{"id":9039,"date":"2019-11-28T17:33:09","date_gmt":"2019-11-28T12:03:09","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9039"},"modified":"2021-07-02T15:44:30","modified_gmt":"2021-07-02T10:14:30","slug":"1st-puc-chemistry-question-bank-chapter-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-6\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 6 Thermodynamics"},"content":{"rendered":"

You can Download Chapter 6 Thermodynamics Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 6 Thermodynamics<\/h2>\n

1st PUC Chemistry Thermodynamics One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhat is thermodynamics ?
\nAnswer:
\nThermodynamics is a branch of chemistry that deals with the study of interconversion of with other forms energy during physical and chemical changes.<\/p>\n

Question 2.
\nWhat you mean by enthalpy?
\nAnswer:
\nThe total heat content of a system is called enthalpy.<\/p>\n

Question 3.
\nWrite the relationship between \u2206H and \u2206E.
\nAnswer:
\n\u2206H = \u2206E + RT\u2206n, \u2206H = Enthalpy change, \u2206E = Internal energy change
\n\u2206n = Number of gaseous product number of gaseous reactant or number of moles.<\/p>\n

Question 4.
\nDefine (Heat of reaction) Enthalpy of reaction.
\nAnswer:
\nHeat of a reaction is the change in enthalpy produced when the number of moles of the reactants as represented in the balanced chemical equation have completely reacted with each other.<\/p>\n

Question 5.
\nHeat (enthalpy) of formation.
\nAnswer:
\nHeat of formation of a compound is the change in enthalpy produced when one mole of the compound is formed from its elements denoted as \u2206Hf<\/sub>.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhat is Heat Capacity ?
\nAnswer:
\nHet capacity of a system is defined as quantity of heat required to raise temperature by 1\u00b0.<\/p>\n

Question 7.
\nWhat is specific heat capacity ?
\nAnswer:
\nIt is defined as the quantity of heat required to raise temperature of 1 gram of a substance by 1\u00b0 C or 1K<\/p>\n

Question 8.
\nWhat is the relation between CP<\/sub> and Cv<\/sub> for an ideal gas.
\nAnswer:
\nCP<\/sub>– Cv<\/sub> = R<\/p>\n

Question 9.
\nWhat is entropy ?
\nAnswer:
\nA measure of degree of disorder of a system.<\/p>\n

Question 10.
\nWhat is the unit of entropy ?
\nAnswer:
\nJ \/ K (Joule per Kelvin)<\/p>\n

Question 11.
\nWhat is Gibbs-Helmholtz equation ?
\nAnswer:
\n\u2206G\u00b0 = \u2206H\u00b0 – T\u2206S\u00b0.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nGive an expression for the work done in a reversible isothermal expansion of an ideal gas.
\nAnswer:
\nw = -2.303nRT log \\(\\frac{V_{2}}{V_{1}}\\)<\/p>\n

Question 13.
\nExplain : Heat of formation a acetylene is + 54 kcal.
\nAnswer:
\nWhen 1 mole of gaseous acetalylene is formed from its elements, carbon (s) and hydrogen (g), 54 kcal of heat is absorbed.<\/p>\n

Question 14.
\nWhite phosphorous is less stable than red phosphorous. Mention whether the process of conversion of white phosphorous to red phosphorous is exothermic or endothermic reaction.
\nAnswer:
\nExothermic reaction.<\/p>\n

Question 15.
\nWhat is the value of the standard enthalpy of formation of an element ?
\nAnswer:
\nZero<\/p>\n

Question 16.
\nThe standard enthalpies of formation of Hydrogen fluoride and Hydrogen chloride are -268.6 kJ and -92.3 kJ respectively. Which is between the two is stable?
\nAnswer:
\nHydrogen fluoride is more stable.<\/p>\n

Question 17.
\nGive an example for a reaction in which AH = AE
\nAnswer:
\nC(s) + O2<\/sub>(g) \u2192 CO2<\/sub>(g)<\/p>\n

Question 18.
\nWhat is the meaning of the statement the enthalpy of formation of PCl3<\/sub> is -373 kJ ?
\nAnswer:
\nWhen 1 mole of PCl5<\/sub> is formed from its elements, P(s) and Cl2<\/sub>(g), 373 kJ of heat is liberated.<\/p>\n

Question 19.
\nDefine ‘standard enthalpy of formation’.
\nAnswer::
\nStandard enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states under standard conditions i.e., at 298 K and 101.3 kPa pressure.<\/p>\n

Question 20.
\nDefine enthalpy of combustion.
\nAnswer:
\nEnthalpy of combustion Of a substance is the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.<\/p>\n

Question 21.
\nWrite the thermochemical equation for the combustion of glucose.
\nAnswer:
\nC6<\/sub>H12<\/sub>O6(s)<\/sub> + 6O2(g)<\/sub> \u2192 6CO2(g)<\/sub> + 6H2<\/sub>O(l)<\/sub> \u0394H = -xkJ<\/p>\n

Question 22.
\nThe enthalpy of combustion of graphite and diamond are -393.5 kJ and -395 kJ respectively. Which of them is more stable ?
\nAnswer:
\nGraphite.<\/p>\n

Question 23.
\nExplain : Heat of combustion of methane is -890 kJ.
\nAnswer:
\nHeat of combustion of methane is -890 kJ means when one mole of methane (CH4)
\ni. e., 16 grams of methane is completely burnt in oxygen, 890 kJ of heat is liberated.<\/p>\n

Question 24.
\nWrite the thermochemical equation for the reaction involving the burning of hydrogen gas in excess of air forming water and liberating heat.
\nAnswer:
\n2H2<\/sub>(g) + O2<\/sub>(g) \u2192 2H2<\/sub>O(l); \u0394H = -xkJ<\/p>\n

Question 25.
\nThe enthalpy of combustion of ethyl alcohol is – 1360 kJ \/ mol. Write the thermochemical equation of the reaction.
\nAns:
\nC2<\/sub>H5<\/sub>OH(l) + 3O2<\/sub>(g) \u2192 2CO2<\/sub>(g) + 3H2<\/sub>O(l) \u0394H = -1360 kJ \/ mole.<\/p>\n

Question 26.
\nDefine heat of enthalpy of a soultion.
\nAnswer:
\nHeat of a solution is the change in enthalpy produced when one mole of the solute is dissolved in excess of a solvent so that further dilution does not produce any heat exchange.<\/p>\n

\"KSEEB<\/p>\n

Question 27.
\nGive one difference between isolated system and closed system.
\nAnswer:
\nIsolated system can neigher exchange matter nor energy with the surroundings. Closed system can exchange energy but not matter with the surroundings.<\/p>\n

Question 28.
\nWhich of the following is an intensive property : Surface tension, mass, volume, enthalpy, density ?
\nAnswer:
\nSurface tension and density are intensive properties.<\/p>\n

