What is its equilibrium constant. Zeroth law of thermodynamics. $-\left[\frac{1}{2} \Delta_{f} H^{o}\left(H_{2}\right)+\frac{1}{2} \Delta_{f} H^{o}\left(C l_{2}\right)\right]$ 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) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. ... THERMODYNAMICS Interview Questions And Answers <—- CLICK HERE. 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 $\Delta E_{k}=3 / 2 R(T+1)-3 / 2 R T=3 / 2 R$ $=[-553.58+(-237.13)+(-394.36)]-[-1206.9+0]$ Now $C_{v}=\frac{\Delta U}{\Delta T}$ and $\Delta T=1^{\circ} \mathrm{C}$ Electricity key facts (1/9) • Electric charge 𝑄𝑄is an intrinsic property of the particles that make up matter, and can be The standard Gibb’s energy of reactions at $1773 \mathrm{K}$ are given $\mathbf{a} \mathbf{S}$ $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, $q=125 g \times 4.18 J / g \times(286.4-296.5)$, $q=125 g \times 4.18 \times 10^{-3} \mathrm{kJ} / \mathrm{g} \times-10.1=-5.28 \mathrm{kJ}$, Q. \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. Please use the purchase button to see the entire solution. Why is $\Delta E=0,$ for the isothermal expansion of ideal gas? Give suitable examples. So, molar heat capacity of these elements can be obtained by multiplying specific heat capacity by atomic mass. Molar heat capacity of $N a(s)=1.23 \times 23=28.3 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ $-C l$ bond in $C C l_{4}(g)$ Proceeding from $\Delta_{f} H^{\circ} \mathrm{CO}_{2}=-393.5 \mathrm{kJ} \mathrm{mol}^{-1}$ and ther mo- chemical equation: Moreover, Class 11 Chemistry Thermodynamics solutions are available in PDF format for easy download. $\Delta H=\Delta U+P \Delta V$ Under what condition $\Delta H$ becomes equal to $\Delta E ?$. Hence this fundamental equation is known as Clapeyron-Clausius Equation. R=8.31 \mathrm{JK}^{-1} \mathrm{mol}-^{-1} . The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. (Given that power $=$ energy/time and $\left.1 W=1 J s^{-1}\right)$ $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, (i) $\quad 4 F e(s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)$, $S^{\circ}(F e(s))=27.28 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $S^{\circ}\left(O_{2}(g)\right)=205.14 J K^{-1} m o l^{-1}$, $S^{\circ}\left(F e_{2} O_{3}(s)\right)=87.4 \quad J K^{-1} m o l^{-1}$, (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$, $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$, $S^{\circ}\left[H_{2} O(l)\right]=69.91 J K^{-1} m o l^{-1}$, $S^{\circ} \mathrm{Ca}(\mathrm{OH})_{2}(a q)=-74.50 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$, $=\left(2 S_{F e_{2} O_{3}}^{\circ}\right)-\left(4 S_{F e(s)}^{\circ}+3 S_{O_{2}}^{\circ}(g)\right)$, $=(2 \times 87.4)-(4 \times 27.28+3 \times 205.14)$, $=(174.8)-(724.54)=-549.74 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, (ii) $\quad \Delta S_{\text {reaction}}^{\circ}=\Sigma S^{\circ}(\text { products })-\Sigma S^{\circ}(\text { reactans })$, $=\left(S_{C a(O H)_{2}}^{\circ}(a q)+S_{H_{2}(g)}^{\circ}-\left(S_{C a(s))}^{\circ}+2 S_{H_{2} O(l)}^{\circ}\right)\right.$, $=\{-74.50+130.68\}-\{41.42+(2 \times 69.91)\}$, $\Rightarrow 56.18-(41.42+139.82) \Rightarrow-125.06 J K^{-1} \mathrm{mol}^{-1}$, Q. $\Delta G^{\circ}=-2.303 R T \log K$ $\Delta G_{f}^{\circ} C a^{2+}(a q)=-553.58 k J m o l^{-1}$ Thus, entropy increases. Get help with your Thermodynamics homework. because $\Delta G_{r}$ is $+22500 \mathrm{kJ} .$ When this process is coupled with, combustion of $C$ to $C O_{2} .$ The net free energy change is calculated, Hence it is non-spontaneous. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (ii) $\quad \Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} \mathrm{O}(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ As no heat is absorbed by the system, the wall is adiabatic. In this no mass (water) cross the boundary. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Textbook Authors: Moore, John W.; Stanitski, … $I_{2}$ molecules upon dissolution. Calculate the standard enthalpy of formation of $C H_{3} O H(l)$ from the following data: A piston exerting a pressure of 1 atm rests on the surface of water at $100^{\circ} \mathrm{C} .$ The pressure is reduced to smaller extent when $10 g$ of water evaporates and $22.2 \mathrm{kJ}$ of heat is absorbed. Which of the following are open close or nearly isolated system ? SHOW SOLUTION $\Delta_{r} H^{\circ}=-726 \mathrm{kJmol}^{-1}$ 5 Penn Plaza, 23rd Floor Under what conditions will the reaction occur spontaneously? $\Delta_{R} H^{\circ}=-557.5 \mathrm{kJ} .$ Calculate the heat of the formation of $=2.303\left(18 \mathrm{cal} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \log \frac{273}{373}$ For the reaction All HL items are old, recycled materials and are therefore not original. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$ Calculate $\Delta S$ for the conversion of: As heat is taken out, the system must be having thermally conducting walls. Sorry, there was a problem with your payment. $-$ (i) $\quad S=+v e$ because liquid changes to more disordered gaseous state. $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$ Upload a file $=-0.56 \times 8.314 \times 10^{-3} \times 373.15=-1.737 k J$ Q. Silane $\left(S i H_{4}\right)$ burns in air as: \[ =\Delta_{d i s s} H^{o}=-75.2 k J m o l^{-1} \] $\Delta H^{\circ}=\Delta_{a} H^{\circ}(C)+4 \Delta_{a} H^{\circ}(C l)-\Delta_{f} H^{\circ}\left(C C l_{4}\right)(g)$ $\Delta_{a} H^{o}(C l)=\frac{1}{2} \times 242=121 k J m o l^{-1}$ are $30.56 \mathrm{kJ} \mathrm{mol}^{-1}$ and $66.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ respectively. $\Delta_{r} G^{\circ}=491.18 k J \mathrm{mol}^{-1}-298 \mathrm{Kx}$ 4. The change in internal energy is a state function and it depends upon the temperature only. $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, $X \Longrightarrow Y$ if the value of $\Delta H^{\circ}=28.40 \mathrm{kJ}$ and equilibrium, constant is $1.8 \times 10^{-7}$ at $298 \mathrm{K} ?$, $=-2.303 \times 8.314 \times 298 \times \log \left(1.8 \times 10^{-7}\right)=38484.4$, Now, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$, $\therefore \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}=\frac{28400-38484.4}{298}$, $=-33.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$, Q. Compare it with entropy decrease when a liquid sample is converted into a solid. Given that power $=\frac{\text { Energy }}{\text { time }} \Rightarrow$ time $=\frac{\text { energy }}{\text { power }}$ $\Rightarrow \Delta H$ during vapourisation of $28 g=6.04 \mathrm{kJ}$ (ii) Calculation of $w$ The enthalpy change $(\Delta H)$ for the reaction: Molar mass of phosphorus $=30 \mathrm{gmol}^{-1}$, Moles of $P=\frac{10.32}{31}=0.333 \mathrm{mol}$, Enthalpy change for 2 mole of $P=-243 \mathrm{kJ}$, Enthalpy change for 0.333 mole of $P=-\frac{243}{2} \times 0.333$, Q. $C(s)+H_{2} O(g) \rightleftarrows C O(g)+H_{2}(g)$ 250+ Thermodynamics Interview Questions and Answers, Question1: State the third law of thermodynamics. SHOW SOLUTION }}{T_{b}}=\frac{\left(26000 \mathrm{J} \mathrm{mol}^{-1}\right)}{(308 \mathrm{K})}=84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \mathrm{since}$ $\therefore \quad \Delta H=+22.2 k_{0} J$ (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. Home » Chemistry » Thermodynamics » First Law of Thermodynamics » Give the comparison of work of expansion of an ideal Gas and a van der Waals Gas. Given : Average specific heat of liquid water $=1.0 \mathrm{cal} K^{-1} g^{-1} .$ Heat of vaporisation at boiling point $=539.7$ cal $g^{-1} .$ Heat of fusion at freezing point $=79.7 \mathrm{cal} g^{-1}$, Download or view Key Concepts of Thermodynamics & Thermochemistry. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. What type of system would it be ? (ii) Due to settling of solid $A g C l$ from solution, entropy decreases. (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$ $K_{p}$ for this conversion is $2.47 \times 10^{-29}$ $\Delta G_{f}^{o} \mathrm{CO}_{2}(g)=-394.36 \mathrm{kJ} \mathrm{mol}^{-1}$ $=[-394.36+\{2 \times(-228.57)\}-[-50.72+0]$ $S(C a(s))^{\circ}=41.42 \quad J K^{-1} m o l^{-1}$ $\left[\text { If } \Delta G_{f}^{o} N O_{2}=51.3 \Delta G_{f}^{o}(N O)=86.55\right]$, $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$, since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is, Q. Thermodynamics is the study of energy transformations. We also acknowledge previous National Science … (ii) $\operatorname{CaCQ}(s)+2 H^{+}(a q) \rightarrow C a^{2+}(a q)+H_{2} O(l)+C O_{2}(g)$ 1. You will get here all the important questions with answers for class 11 Chemistry Therodynamics and  chapters. $\Delta S_{\text {condensation }}=-84.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$. If not at what temperature, the reaction becomes spontaneous. $=21.129 \mathrm{kJ} \mathrm{mol}^{-1}$, $\begin{aligned} \therefore \text { Total heat capacity of calorimeter } &=125 \times 4.184 \\ &=523 \mathrm{JK}^{-1} \end{aligned}$, $\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$, \[ \Rightarrow q=523 \times(-10.1)=-5282.3 \mathrm{J}=-5.282 \mathrm{kJ} \], Thus, calorimeter loses $5.282 \mathrm{kJ}$ of heat during dissolution, of $20 g$ of $N H_{4} N O_{3}$ in $125 g$ water Molecular mass of, $=(2 \times 14)+(4 \times 1)+(3 \times 16)=80$, Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$, Q. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. 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. (iv) because graphite has more disorder than diamond. For the water gas reaction $\left.\mathrm{Q}_{5} \text { ) }+\mathrm{H}_{2} \mathrm{Q}(g) \rightleftharpoons \mathrm{C} Q_{g}\right)+\mathrm{H}_{2}(\mathrm{g})$ How many times is molar heat capacity than specific heat capacity of water ? We have provided Thermodynamics Class 11 Chemistry MCQs Questions with Answers … (i) $\quad \Delta H=\Delta U+\Delta n R T$ It is mainly based on three laws of thermodynamics. $\Delta H_{f}=\left(79.7 \mathrm{cal} g^{-1}\right)\left(18.0 \mathrm{g} \mathrm{mol}^{-1}\right)=1435 \mathrm{cal} \mathrm{mol}^{-1}$ SHOW SOLUTION Q. NCERT Class 11 Chemistry Thermodynamics PDF. 5.1 Test (mark scheme) More Exam Questions on 5.1 Thermodynamics (mark scheme) 5.1 Exercise 1 - calculating approximate enthalpy changes 5.1 Exercise 2 - Born … (Files = Faster Response). Internal energy : The energy of a thermodynamic system under given conditions is called internal energy. Treat heat capacity of water as the heat capacity of calorimeter and its content). Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Professionals, Teachers, Students and Kids Trivia Quizzes to test your knowledge on the subject. An exothermic reaction $X \rightarrow Y$ is spontaneous in the back direction. Thermodynamics key facts (4/9) ... • Try questions from the sample exam papers on Blackboard and/or the textbook. 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. Calculate the standard molar entropy change for the following reactions at $298 K$ SHOW SOLUTION Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2 R. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. (i) $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$ We know Explain both terms with the help of examples. 