$-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$ $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. What will be the direction of the reaction at this temperature and below this temperature and why? Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam. 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. 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. The enthalpy change for the reaction: Calculate Gibbs energy change for the reaction is spontaneous or not. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. Multiply eqn. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: This is only a preview of the solution. $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$ $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$ Email: help@24houranswers.com $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ Sorry, there was a problem with your payment. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. For a reaction both $\Delta H$ and $\Delta S$ are positive. $\Delta H$ and $\Delta S$ for the reaction: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta E$ 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. $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$ 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 $24 J m o l^{-1} K^{-1}$. $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $\Delta H$ and $\Delta U$ are related as Treat heat capacity of water as the heat capacity of calorimeter and its content). These important questions will play significant role in clearing concepts of Chemistry. $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ (i) $\quad \Delta S\Delta H)$, Q. $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ (2) Answer: Heat absorbed by the system, (q) = + 701 J Work done by the system … (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) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$ (iii) by 2 and add to eqn. Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$ $\Delta n=2-4=-2$ Predict the sign of entropy change for each of the following changes of state: Q. Calculate the standard entropy change for the reaction Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Continue without uploading, Attachhomework files Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ Clapeyron discovered an important fundamental equation which helps in finding the extensive application in one component, two phase system and by Clausius from second law of thermodynamics. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Here, we are given, $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$, Q. $A+B \rightarrow C+D$ As no heat is absorbed by the system, the wall is adiabatic. 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