Question 29.
\nWhat happens to the internal energy of the system if:
\n(a) Work is done on the system ? (b) Work is done by the system ?
\nAnswer:
\n(a) If work is done on the system, internal energy will increase, (b) If work is done by the system, internal energy will decrease.<\/p>\n

Question 30.
\nFor the reaction, N2<\/sub>(g) + 3H2<\/sub>(g) \u2192 2NH3<\/sub>(g) predict whether the work is done on the system or by the system.
\nAnswer:
\nVolume is decreasing therefore, work is done on the system.<\/p>\n

Question 31.
\nWhat is the limitation of first law of thermodynamics ?
\nAnswer:
\nIt cannot tell us the direction of the process.<\/p>\n

Question 32.
\nWhich of the following is an extensive property ?
\n(a) Volume, (b) Surface tension, (c) Viscosity, (d) Density
\nAnswer:
\n(a) Volume is an extensive property.<\/p>\n

Question 33.
\nWrite the relation between standard free energy change and equilibrium constant Kp for a reversible reaction.
\nAnswer:
\nRelationship between standard free energy change \u2206G\u00b0 and the equilibrium constant \u2206G\u00b0 = -2.303 RT log Kp<\/sub>, where R is the gas constant = 8.314 JK-1<\/sup>. T is the temperature in Kelvin.<\/p>\n

\"KSEEB<\/p>\n

1st PUC Chemistry Thermodynamics Two Marks Questions and Answers<\/h3>\n

Question 1.
\nWhat is exothermic reaction ? Give example.
\nAnswer:
\nThe reaction in which heat is evolved is called exothermic reaction.
\nExample : C(s)<\/sub> + O2(g)<\/sub> \u2192 CO2(g)<\/sub> \u2206H = -393.5 kJ<\/p>\n

Question 2.
\nWhat is endothermic reaction ? Give example.
\nAnswer:
\nThe reaction in which heat is absorbed is called endothermic reaction.
\nExample : N2(g)<\/sub> + O2(g)<\/sub> \u2192 2NO(g)<\/sub> \u2206H = +180.8kJ<\/p>\n

Question 3.
\nMention different types of process.
\nAnswer:<\/p>\n

    \n
  1. Isothermal process<\/li>\n
  2. Adiabatic process<\/li>\n
  3. Isoehoric process<\/li>\n
  4. Isobasic process<\/li>\n
  5. Reversible process<\/li>\n
  6. Irriversible process<\/li>\n
  7. Cyclic process<\/li>\n<\/ol>\n

    Question 4.
    \nWhat is open system ? Give example.
    \nAnswer:
    \nAn open system is one in which there is an exchange of both matter and energy with its surroundings.
    \nExample : Water in an open beaker.<\/p>\n

    Question 5.
    \nWhat is Extensive propery ? Give example.
    \nAnswer:
    \nIt is a property of which depends on the amount of the substance present in the system.
    \nExample : Mass, Volume, Energy<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nWrite any two difference between Isothermal and Adiabatic process.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 7.
    \nMention the factors affecting enthalpy of reaction.
    \nAnswer:<\/p>\n

      \n
    1. Physical state of reactant and product.<\/li>\n
    2. Amount of reactant and product.<\/li>\n
    3. Temperature of the reaction.,<\/li>\n
    4. Allotropic ferrous.,<\/li>\n
    5. Condition of constant volume or constant pressure.,<\/li>\n
    6. Reaction stoichiometry.<\/li>\n<\/ol>\n

      Question 8.
      \nState and illustrate Hess’s law.
      \nAnswer:
      \nHess’s law : Whether a chemical reaction takes place in a single step or in several steps, the total change in enthalpy remains the same.
      \nConsider the formation of CO2<\/sub> from C.
      \n\"1st<\/p>\n

      Question 9.
      \nExplain the meaning of thermochemical equation for the reaction.
      \n2H2<\/sub>(g) + O2<\/sub>(g) \u2192 2H2<\/sub>O(l) +136 kcal.
      \nAnswer:
      \nWhen two moles i.e., 4 grams of gaseous H2<\/sub> reacts with 1 mole (32g) of gaseous O2<\/sub> to give 2 moles of liquid water i.e., 36 gram of liquid water, 136 kcal of heat is liberated.<\/p>\n

      \"KSEEB<\/p>\n

      Question 10.
      \nExplain heat of neutralisation with an example.
      \nAnswer:
      \nHeat of neutralisation is the change in enthalpy produced when one gram equivalent weight of an acid is neutralised by one gram equivalent weight of a base in dilute solution.
      \nExample : HCl(aq) + NaOH(aq) \u2192 NaCl(aq) + H2<\/sub>O;= \u0394H – 57.3 kJ<\/p>\n

      Question 11.
      \nWith example explain the term ‘Heat of transition’.
      \nAnswer:
      \nHeat of transition is the charge in enthalpy produced when one mole of one allotropic form of an element is converted into another allotropic form.
      \nExample : C(diamond) > C(graphite); \u0394H = -X1<\/sub> kJ When 1 mole of diamond is coverted into graphite x kJ of heat is liberated, -x kJ is the heat of transition in this case.<\/p>\n

      Question 12.
      \nDefine heat (enthalpy) of transition.
      \nAnswer:
      \nIt is enthalpy change when one mole of an element is changed from one equilibrium to another.<\/p>\n

      Question 13.
      \nWhat do you understand by the statements :
      \n(i) enthalpy of formation of nitric oxide is +90.7 kJ.
      \n(ii) enthalpy of combustion of carbon disulphide is =1065.2 kJ
      \nAnswer:
      \n(i) When 1 mole of NO(g) is formed from its elements N2<\/sub>(g). and O2<\/sub>(g), 90.7 kJ of heat is absorbed.
      \n(ii) When 1 mole of CS2<\/sub> (l) is completely burnt in O2<\/sub>,1065.2 kJ of heat is liberated.<\/p>\n

      Question 14.
      \nEnthalpy of neutralisation of NH4<\/sub>OH with HC is 51.45 kJ. Calculate the enthalpy of ionization of ammonium hydroxide.
      \nAnswer:
      \nEnthalpy of ionzators of NH4<\/sub>OH = -51-41 – (57.3) = 5.8500 J<\/p>\n

      \"KSEEB<\/p>\n

      Question 15.
      \nExplain: “The enthalpy of formation of nitric oxide is +90.4 kJ”.
      \nAnswer:
      \n\\(\\frac { 1 }{ 2 }\\)N2(g)<\/sub> + \\(\\frac { 1 }{ 2 }\\) O2(g)<\/sub> \u2192 ANO(g)<\/sub>\u0394H = +90.4kJ When 1 mole of nitric oxide is formed from its elements 90.4 kJ of heat is absorbed. (Actually \u0394H\\+90.4kJ Error in the question)<\/p>\n

