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Chemical Themodynamics, EXERCISES [PAGES 86 - 89], Exercises | Q 1.01 | Page 86, Select the most appropriate option., The correct thermodynamic conditions for the spontaneous reaction at all temperatures, are _______., 1. ΔH < 0 and ΔS > 0, 2. ΔH > 0 and ΔS < 0, 3. ΔH < 0 and ΔS < 0, 4. ΔH < 0 and ΔS = 0, Solution: The correct thermodynamic conditions for the spontaneous reaction at all, temperatures are ΔH < 0 and ΔS > 0., Exercises | Q 1.02 | Page 86, Select the most appropriate option., A gas is allowed to expand in a well-insulated container against a constant external, pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change, in internal energy, ΔU of the gas will be _______., 1. –500 J, 2. + 500 J, 3. –1013 J, 4. + 1013 J, Solution:, - 500 J, Explanation:, Since the container is insulated, this is an adiabatic process., For adiabatic process,, ΔU = +W = - Pext Δ V = - Pext (V2 - V1), Initial volume (V1) = 2.5 L = 2.5 dm3, Final volume (V2) = 4.5 L = 4.5 dm3, External pressure (Pext) = 2.5 bar

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ΔU = W = - 2.5 bar × (4.5 dm3 - 2.5 dm3), , Exercises | Q 1.03 | Page 86, Select the most appropriate option., In which of the following, entropy of the system decreases?, 1. Crystallization of liquid into solid, 2. Temperature of crystalline solid is increased from 0 K to 115 K, 3. H2(g) → 2H(g), 4. 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g), Solution: Crystallization of liquid into solid, Exercises | Q 1.04 | Page 87, Select the most appropriate option., The enthalpy of formation for all elements in their standard states is _______., 1. unity, 2. zero, 3. less than zero, 4. different elements, Solution: The enthalpy of formation for all elements in their standard states is zero., Exercises | Q 1.05 | Page 87, Select the most appropriate option., Which of the following reactions is exothermic?, 1. H2(g) → 2H(g), 2. C(s) → C(g), 3. 2Cl(g) → Cl2(g), 4. H2O(s) → H2O(l), Solution: 2Cl(g) → Cl2(g), Hint:, Bond is formed between two Cl atoms and hence, energy is released., Exercises | Q 1.06 | Page 87

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Select the most appropriate option., 6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of vaporization of, ethanol will be _______., 1. 43.4 kJ mol–1, 2. 60.2 kJ mol–1, 3. 38.9 kJ mol–1, 4. 20.4 kJ mol–1, Solution: 6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of, vaporization of ethanol will be 43.4 kJ mol–1., Explanation:, Molar mass of ethanol (C2H6O) = 46 g mol–1, , Exercises | Q 1.07 | Page 87, Select the most appropriate option., If the standard enthalpy of formation of methanol is –238.9 kJ mol–1 then entropy, change of the surroundings will be _______., 1. –801.7 J K–1, 2. 801.7 J K–1, 3. 0.8017 J K–1, 4. –0.8017 J K–1, Solution:, If the standard enthalpy of formation of methanol is –238.9 kJ mol–1 then entropy, change of the surroundings will be 801.7 J K–1., Explanation:, For standard state, temperature = 298 K

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Exercises | Q 1.08 | Page 87, Select the most appropriate option., Which of the following are not state functions?, Q+W, Q, W, H - TS, 1. 1, 2 and 3, 2. 2 and 3, 3. 1 and 4, 4. 2, 3 and 4, Solution: 2 and 3 i.e., Q and W, Exercises | Q 1.09 | Page 87, Select the most appropriate option., For vaporization of water at 1 bar, ΔH = 40.63 kJ mol–1 and ΔS = 108.8 J K–1 mol–1 . At, what temperature, ΔG = 0?, 1. 273.4 K, 2. 393.4 K, 3. 373.4 K, 4. 293.4 K, Solution: 373.4 K, Explanation:, Temperature at which reaction is at equilibrium (ΔG = 0) is,

