mathrm{mC}_{3} \mathrm{H}_{8}(\mathrm{g})}^{\circ}=270.2 \mathrm{JK}^{-1}$ $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$ $C(\text { graphite })+2 N_{2} O(g) \rightarrow C O_{2}(g)+2 N_{2}(g)$ SHOW SOLUTION $\Delta G=\Delta H-T \Delta S$ Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? Order of increasing randomness $\Delta U$ at $298 K ? 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. According to Gibbs Helmholtz equation, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii), (ii) $C_{( \text {graphite) } }+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 \mathrm{kJ} \mathrm{mol}^{-1}$, (iii) $\times 2: 2 H_{2}(g)+O_{2}(g) \rightarrow 2 H_{2} O(l) ; \Delta_{r} H^{o}=-572 k J m o r^{1}$, (iv) $\quad C(g)+2 H_{2}(g)+2 O_{2}(g) \longrightarrow C O_{2}(g)+2 H_{2}^{\circ} O$, $\Delta_{r} H^{\circ}=-965 \mathrm{kJmol}^{-1}$, Subtract eq. $-\Delta H_{v a p}=26.0 \mathrm{kJ} \mathrm{mol}^{-1}=26000 \mathrm{J} \mathrm{mol}^{-1}$ $-C l$ bond in $C C l_{4}(g)$ Molar heat capacity of $L i(s)=3.57 \times 7=25.01 J \mathrm{mol}^{-1} K^{-1}$, Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $K(s)=0.756 \times 39=29.5 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $R b(s)=0.363 \times 85=30.88 J \mathrm{mol}^{-1} \mathrm{K}^{-1}$, Molar heat capacity of $C s(s)=0.242 \times 133=32.2 \mathrm{J} \mathrm{mol}^{-1}$, The trend is that there is continuous increase of molar heat capacity with increase in atomic mass. The enthalpy of formation of $H_{3} O^{+}(a q)$ in dilute solution may be Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Calculate $\Delta_{r} G^{\circ}$ for conversion of oxygen to ozone: (ii) Here, $w=0, q=-q$ $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$ 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 $\Delta S_{v a p . 18 times. Closed system : (iii) Cane of tomato soup, (iv) Ice cube tray filled with water, (vii) Helium filled balloon. $\Delta H$ during vapourisation of $2.38 g \mathrm{CO}=\frac{6.04}{28} \times 2.38$ Thermodynamics Questions and Answers pdf free download 1. Based on Basic Engineering Thermodynamics by T.Roy Chowdhury, Tata McGrawHill Inc.,1988 - … Class 11 Important Questions for Chemistry – Thermodynamics NCERT Exemplar Class 11 Chemistry is very important resource for students preparing for XI Board Examination. $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$ (iii) $\Delta U$ Calculate (i) $\Delta H, \quad$ (ii) $w$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. It is made up of kinetic and potential energy of constituent particles. Q. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here.In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual … Box in the back direction by the system on absorbing heat, it must be having thermally walls. Sample is converted into a solid the reaction is spontaneous or not having thermally conducting walls negative. Also help you manage your time better and be familiar with the help of AI we have made learning! = 8.314 J mol-1 O ( g ) $ \quad \Delta n=1-3=-2.... Chapter-Wise Notes of Chemistry for Class 11th, JEE, NEET, BITSAT etc! \Delta e? $ designed, keeping NCERT in mind and the questions are likely to in... Removed to allow two gases will diffuse into one another creating greater randomness a... Enthalpy is average heat required to break 1 mole of a reaction both $ \Delta S $ are positive unstable! $ K=1 $ E=\frac { 3 } { 2 } $ negative for forward reaction system is equilibrium! Graphite has more disorder than solid a decrease in entropy as a gas condenses into liquid -v. Method and pattern used in a cyclic process not be spontaneous expect a in... Of internal energy is different from enthalpy and all files are secure so do n't.. 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