      Question 16.
      \nDistinguish between exothermic and endothermic reactions.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 17.
      \nState and explain the first law of thermochemistry.
      \nAnswer:
      \nIf enthalpy of formation of a compound is -x kJ, enthalpy of its decomposition intoits elements is + xkJ.
      \nExample : C(s) + O2<\/sub>(g) \u2192 CO2<\/sub>; \u0394H = -xkJ According to the law (Lavoisier & Laplae law) CO2<\/sub> (g) \u2192 C(s) + O2<\/sub> (g); \u0394H = +x kJ<\/p>\n

      Question 18.
      \nWhich of the following has highest heat of combustion out of the following and why ?
      \n(a) C2<\/sub>H6<\/sub> (b) C2<\/sub>H4<\/sub> (c) C2<\/sub>H2<\/sub> (d) CH4<\/sub>
      \nAnswer:
      \nC2<\/sub>H6(g)<\/sub> will have highest heat of combustion because it has highest molecular weight
      \nand second highest calorific value.
      \n\u0394Hc<\/sub> = calorific value of kJ\/g \u00d7 mol. wt. = 52 \u00d7 30 = 1560 kJ\/mol.<\/p>\n

      \"KSEEB<\/p>\n

      Question 19.
      \nWhat do you understand by state functions ? Neither q nor w is a state function but q + w is a state function ? Explain.
      \nAnswer:
      \nState function is a property whose value depends only uopon the state of system an is independent of the path, q and w are not state functions because they depend upon path. But q + w = AE which is a state function. AE does not depend upon path.<\/p>\n

      Question 20.
      \nAcetic acid and hydrochloric acid react With KOH solution. The enthalpy of neutralization of acetic acid is -55.8 kJ mol-1<\/sup> while that of hydrochloric acid is-57.3 kJ mol-1<\/sup>. Why?
      \nAnswer:
      \nIt is because HCl is strong acid, it ionises completely in aqueous solution whereas acetic acid is weak acid which does not ionise completely and some energy is used in its complete ionisation.<\/p>\n

      Question 21.
      \nTaking a specific example show that \u0394Stotal<\/sub> is a criterion for spontaneity of a change.
      \nAnswer:
      \n\u0394G = -T\u0394Stotal<\/sub>
      \nIf \u0394Stotal<\/sub> is +ve, \u0394G will be -ve, reaction will be spontaneous.
      \nIf \u0394Stotal<\/sub> is -ve, \u0394G will be +ve, reaction will be non-spontaneous,
      \n\u0394Stotal<\/sub> = \u0394Ssyst<\/sub>\u00a0+ \u0394Ssurroundings<\/sub>
      \nFor example in case of freezing of water at 273 K, TAStotai is +0.08 J\/K\/mol whereas at 274 K it is -0.080 J\/K\/mol. Since the \u0394S total is +ve at 273 K, freezing of water will take place. On the other hand at 274 K, the total entropy change is negative, the freezing will not occur.<\/p>\n

      Question 22.
      \nWhat is meant by entropy driven reaction ? How can a reaction with positive changes of enthalpy and entropy be made entropy driven ?
      \nAnswer:
      \nThose reactions which are endothermic and entropy is increasing such that T\u0394S > \u0394H are entropy driven reactions. If \u0394S total is +ve, the reaction will be spontaneous. If \u0394S is +ve and is small, but T is large, the reaction will be spontaneous.
      \nA reaction which is endothermic, i.e., \u0394H is +ve and \u0394S is +ve, T must be large.
      \n\u0394G = \u0394H – T\u0394S ; \u0394G will be -ve if T\u0394S > \u0394H, i.e., the reaction will be spontaneous at higher temperature because entropy will increase with increase in temperature which will make this reaction spontaneous.<\/p>\n

      \"KSEEB<\/p>\n

      Question 23.
      \nJustify the following statements:
      \n(a) An exothermic reaction is always thermodynamically spontaneous.
      \n(b) The entropy of a substance increases on going from liquid to vapour state at any temperature.
      \nAnswer:
      \n(a) It is false. Exothermic process are not always spontaneous. If \u0394S = -ve and T\u0394S > \u0394H, the process will be non spontaneous even if it is exothermic.
      \n(b) The entropy of vapour is more than that of liquid therefore entropy increases during vaposisation.<\/p>\n

      Question 24.
      \nEvaporation of water is an endothermic process but spontaneous. Explain.
      \nAnswer:
      \nIt is because entropy increases -during this process because water vapour have more entropy than liquid water. \u0394G become -ve because T\u0394S > \u0394H.<\/p>\n

      Question 25.
      \nWhat is meant by toral entropy change of a process ? Assuming the thermodynamic relationship \u0394G = \u0394H – T\u0394S, derive the relationship \u0394G = T\u0394Stotal<\/sub> for a system.
      \nAnswer:
      \nTotal entropy change of a process if entropy change taking place in universe, i.e.
      \n\u0394Ssystem<\/sub> + \u0394Ssurrounding<\/sub> = \u0394STotal<\/sub>
      \n\u0394STotal<\/sub> = \u0394Ssys<\/sub> – \\(\\frac{q}{T}\\)= \u0394SSys<\/sub> – \\(\\frac{\\Delta \\mathrm{H}}{\\mathrm{T}}\\)
      \nT\u0394STotal<\/sub> = TASSys<\/sub> – \u0394H [\u2235 \u0394G = \u0394H – T\u0394S]
      \nT\u0394STotal<\/sub> = -\u0394G
      \n\u0394G = -T\u0394STotal<\/sub><\/p>\n

      Question 26.
      \nAt a certain temperature ‘T’ endothermic reaction A \u2192 B proceeds virtually to end. Determine the sign of \u2206S for the reaction A \u2192 B and \u2206G for the reverse reaction B \u2192A.
      \nAnswer:
      \n\u2206S – +ve \u2206H = +ve, \u2206G = -ve the reaction proceeds virtually to end in reverse reaction B \u2192 A, \u2206S = -ve, \u2206G = +ve<\/p>\n

      \"KSEEB<\/p>\n

      Question 27.
      \nStarting with the thermodynamics relationship G = H – TS, derive the following relationship \u2206G = -T\u2206STotal<\/sub>
      \nAnswer:
      \nAt initial state of the system Gi<\/sub> = Hi<\/sub> – T Si<\/sub>
      \nAt final state of the system Gf<\/sub> = Hf<\/sub> – TSf<\/sub>
      \n\u2234 Change is free energy \u2206G = Gf<\/sub> – Gi<\/sub> = (Hf<\/sub> – Hi<\/sub>) – T (St<\/sub> – Si<\/sub>)
      \n\u2206G = \u2206H – T\u2206S<\/p>\n

      Question 28.
      \nExplain the help of example, the difference between bond dissociation energy and bond energy.
      \nAnswer:
      \nBond disoociation energy is energy required to break 1 mole of bonds e.g.
      \nH – H(g)) \u2192 2H(g) \u2206H = 436 kJ mol-1<\/sup>
      \nBond energy is energy released when 1 mole of bonds are formed example
      \n2H \u2192H2<\/sub>(g) \u2206H = -436kJ mol-1<\/sup>
      \nIn diatomic molecule both bond dissociation energy and bond energy are equal in magnitude but opposite in sign.<\/p>\n