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1 J = 1 kg m2 s-2 = 1 Pa m3, 1 Pa = 1 kg m-1 s-2, From the equation, W = – Pext∆V, If the pressure is expressed in bar and ΔV in dm3, then the work has the units of bar, dm3., 1 bar = 105 Pa = 105 kg m–1 s–2, 1 dm3 bar = dm3 × 105 kg m-1 s-2, = m3 × 10-3 × 105 kg m-1 s-2, = 100 kg m2 s-2 = 100 J, Exercises | Q 2.2 | Page 87, Answer the following in one or two sentences., State the first law of thermodynamics., Solution:, According to the first law of thermodynamics, “the total energy of a system and, surroundings remains constant when the system changes from an initial state to final, state.”, Exercises | Q 2.3 | Page 87, Answer the following in one or two sentences., What is enthalpy of fusion?, Solution:, Enthalpy change that occurs when one mole of a solid is converted into liquid without a, change in temperature at constant pressure is the enthalpy of fusion., Exercises | Q 2.4 | Page 87, Answer the following in one or two sentences., What is standard state of a substance?, Solution:, 1. The standard state of a substance is the form in which the substance is most stable at a, pressure of 1 bar and at temperature 298 K., 2. If the reaction involves species in solution its standard state refers to 1 M concentration., e.g. Standard states of certain elements and compounds are (at 1 bar and 25 °C); H 2(g),, Hg(l), Na(s), C(graphite), C2H5OH(l), CaCO3(s), CO2(g), C2H5OH(l), H2O(l), CaCO3(s), CO2(g)., Exercises | Q 2.5 | Page 87, Answer the following in one or two sentences.

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State whether ΔS is positive, negative or zero for the reaction 2H(g) → H2(g). Explain., Solution:, ΔS is negative. Two moles of gaseous H atoms are converted into 1 mole H2 gas. Thus,, a disorder of the system decreases, and hence, entropy increases or ∆S is positive., Exercises | Q 2.6 | Page 87, Answer the following in one or two sentences., State second law of thermodynamics in terms of entropy., Solution:, Statement: “The second law of thermodynamics states that total entropy of a system, and its surroundings increases in a spontaneous process.”, For the process to be spontaneous,, , Exercises | Q 2.7 | Page 87, Answer the following in one or two sentences., If the enthalpy change of a reaction is ∆H how will you calculate the entropy of, surroundings?, Solution:, If ∆H is the enthalpy change accompanying a reaction (system) the enthalpy change of, the surroundings is then –∆H. The entropy change of surroundings can be calculated, using the following expression:\, , Exercises | Q 2.8 | Page 87, Answer the following in one or two sentences., Comment on the spontaneity of reactions for which ∆H is positive and ∆S is negative., Solution:, When ∆H positive and ∆S is negative, then ∆G is positive regardless of temperature., Such reactions are nonspontaneous at all temperatures., Exercises | Q 3.1 | Page 87, Answer in brief., Obtain the relationship between ∆G° of a reaction and the equilibrium constant., Solution:

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1. Gibbs energy change for a chemical reaction is given by, ΔG = ΔG° + RT ln Q, , …(1), , where, ΔG° is standard Gibbs energy change that is, the Gibbs energy change when, the reactants and products in a reaction are in their standard states. Q is called reaction, quotient. Q is analogous to that of the equilibrium constant and involves nonequilibrium, concentrations or partial pressures in case of a gaseous reaction., 2. Consider the reaction, aA + bB → cC + dD, From equation (1),, ΔG = ΔG° + RT lnQC, , ΔG = ΔG° + RT lnQP, , or, , 3. When the reaction reaches equilibrium, ΔG° = 0 and QC and QP become KC and KP,, respectively., Thus,, ∴ 0 = ΔG° + RT lnKC, , or, , 0 = ΔG° + RT lnKP, , ∴ ΔG° = –RT ln KC, , or, , ΔG° = –RT ln KP, , ∴ ΔG° = –2.303 RT log10KC or ΔG° = –2.303 RT log10KP, Exercises | Q 3.2 | Page 87, Answer in brief., What is entropy? Give its units., Solution:, 1. Entropy is a measure of molecular disorder or randomness., 2. An entropy change of a system is equal to the amount of heat transferred (Qrev) to it in a, reversible manner divided by the temperature (T) in Kelvin at which the transfer takes, place. Thus,, , 3. Units of entropy: J K–1, Note: Entropy or its change ∆S is a state function and depends on the initial and final, states of the system and not on the path connecting two states, Exercises | Q 3.3 | Page 87, Answer in brief., How will you calculate reaction enthalpy from data on bond enthalpies?