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. (i) At what temperature the reaction will occur spontaneously from left to right? $S^{\circ}\left(H_{2}\right)(g)=130.68 J K^{-1} m o l^{-1}$ On adding, $2 P b O+C \longrightarrow 2 P b+C O_{2}$ Our 1000+ Thermodynamics questions and answers focuses on all areas of Thermodynamics covering 100+ topics. SHOW SOLUTION $\frac{3}{2} O_{2}(g) \rightarrow O_{3}(g)$ at $298 K$ Heat released for the formation of $35.2 g$ of $C O_{2}$ Energy required for vapourising $28 g$ of $C O=6.04 k_{U}$ Treat heat capacity of water as the heat capacity of calorimeter and its content). Question3: Explain the following terms: Isolated system, Open system and closed system and give example where … (ii) $\Delta G_{r}^{\circ}=\left[\Delta G_{f}^{\circ} \mathrm{C} a^{2+}(a q)+\Delta G_{f}^{\circ} H_{2} O(l)+\Delta G_{f}^{\circ} \mathrm{CO}_{2}(g)\right]$ Internal energy change is measure at constant volume. $\Delta G=120-380=-260 k J$ Here is a list of Thermodynamics MCQs with Answers (Multiple Choice Questions) is given below. (ii) $\quad A g N O_{3}(s) \rightarrow A g N O_{3}(a q)$ $=(174.8)-(109.12+615.42)$ }=\frac{\Delta H_{v a p . THERMODYNAMICS Mechanical Interview Questions And Answers pdf free download for gate,objective questions,mcqs,online test quiz bits,lab viva manual. Thermodynamics … (ii) $\quad C a(s)+2 H_{2} O(l) \rightarrow C a(O H)_{2}(a q)+H_{2}(g)$ Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. SHOW SOLUTION $C(s)+2 C l_{2}(g) \rightarrow C C l_{4}(g) \Delta H^{\circ}=-135.5 k J m o l^{-1}$ What is meant by average bond enthalpy ? Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. since, the value of $\Delta G_{r}^{\circ}$ is negative, therefore, the reaction is First law of thermodynamics. No, there is no enthalpy change in a cyclic process because the system returns to the initial state. (i) Write the relationship between $\Delta H$ and $\Delta U$ for the process at constant pressure and temperature. $=[2 \times 51.3]-[2 \times 86.55+0]$ $\Delta U=-92380+4955=-87425 J=-87.425 k J$, $N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$, $\Delta H-92.38 k J=-92380 J, R=8.314 J K^{-1} \mathrm{mol}^{-1}$, $-92380=\Delta U-2 \times 8.314 \times 298$, $\Delta U=-92380+4955=-87425 J=-87.425 k J$, Q. The standard free energy of a reaction is found to be zero. The reaction is spontaneous in the backward direction, therefore, $\Delta G$ is positive in the forward direction. Here, $C_{m}=24.0 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; n=\frac{60}{27}=2.22 \mathrm{mol}$ since process occurs at constant $T$ and constant $P$ Thus heat change refers to $\Delta H$ (iii) $\quad \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$ (iii) $\quad I_{2}(g) \rightarrow I_{2}(s)$ Click Here for Detailed Notes of any chapter. $\therefore$ Heat change $(q)$ for calorimeter $=$ Heat capacity $\times \Delta T$ 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. 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 $F r$ $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$ Under what conditions will the reaction occur spontaneously? $\Delta H^{\circ}=-168.0 \mathrm{kJ} \mathrm{mol}^{-1}$ $\Delta_{a} H^{\circ}\left(C l_{2}\right)=242 k J m o l^{-1}$ Let us calculate $T$ at which $\Delta_{r} G^{\circ}$ becomes zero Open system : (i) Human being (ii) The earth (v) A satellite in orbit, Enthalpy is defined as heat content of the system $H=U+P V$, Enthalpy change is measured at constant pressure, Q. 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 SHOW SOLUTION $\Delta_{r} H^{\circ}=-239 k J m o l^{-1}$, $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$, $\Delta_{r} H^{\circ}=-726 k J m o l^{-1}$, $C(g)+O_{2}(g) \rightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o l^{-1}$, $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l) ; \Delta_{r} H^{\circ}=-286.0 k J m o l^{-1}$, $C(\text { graphite })+2 H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow C H_{3} O H(l)$, (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$, (ii) $\quad C(g)+O_{2}(g) \longrightarrow C O_{2}(g) ; \Delta_{r} H^{\circ}=-393 k J m o r^{1}$, (iii) $H_{2}(g)+\frac{1}{2} O_{2}(g) \longrightarrow H_{2} O(\eta) ; \Delta_{r} H^{\circ}=-2860 \mathrm{kJ} \mathrm{mol}^{-1}$, Multiply eqn. 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