      Question 29.
      \nWhat is meant by the free energy of a system ? What will be the direction of chemical reaction when (i) \u2206G = 0 (ii) \u2206G > 0 (iii) \u2206G < 0 Answer: Free energy is defined as energy which can be converted into useful work (i) \u2206G = 0, the reaction will be in equilibrium (ii) \u2206G > 0, the reaction will not take place
      \n(iii) \u2206G < 0, the reaction will be spontaneous<\/p>\n

      Question 30.
      \nWhy most of the exothermic process (reactions) spontaneous ?
      \nAnswer:
      \n\u2206G = \u2206H – \u2206H – T\u2206S; For exothermic reactions \u2206H is -ve. For a spontanec s process AG is to be -ve.
      \nThus decrease in enthalpy (-\u2206H) contributes significantly to the driving force (to make AG negative).<\/p>\n

      Question 31.
      \nThe enthalpy of combustion of sulphur is 297 kJ. Write the thermochemical equation for combustion of sulphur. What is the value of \u0394f<\/sub>H of SO2<\/sub> ?
      \nAns wer:
      \nS(s) + O2<\/sub>(g) \u2192 SO2<\/sub>(g); \u0394H = -297kJ and \u0394f<\/sub>H of SO2<\/sub> = -297 kJ mol-1<\/sup>.<\/p>\n

      Question 32.
      \nWhat is the most important condition for a process to be reversible in . thermodynmics ?
      \nAnswer:
      \nThe process should be carried out infinitesimally slowly of the driving force should be infinitesimally greater than the opposing force.<\/p>\n

      \"KSEEB<\/p>\n

      Question 33.
      \nWhy absolute value of enthalpy cannot be determined ?
      \nAnswer:
      \nAs H = E = +PV
      \nAbsolute value of E – the interned energy cannot be determined as it depends upon various factors whose value cannot be determined.
      \n\u2234 Absolute value of H cannot be determined.<\/p>\n

      Question 34.
      \nWhat are the applications of Hess’s Law of constant heat summation ?
      \nAnswer:<\/p>\n

        \n
      • It helps to calculate the enthalpies of formation of those compounds which cannot be determined experiementally.<\/li>\n
      • It helps in determine the enthalpy of allotropic transformation like C (graphite) \u2192 C (diamond)<\/li>\n
      • It helps to calculate the enthalpy of hydration.<\/li>\n<\/ul>\n

        Question 35.
        \nTwo litres of an ideal gas at a pressure of 10 atm expands isothermally into a vaccum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
        \nAnswer:
        \nWe have q = – w : pext<\/sub> (10 – 2) = 0 \u00d7 8 = 0
        \nNo. work done; No heat, is absorbed.<\/p>\n

        Question 36.
        \nDefine (i) Molar enthalpy of fusion (ii) Molar enthalpy of vaporization ?
        \nAnswer:
        \n(i) The enthalpy change that accompanies melting of one mole of a solid substance at its melting point is called molar enthalpy of fusion.
        \n(ii) Amount of heat required to vaporize one mole of a liquid at constant temperature and understand pressure (1 bar) is called molar enthalpy of vapourisaton.<\/p>\n

        \"KSEEB<\/p>\n

        Question 37.
        \nDefine : (a) Enthalpy of atomization (b) Lattice enthalpy.
        \nAnswer:
        \nEnthalpy of atomization : It is defined as the enthalpy change accompyning the breaking of one mole of a substance completely into its atoms in the gas phase.
        \nH2<\/sub>(g) \u2192 2H(g); \u2206a<\/sub>H\u00b0 = 435.0kJ mol-1<\/sup>
        \nLattice enthalpy ; Lattice of an comp; the enthalpy change which occure when one mole of an ionic compound dissociates into its gaseous ionic state.
        \nNaCl(s) \u2192 Na+<\/sup> (g) + Cl–<\/sup> (g); \u2206H = +788 ks \/ mol<\/p>\n

        Question 38.
        \nExplain Laplace- Lavoisier law.
        \nAnswer:
        \nThe quantity of heat that must be supplied to decompose a compound into its elements is equal and opposite to the heat evolved when the same compound is formed from its elements.
        \nC(s) + O2(g)<\/sub> \u2192 CO2(g)<\/sub> \u2206H = -393.5 kJml-1<\/sup>
        \nCO2(g)<\/sub> \u2192 C(s)<\/sub> + O2(g)<\/sub> \u2206H = +393.5 kJml-1<\/sup><\/p>\n

        Question 39.
        \nIf the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ moh1. What is the enthalpy change for the C (graphite) \u2192 C (diamond) ?
        \nAnswer:
        \nC(diamond) + O2<\/sub>(g) \u2192 CO2<\/sub>(g) \u2206H =-395.4 kJmol-1<\/sup> …(1)
        \nC(graphite) + O2<\/sub>(g) \u2192 CO2<\/sub>(g) \u2206H = -393.6kJmol-1<\/sup> … (2)
        \nC(graphite) \u2192 C(diamond ) substracting (1) from (2), we get
        \nC(graphite) \u2192 C(diamond) \u2206H = -393.6 kJ – (-395.4 kJ), \u2206H = +1.8 kJ mol-1<\/sup><\/p>\n

        Question 40.
        \nH2<\/sub>(g) + \\(\\frac { 1 }{ 2 }\\) O2<\/sub>(g) \u2192 H2<\/sub>O(g) \u0394f<\/sub>H = -242kJ mol-1<\/sup>. Bond energy of H2<\/sub> and O2<\/sub> are 436 mol-1<\/sup> and 500 mol-1<\/sup> respectively. What is bond energy of O – H bond ?
        \nAnswer:
        \nH2<\/sub>(g) + \\(\\frac { 1 }{ 2 }\\) O2<\/sub>(g) \u2192 H2<\/sub>O(g); \u0394f<\/sub>H\u00b0 = -242kJ mol-1<\/sup>
        \nH = Bond energy of reactions – Bond energy of products
        \n\u0394H = BH-H<\/sub> + \\(\\frac { 1 }{ 2 } \\)BO – O<\/sub>\u00a0+ BO – H<\/sub> \u00a0; 242kJ = 436 + \\(\\frac { 1 }{ 2 }\\) (500) – 2BO – H<\/sub> \u21d4 -928kJ = -2BO – H<\/sub>; BO – H<\/sub> = \\(\\frac { 928 }{ 2 }\\) = 464kJ mol-1<\/sup><\/p>\n

        \"KSEEB<\/p>\n

        Question 41.
        \nWhat is spontaneous process ? Give example.
        \nAnswer:
        \nProcess which takes place itself, without any external aid under the given condition is called spontaneous process.
        \nExample : Flow of heat from higher to lower temperature.<\/p>\n

        Question 42.
        \nDefine non-spontaneous process.
        \nAnswer:
        \nProcess which does not takes place itself or on its own but with the help of external aid under the given condition is called non-spontaneous process.
        \nExample : Flow of heat energy from lower to higher temperatures.<\/p>\n