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Solution:, Reaction and bond enthalpies:, 1. In a chemical reaction, bonds are broken and formed., 2. The enthalpies of reactions involving substances having covalent bonds are calculated, by knowing the bond enthalpies of reactants and those in products., 3. The calculations assume all the bonds of a given type are identical., 4. Enthalpy change of a reaction can be calculated using the following expression:, ΔrH° = ∑ ΔH° (reactant bonds) - ∑ ΔH° (product bonds), e.g. Consider the reaction, H2(g) + I2(g) → 2HI(g), The enthalpy is given by, ΔrH° = [ΔH° (H - H) + ΔH° (I - I)] - [2ΔH° (H - I)], Note:, i. If reactants and products are diatomic molecules, the above equation gives accurate, results. The bond enthalpies are known accurately., ii. For reactions involving polyatomic molecules, the reaction enthalpies calculated using, above equation would be approximate and refer to average bond enthalpies., Exercises | Q 3.4 | Page 87, Answer in brief., What is the standard enthalpy of combustion? Give an example., Solution:, 1. The standard enthalpy of combustion of a substance is the standard enthalpy change, accompanying a reaction in which one mole of the substance in its standard state is, completely oxidised., 2. Consider the reaction,, C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l), Δr H° = - 1300 kJ, In the above reaction, the standard enthalpy change of the oxidation reaction, –1300 kJ, is the standard enthalpy of combustion of C2H2(g)., Exercises | Q 3.5 | Page 87, Answer in brief., What is the enthalpy of atomization? Give an example., Solution:, 1. The enthalpy change accompanying the dissociation of one mole of gaseous substance, into atoms is called enthalpy of atomization., 2. For example, Cl2(g) → Cl(g) + Cl(g); ΔatomH = 242 kJ mol–1, Exercises | Q 3.6 | Page 87, Answer in brief., Obtain the expression for work done in chemical reaction.

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1., 2., , 3., , 4., , Solution:, The work done by a system at constant temperature and pressure is given by, W = Pext ΔV, ....(1), Assuming Pext = P,, W = - PΔV, = - P (V2 - V1), W = - PV2 + PV1, .....(2), If the gases were ideal, at constant temperature and pressure.,, PV1 = n1RT and PV2 = n2RT, ....(3), Substitution of equation (3) into equation (2) yields, W = - n2RT + n1RT, = - (n2 - n1) RT, = - Δng RT, .....(4), The equation (4) gives the work done by the system in chemical reactions, Exercises | Q 3.7 | Page 88, Answer in brief., Derive the expression for PV work., Solution:, Pressure-volume work, , 1. Consider a certain amount of gas at constant pressure P is enclosed in a cylinder fitted, with a frictionless, rigid movable piston of area A. Let the volume of the gas be V 1 at, temperature T. This is shown in the adjacent diagram., 2. On expansion, the force exerted by a gas is equal to area of the piston multiplied by, pressure with which the gas pushes against piston. This pressure is equal in magnitude, and opposite in sign to the external atmospheric pressure that opposes the movement, and has its value - Pext., Thus,, f = - Pext × A, .....(1), where, Pext is the external atmospheric pressure., 3. If the piston moves out a distance d, then the amount of work done is equal to the force, multiplied by distance., W=f×d, .....(2)

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Substituting equation (1) in (2) gives, W = - Pext × A × d, .....(3), 4. The product of area of the piston and distance it moves is the volume change (ΔV) in, the system., ΔV = A × d, .....(4), Combining equation (3) and (4), we get, W = - Pext ΔV, W = - Pext (V2 - V1), Where V2 is the final volume of the gas., Exercises | Q 3.8 | Page 88, Answer in brief., What are intensive properties? Explain why density is an intensive property., Solution:, 1. A property which is independent of the amount of matter in a system is called intensive, property., 2. Density is a ratio of mass to volume. Mass and volume are extensive properties. Since, density is a ratio of two extensive properties, it is an intensive property. Thus, density is, independent of the amount of matter present., Exercises | Q 3.9 | Page 88, Answer in brief., How much heat is evolved when 12 g of CO reacts with NO2? The reaction is:, 4CO(g) 2NO2(g) → 4CO2(g) + N2(g), ΔrH° = - 1200 kJ, Solution:, Given: ΔrH° = - 1200 kJ, Mass of CO = 12 g, To find: Heat evolved when 12g of CO reacts with NO2, Calculation:, According to the given reaction, 1200 kJ of heat is evolved when 4 moles of CO react, with NO2. So heat evolved per mole is 1200kJ/4 mol = 300 kJ mol-1, Molar mass of CO = 12 + 16 = 28 g mol–1, , So, heat evolved when 0.4286 moles of CO reacts, = 0.4286 mol × 300 kJ mol-1 = 128.58 kJ, The heat evolved when 12 g of CO reacts with NO2 is 128.58 kJ., Exercises | Q 4.01 | Page 88