        1st PUC Chemistry Thermodynamics Three Marks Questions and Answers<\/h3>\n

        Question 1.
        \nClassify the following processes as reversible or irreversible :
        \n(i) Dissolution of sodium chloride
        \n(ii) Evaporation of water at 373 K and 1 atm. pressure
        \n(iii) Mixing of two gases by diffusion
        \n(iv) Melting of ice without rise in temperature
        \n(b) When an ideal gas expands is vaccum, there is neither absorption nor revalutions of what ? why ?
        \nAnswer:
        \n(a)(i) Irreversible, (ii) Irreversible, (iii) Irreversible, (iv) Reversible
        \n(b) It is because no work is done,
        \ni.e., w = 0 w = -pext<\/sub> \u00d7 \u2206V = 0 \u00d7 \u2206V = 0
        \nIn \u2206U = q + w
        \nq = 0 because gas chamber is insulated \u2206U = 0 + 0 = 0<\/p>\n

        \"KSEEB<\/p>\n

        Question 2.
        \nJustify the following statements :
        \n(a) Reaction with \u2206G\u00b0 < 0 always have an equilibrium constant greater thanl.
        \n(b) Many thermodynamically feasible reaction do not occur under ordinary conditions.
        \n(c) At low temperatures enthalpy change dominates the AG expression and at high temperature it is the entropy which dominates the value of AG.
        \nAnswer:
        \n(a) \u2206G\u00b0 = -2.303RT logK If K > 1, \u2206G\u00b0 will be less than zero because products formed are more than that of reactants, i.e., process is spontaneous in forward direction.
        \n(b) It is because heat energy is required to overcome activation energy.
        \n(c) \u2206G = \u2206H – T\u2206S, At low temperatures \u2206H > T\u2206S whereas at high temperature T\u2206S >\u2206H
        \n\u2234 \u2206G decreases, i.e., becomes negative.<\/p>\n

        Question 3.
        \nIf for the reaction PbO2<\/sub> \u2192 PbO, \u0394G\u00b0 < 0
        \nAnd for the reaction SnO2<\/sub> \u2192 SnO, \u0394G\u00b0 > 0
        \nWhat are the most probable oxidation states of Pb and Sn ?
        \nAnswer:
        \nOxidation states of the elements in the following compounds are
        \n\"1st
        \nSince PbO2<\/sub> \u2192 PbO change is accompanied by decrease in free energy it is a spontaneous change hence Pb in +2 state is more stable than in +4 state.<\/p>\n

        Question 4.
        \n(i) What is the value \u0394r<\/sub>H\u00b0 of the following reaction :
        \nH+<\/sup>(aq) + OH–<\/sup>(aq) \u2192H2<\/sub>O(l).
        \n(ii) Give an example in which enthalpy change is equal to internal energy ?
        \nAnswer:
        \n(i) \u0394r<\/sub>H\u00b0 = -57.1kJmol-1<\/sup>
        \n(ii) H2<\/sub>(g) + I2<\/sub>(g) \u2192 2HI(g) here \u0394H = \u0394U<\/p>\n

        Question 5.
        \nConsider the reaction. A + B \u2192 C + D
        \n(i) If the reaction is endothermic and spontaneous in the direction indicated, comment on the sign of \u0394G and \u0394S.
        \n(ii) If the reaction is exothermic and spontaneous in the direction indicated, can you comment on the sign of G and S?
        \n(iii) If the reactin is exothermic and spontaneous only in the direction opposite to the indicated, comment on the sign of AG and AS for the direction indicated in the equation.
        \nAnswer:
        \n(i) \u0394G = -ve, \u0394S = +ve
        \n(ii) \u0394G = -ve, \u0394S = -ve (if \u0394H > T\u0394S)
        \n(iii) \u0394G = \u0394S = -ve<\/p>\n

        Question 6.
        \nPredict in which of the following entropy increases \/ decreases.
        \ni) Temperature of a crystalline solid is raised from 0 K to 115 K.
        \nii) 2NaHCO3<\/sub> \u2192 Na2<\/sub>CO3<\/sub>(s) +CO2<\/sub> (g) + H2<\/sub>O(g)
        \niii) H2<\/sub>(g) \u2192 2H(g
        \nAnswer:
        \n(i) When temperature is raised, disorder in molecules increases and therefore entropy increases.
        \n(iii) Reactant is a solid and hence has low entropy. Among the products there are two gases and one solid so products represent a condition of higher entropy.
        \n(iv) Here 2 moles of H atoms have higher entropy than one mole of hydrogen molecule.<\/p>\n

        Question 7.
        \nWhich of the following is \/ are exothermic and which are endothermic ?
        \n(i) Ca(g) \u2192 Ca2+<\/sup>(g) + 2e–<\/sup>
        \n(ii) O(g) + e–<\/sup> \u2192 O–<\/sup> (g)
        \n(iii) N2-<\/sup>(g) + e–<\/sup> \u2192 N3-<\/sup>(g)
        \nAnswer:
        \n(i) Endothermic (ionisation energy is required)
        \n(ii) Exothermic (first electron affinity – energy is released)
        \n(iii) Endothermic (higher electron affinities energies are required)<\/p>\n

        \"KSEEB<\/p>\n

        1st PUC Chemistry Thermodynamics Numerical Problems and Answers<\/h3>\n

        Question 1.
        \nCalculate the heat of reaction of the following reaction :<\/p>\n

        C6<\/sub>H12<\/sub>O6<\/sub>(s) + 6O2(g)<\/sub> \u2192 6CO2(g)<\/sub>+ 6H2<\/sub>O(g); \u0394H = ?
        \nC(graphite) + O2<\/sub>(g) \u2192 CO2<\/sub>(g); DH = -395.0 kJ … (1)
        \nH2<\/sub> (g) + \\(\\frac { 1 }{ 2 }\\)O2<\/sub>(g) \u2192 H2<\/sub>O(l); DH = -269.5 kJ … (2)
        \n6C(graphit.e) + 6H2<\/sub>(g) + O2<\/sub>(g) \u2192 C6<\/sub>H12<\/sub>O6<\/sub>(s); DH = -1169.8 kJ … (3)
        \nAnswer:
        \nMultiplying equation (1) and (2) each by 6 reversing (3), we get,
        \n6C(graphite)<\/sub> + 6O2(g)<\/sub> \u2192 6CO2(g)<\/sub>; \u0394H = -2370 kJ … (4)
        \n6H2(g) <\/sub>+ 3O2(g)<\/sub>\u00a0\u2192 6H2<\/sub>O(l)<\/sub>; \u0394H = -1616.4 kJ … (5)
        \nC6<\/sub>H12<\/sub>O6(s)<\/sub> \u2192 6C(graphite)<\/sub> + 6H2(g)<\/sub> + 3O2(g)<\/sub> ; \u0394H = +1169-8 kJ …. (6)
        \nAdding (4), (5) and (6), C6<\/sub>H12<\/sub>O6(s)<\/sub> + 6O2(g)<\/sub> \u2192 6CO2(g)<\/sub> + 6H2<\/sub>O(g)<\/sub> ;
        \nAH(C6<\/sub>H12<\/sub>O6<\/sub>) = -2370.0 – 1616.4 + 1169,8 = -2816.6 kJ<\/p>\n

        Question 2.
        \nCalculate the heat of reaction of the following reaction :
        \nCO2<\/sub>(g) + H2<\/sub>(g) \u2192 CO(g) + H2<\/sub>O(g)
        \nGiven that the \u0394f<\/sub>H\u00b0CO(g) =-110.5kJ, \u0394f<\/sub>H\u00b0CO2<\/sub>(g) =-393.8kJ,
        \n\u0394f<\/sub>H\u00b0H2<\/sub>O(g) = -241.8kJ respectively.
        \nAnswer:
        \nThe required equation is CO2<\/sub>(g) + H2<\/sub>(g) \u2192 CO(g) + H2<\/sub>O(g)
        \n\u0394H – \u03a3 \u0394Hf(products)<\/sub> \u2014 \u03a3 \u0394Hf(reactants)<\/sub>
        \n= \u0394Hf2<\/sup><\/sub>CO(g) + \u0394Hf2<\/sup><\/sub>H2<\/span><\/sub>O(g) – \u0394Hf2<\/sup><\/sub>CO2(g)<\/span><\/sub> – \u0394Hf2<\/sup><\/sub>H2(g)<\/span><\/sub>
        \n= -110.5 kJ – 241.8 kJ – (-398.3 kJ) – 0 = -352.3 kJ + 393.8 kJ = 41.5 kJ.<\/p>\n

        Question 3.
        \n1 m3<\/sup> of C2<\/sub>H4<\/sub> at STP is burnt in oxygen, according to the thermochemical
        \nreaction: C2<\/sub>H4<\/sub>(g) + 3O2<\/sub>(g) \u2192 2CO2<\/sub>(g) + 2H2<\/sub>O(l); \u0394H = -1410 kJ mol-1<\/sup>
        \nAssuming 70% efficiency, determine how much of useful heat is evolved in the reaction.
        \nAnswer:
        \n22.4 L of C2<\/sub>H4<\/sub>\u00a0at STP produces 1410 kJ of energy.
        \n\"1st<\/p>\n

        \"KSEEB<\/p>\n

        Question 4.
        \nWith the help of thermochemical equation, calculate \u0394f<\/sub>H\u00b0 at 298 K for the following reactions:
        \nC(graphite) + O2<\/sub>(g) \u2192 CO2<\/sub>(g) ; \u0394f<\/sub>H\u00b0 =-393.5kJ\/mol
        \nH2<\/sub> (g) + \\(\\frac { 1 }{ 2 }\\)O2<\/sub>(g) \u2192 H2<\/sub>O(l) ; \u0394f<\/sub>H\u00b0 = -285.8kJ\/mol
        \nCO2<\/sub> (g) + 2H2<\/sub>O(l) \u2192 CH4<\/sub> (g) + 2O2<\/sub>(g) ; \u0394f<\/sub>H\u00b0 = +890.3kJ\/mol
        \nAnswer:
        \nC(graphite) + O2<\/sub>(g) \u2192 CO2<\/sub>(g) ; \u0394f<\/sub>H\u00b0 = -393.5kJ\/mol …(1)
        \nH2<\/sub>(g) + \\(\\frac { 1 }{ 2 }\\)O2<\/sub>(g) \u2192 H2<\/sub>O(l) ; \u0394f<\/sub>H\u00b0 = -285.8kJ\/mol …(2)
        \nCO2<\/sub>(g) + 2H2<\/sub>O(l) \u2192 CH4<\/sub>(g) + 2O2<\/sub>(g) \u0394f<\/sub>H\u00b0 = +890.3kJ\/mol …(3)
        \nHere we want one mole of C(graphite) as reactant, so we write down equation (1) as such, we want two moles of H2<\/sub>(g) as reactant, so we multiply equation (2) by 2, we want<\/p>\n

        C(graphite) + O2<\/sub> (g) \u2192 CO2<\/sub>(g) ; \u0394f<\/sub>H\u00b0 = -393.5kJ \/ mol …(1)
        \n2H2<\/sub>(g) + O2<\/sub>(g) \u2192 2H2<\/sub>O(1) ; \u0394f<\/sub>H\u00b0 = 2(-285.8kJ\/mol) …(2)
        \nCO2<\/sub>(g) + 2H2<\/sub>O(l) \u2192 CH4<\/sub>(g) + 2O2<\/sub>(g) ; \u0394f<\/sub>H\u00b0 =+890.3kJ\/mol . … (3)
        \nAdding we obtain: .
        \nC(graphite) + 2H2<\/sub>(g) \u2192 CH4<\/sub>(g); \u0394f<\/sub>H\u00b0 =-74.8kJ\/mol<\/p>\n

        Question 5.
        \nThe heat of combustion of C2<\/sub>H6<\/sub> is -368.4 kcal. Calculate heat of combustion of C2<\/sub>H4<\/sub>, heat of combustion of H2<\/sub> is 68.32 kcal mol-1<\/sup>.
        \nC2<\/sub>H4<\/sub>(g) + H2<\/sub>(g) \u2192 C2<\/sub>H6<\/sub>(g) ; \u0394H = -37.1 kcl
        \nAnswer:
        \nC2<\/sub>H4<\/sub>(g) + H2<\/sub>(g) \u2192 C2<\/sub>H6<\/sub>(g); \u0394H = -37.1 k cal.
        \n\u0394Hc<\/sub>C2<\/sub>H6<\/sub> = -368.4 k cal, \u0394Hc<\/sub>C2<\/sub>H2<\/sub> = ?; \u0394Hc<\/sub>H2(g) <\/sub>= -68.32 k cal.<\/p>\n

        \u0394H\u00a0 = \u03a3\u0394Hc(reactanta)<\/sub> – \u03a3\u0394Hc(products)<\/sub>
        \n\u0394H = \u0394Hc<\/sub>C2<\/sub>H4<\/sub>+\u0394Hc<\/sub>H2(g) <\/sub>– \u0394Hc<\/sub>C2<\/sub>H6(c)<\/sub>
        \n-37.1k cal = \u0394Hc<\/sub>C2<\/sub>H4<\/sub>-68.32-(-368.4)
        \n\u0394Hc<\/sub>C2<\/sub>H6<\/sub> =-337.18 kcal.<\/p>\n

        \"KSEEB<\/p>\n

        Question 6.
        \nThe following thermochemical equations represent combustion of ammonia and hydrogen:<\/p>\n

        4NH3<\/sub>(g) + 3O2<\/sub>(g) \u2192 6H2<\/sub>O(l) + 2N2<\/sub>(g); \u0394H = -1516 kJ
        \n2H2<\/sub>(g) + O2<\/sub>(g) \u2192 2H2<\/sub>O(l); \u0394H = -572kJ
        \nCalculate enthalpy of formation of ammonia.
        \nAnswer:
        \nRequired equation is\u00a0 \\(\\frac { 1 }{ 2 }\\)N2(g)<\/sub> + \\(\\frac { 3 }{ 2 }\\)H2(g)<\/sub> \u2192 NH3(g) <\/sub>\u0394H = ?
        \n4NH3(g)<\/sub> + 3O2(g)<\/sub> \u2192 2N2(g)<\/sub> + 6H2<\/sub>O2(l) <\/sub>\u0394H = -1516 kJ …(l)
        \n2H2(g) <\/sub>+ O2(g)<\/sub> \u2192 2H2<\/sub>O(l)<\/sub> \u0394H = -572 kJ …(2)
        \nReserving equation (1) and dividing by 4 we get:
        \n\\(\\frac { 1 }{ 2 }\\)N2(g)<\/sub> + \\(\\frac { 3 }{ 2 }\\)2H2<\/sub>O2(l)<\/sub> \u2192 NH3(g)<\/sub>+ \\(\\frac { 3 }{ 4 }\\)O2(g)<\/sub> \u0394H = +379 kJ …(3)
        \nMultiplying equation (2), by 3\/4 we get,
        \n\\(\\frac { 3 }{ 2 }\\)H2(g) <\/sub>+ \\(\\frac { 3 }{ 4 }\\)O2<\/sub> \u2192 \\(\\frac { 3 }{ 2 }\\)H2<\/sub>O(l)<\/sub> \u0394H = -572 \u00d7 \\(\\frac { 3 }{ 4 }\\) = -429kJ
        \nAdding (3) and (4) we get ,
        \n\\(\\frac { 1 }{ 2 }\\)N2(g) <\/sub>+ \\(\\frac { 3 }{ 2 }\\)H2(g) <\/sub>\u2192 NH3(g) <\/sub>;\u0394H = +379 – 429 = -50 kJ\/mol<\/p>\n

        Question 7.
        \nThe equilibrium constant at 25\u00b0 C for the process.
        \nCO3+<\/sup>(aq) + 6NH3<\/sub>(aq) \u21cc [CO(NH3<\/sub>)6<\/sub>]3<\/sup>(aq) is. 2.5 \u00d7 106<\/sup>. Calculate the value of \u0394G\u00b0 at 25\u00b0 C. (R = 8.314 JK1 mol-1<\/sup>). In which direction is the reaction spontaneous under standard conditions ?
        \nAnswer:
        \n\u0394G\u00b0 =-2.303 RT log K .
        \n= -2.303 \u00d7 3.134 \u00d7 298 log (2.5 x 106<\/sup>)
        \n= -5705.8 [0.3980 + 6.0000] = -5705.8 \u00d7 6.3980 = -36.505 k\/mol.
        \nThe reaction is spontaneous in forward direction under standard conditions.<\/p>\n

        Question 8.
        \nWhat is the value of equilibrium constant for the following reaction at 400 E?
        \n2NOCl(g) \u21cc 2NO(g) + Cl2<\/sub>(g)
        \n\u0394H0<\/sup> = 77.5kJmol-1<\/sup>, R = 8.3124 J Mol-1<\/sup>K-1<\/sup>, AS = 135 J K-1<\/sup>mol-1
        \n<\/sup>Answer:
        \n\u0394G = \u0394H – T\u0394S
        \n\u0394G = 77.5 \u00d7 1000 J – 400 K \u00d7 135 J K-1<\/sup> = 77500 J – 54000 J = 23500 J
        \n\u0394G = -2.303 RT log K
        \n23500 J = -2.303 \u00d7 8.314 \u00d7 400 K log K
        \n\"1st
        \nLog k =- 3.068 k = antilog(-3.008)
        \n= Antilog (0.932-4) = Antilog (0.93) \u00d7 10-4 <\/sup>= 8055 \u00d7 10-4<\/sup><\/p>\n

        Question 9.
        \nThe standard free energy change for a reaction is -212.3 kJ mol-1<\/sup>. If the enthalpy of the reaction is -216.7 kJ mol-1<\/sup>, Calculate the entropy change for the reaction.
        \nAnswer:
        \n\u0394G\u00b0 = -212.3 kJ mol-1<\/sup>, \u0394H0<\/sup> = -216.7 kJ mol-1<\/sup>, \u0394S\u00b0 = ?
        \nT = 298 K (Because standard free energy is measured at 298 K)
        \n\u0394G\u00b0 = -212.3 kJ mol-1<\/sup>, \u0394H\u00b0 = 216.7 kJ mol-1<\/sup>, \u0394S\u00b0 = ?
        \n-212.3 kJ = 216.7 kJ – 298 K \u00d7 \u0394S\u00b0
        \n\"1st<\/p>\n

        Question 10.
        \nCalculate the standard free energy change AG\u00b0 for the reaction.
        \n2HgO(s) \u2192 2Hg(l) + O2<\/sub>(g)
        \n\u0394H\u00b0 = OlkJmol-1<\/sup> at 298 K, S\u00b0(Hg0)<\/sub> = 72.0JK-1<\/sup> mol-1<\/sup>
        \nAnswer:
        \n\u0394S = 2S\u00b0Hg<\/sub> + S\u00b0(O2)<\/sub> – 25\u00b0Hgo<\/sub>
        \n= (2 \u00d7 77.4 + 205 – 2 \u00d7 72.0) JK -1<\/sup>mol-1<\/sup> = 215.8 jK-1<\/sup> mol-1<\/sup>
        \n\u0394G\u00b0 = \u0394r<\/sub>H\u00b0 – T\u0394r<\/sub>S\u00b0
        \n\"1st<\/p>\n

        Question 11.
        \nFrom the data given below at 298 K for the reaction :
        \nCH4<\/sub>(g) + 2O2<\/sub>(g) \u2192 CO2<\/sub>(g) + 2H2<\/sub>O(l)
        \nCalculate the enthalpy of formation of CH4<\/sub>(g) at 298 K.
        \nEnthalpy of reaction is = -89.5 kJ
        \nEnthalpy of formation of CO2<\/sub>(g) – -393kJ mol-1<\/sup>
        \nEnthalpy of formation of H2<\/sub>O(l ) = -286.0 kJ mol-1<\/sup>
        \nAnswer:
        \n\u0394H = AHf<\/sub>CO2<\/sub>(g) + 2\u0394Hf<\/sub>H2<\/sub>O(l) – \u0394Hf<\/sub>CH4<\/sub>(g) – \u0394Hf<\/sub>;O2<\/sub>(g)
        \n-890.5 kJ = -393.5 kJ +2 \u00d7 -286 kJ – \u0394 Hf<\/sub>CH4<\/sub>(g) – 0
        \n\u0394Hf<\/sub>CH4 <\/sub>– 75.0kJ<\/p>\n

        Question 12.
        \nCalculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given : Lattice energy of NaCl = – 777.8 kJ mol-1<\/sup>), Hydration energy
        \n= -774.1 kJ mol-1<\/sup> and \u0394S = 0.043 kJK-1<\/sup> mol-1<\/sup> at 298 K.)
        \nAnswer:
        \n\u0394H = Hydration energy – Lattice energy
        \n\u0394H = -774.1 kJ mol-1<\/sup> (-777.8 kJ mol-1<\/sup>) = 3.7 kJ mol-1<\/sup>
        \n\u0394G = \u0394H – T\u0394S = +3.7 kJ- 298 \u00d7 0.043 kJ = +3.7 kJ- 12.81 kJ
        \n\u0394G = -9.11 kJ moH<\/p>\n

        \"KSEEB<\/p>\n

        Question 13.
        \nFor the equilibrium, PCl5<\/sub> (g) \u21cc PCl3<\/sub> (g) + Cl2<\/sub> (g) at 298 K, K = 1.8 \u00d7 10 -7<\/sup>.
        \nCalculate \u0394G\u00b0 for the reaction (R = 8.314JK 1mol 1)
        \nAnswer:
        \nPCl5<\/sub>(g) \u21cc PCl2<\/sub>(g) + Cl2<\/sub>(g), K = 1.8 \u00d7 10-7<\/sup>
        \n\u0394G\u00b0 = -2.303RTlogK = -2.303 \u00d7 8.314JK-1<\/sup>mol-1<\/sup> \u00d7 298K \u00d7 log(1.8 \u00d7 10-7<\/sup>)
        \n= -2.303 \u00d7 8.314 \u00d7 298[log 1.8 + logl0-7<\/sup> ] = -19.147 \u00d7 298 [0.2553 – 7.000]
        \n\"1st<\/p>\n

        Question 14.
        \nCalculate the entropy change involved in conversation of 1 mole of water at 373 K to vapours at the same temperature. Latent heat of vaporation of water = 2.257 kJ g1.
        \nAnswer:
        \nk = 1.8 \u00d7 10-7<\/sup>
        \n\u0394H = 2.257 \u00d7 18kJ= 40.626kJ mol-1<\/sup>
        \n\"1st<\/p>\n

        Question 15.
        \nCalculate \u0394Hf<\/sub> of HCl if bond energy of H – H bond is 437 kJ Cl – Cl bond is 244, and H – C is 433 kJ mol1.
        \nAnswer:
        \n\\(\\frac { 1 }{ 2 }\\)H2(g) + \\(\\frac { 1 }{ 2 }\\) cl2<\/sub>(g) > HCl(g)
        \n\u0394H = \\(\\frac { 1 }{ 2 }\\)BH – H<\/sub> + \\(\\frac { 1 }{ 2 }\\)BO – O<\/sub> = \\(\\frac { 1 }{ 2 }\\) \u00d7 437 + \\(\\frac { 1 }{ 2 }\\) \u00d7 244 -433
        \n= 218.5 kJ + 122 kJ – 433 kJ = -92.5 kJ mol-1<\/sup><\/p>\n

        Question 16.
        \nCalculate bond energy of C – H bond if \u0394Hc of CH4<\/sub> is -891 kJ, \u0394Hc of C (s) is – 394 kJ, \u0394Hc of H2<\/sub>(g) is -286 kJ, heat of sublimation of C(s) is 717 kJ, heat of dissociation of H2<\/sub> is 436 kJ.
        \nAnswer:
        \nCH4<\/sub>(g) + 2O2<\/sub>(g) \u2192 C02<\/sub>(g) + 2H2<\/sub>O(l) ; \u0394H = -891 kJ
        \nC(s) + O2<\/sub>(g) \u2192 CO2<\/sub>(g) ;\u0394H = -394 kJ
        \nH2<\/sub>(g) + \\(\\frac { 1 }{ 2 }\\) O2<\/sub>(g) \u2192 H2<\/sub>O(l) ; \u0394H = -286 kJ
        \nC(s) \u2192 C(g) ;\u0394H = +717 kJ
        \nH2<\/sub>(g) \u2192 2H(g);\u0394 H = +436 kJ
        \nTarget equation is CH4<\/sub> +4H(g)
        \nReversing (ii),Multiply (iii) by 2 and reverse, Multiply eqn. (v) 2 and then adding all together we get:
        \nCH4<\/sub> \u2192 C(g) + 4H(g) \u0394H = -891kJ + 394kJ + 575kJ + 717kJ + 872kJ
        \n\u0394H = 1664 kJ
        \nEnergy required to break 4(C – H) bond = 1664 kJ
        \nEnergy required to break one (C – H) bond =\\frac{1664}{4} = 416 kJ mol-1<\/sup><\/p>\n

        \"KSEEB<\/p>\n

        Question 17.
        \nWhat would be heat released when :
        \n(i) 0.25 mole of HCl in solution is neutralized 0.25 mole of NaOH solution ?
        \n(ii) 0.5 mole of HNO3<\/sub> in solution is mixed with 0.2 mole of KOH solution ?
        \n(iii) 200 cm3<\/sup> of 0.2 M HCl is mixed with 300 cm3 of 0.1 M NaOH solution ?
        \n(iv) 400 cm3<\/sup> of 0.2 M H2<\/sub>SO4<\/sub> is mixed with 300 cm3 of 0.1 M NaOH solution ?
        \nAnswer:
        \n(i) H+<\/sup>(0.25mole) + OH–<\/sup>(0.25mole) \u2192 H2<\/sub>O(0.25mole)
        \nHeat released = 0.25 \u00d7 57.1 = 14.3 kJ
        \n(ii) 0.5 mole HNO3<\/sub> (aq) + 0.2 mole KOH(aq)
        \nThe net reaction is (0.2 mole HNO3<\/sub> reacts with only 0.2 mole KOH)
        \n0.2 mole is limiting reagent.
        \n0.2 mole of H+<\/sup> + 0.2 mole of OH–<\/sup> \u2192 0.2 mole of H2<\/sub>O
        \n0.3 mole of H+<\/sup> will remain unreacted.
        \nHeat evolved = 0.2 \u00d7 57.1 = 11.4 kJ<\/p>\n

        (iii)
        \n\"1st
        \n0.03 mole of OH–<\/sup> is limiting reactant
        \nHeat evolved = 0.03 \u00d7 57.1 = 1.71 kJ<\/p>\n

        (iv)
        \n\"1st
        \n0.03 is limiting reagent Heat evolved = 0.03 \u00d7 57.1 = 1.7 kJ<\/p>\n

        (v) Mass of solution = 200 + 300 = 500 g in (iii) part (Assuming d = 1 g cm 3)
        \nq = m \u00d7 c \u00d7 T 1.71 \u00d7 1000 J = 500 g \u00d7 4.18 \u00d7 T
        \nT = 0.82 K For (iv), T = \\(\\frac{1.7 \\times 1000}{700 \\times 4.18}\\) = 0.58K<\/p>\n

        \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

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