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Answer the following question., Derive the expression for the maximum work., Solution:, , 1. Consider n moles of an ideal gas enclosed in a cylinder fitted with a frictionless, movable rigid piston. It expands isothermally and reversibly from the initial volume V1 to, final volume V2 at temperature T. The expansion takes place in a number of steps as, shown in the figure., 2. When the volume of a gas increases by an infinitesimal amount dV in a single step,, the small quantity of work done, dW = -Pext dV, , ....(1), , 3. As the expansion is reversible, P is greater by a very small quantity dp than P ext., Thus P - Pext = dP or Pext = P - dP, , ....(2), , Combining equations (1) and (2),, dW = - (P - dP)dV = - PdV + dP.dV, Neglecting the product dP.dV which is very small, we get, dW = - PdV, , .....(3), , 4. The total amount of work done during the entire expansion from volume V 1 to, V2 would be the sum of the infinitesimal contributions of all the steps. The total work is, obtained by integration of Equation (3) between the limits of initial and final states. This, is the maximum work, the expansion being reversible., Thus,

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Equations (5) and (6) are expressions for work done in reversible isothermal process., Exercises | Q 4.02 | Page 88, Obtain the relationship between ΔH and ΔU for gas phase reactions.

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Solution:, 1. At constant pressure, ΔH and ΔU are related as, ΔH = ΔU + PΔV, , …(1), , 2. For reactions involving gases, ΔV cannot be neglected. Therefore,, ΔH = ΔU + PΔV, = ΔH + P(V2 – V1), ΔH = ΔU + PV2 – PV1, , …(2), , where, V1 is the volume of gas-phase reactants and V2 that of the gaseous products., 3. We assume reactant and product behave ideally. Applying an ideal gas equation, PV, = nRT. Suppose that n1 moles of gaseous reactants produce n2 moles of gaseous, products. Then,, PV1 = n1RT and PV2 = n2RT, , …(3), , 4. Substitution of equation (3) into equation (2) yields, ΔH = ΔU + n2RT – n1RT, = ΔU + (n2 – n1) RT, = ΔU + Δng RT, , ....(4), , where, Δng is the difference between the number of moles of products and those of, reactants., Exercises | Q 4.03 | Page 88, Answer the following question., State Hess’s law of constant heat summation. Illustrate with an example. State its, applications., Solution:, 1. Hess’s law of constant heat summation:, Hess’s law of constant heat summation states that, “Overall the enthalpy change for a, reaction is equal to sum of enthalpy changes of individual steps in the reaction”., 2. Illustration:, •, , •, , The enthalpy change for a chemical reaction is the same regardless of the path by, which the reaction occurs. Hess’s law is a direct consequence of the fact that enthalpy, is state function. The enthalpy change of a reaction depends only on the initial and final, states and not on the path by which the reaction occurs., To determine the overall equation of reaction, reactants and products in the individual, steps are added or subtracted like algebraic entities.

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•, , Consider the synthesis of NH3,, , 3H2(g) + N2(g) → 2NH3(g), ΔrH° = - 92.2 kJ, The sum of the enthalpy changes for steps (1) and (2) is equal to enthalpy change for, the overall reaction., 3. Application of Hess’s law:, The Hess's law has been useful to calculate the enthalpy changes for the reactions with, their enthalpies being not known experimentally., Exercises | Q 4.04 | Page 88, Answer the following question., Although ΔS for the formation of two moles of water from H2 and O2 is –327J K–1, it is, spontaneous. Explain., (Given ΔH for the reaction is –572 kJ)., Solution:, 1. For the process to be spontaneous, △Stotal=△Ssys+△Ssurr>0., 2. For the reaction, 2H2(g) + O2(g) → 2H2O(l), when 2 moles of H2 and 1 mole of O2 gas, combine to form 2 moles of liquid water, 572 kJ of heat is released which is received by, surroundings at constant pressure (and 298 K)., 3. The entropy change of the surroundings is,, , = - 327 J K-1 + 1919 J K-1, = + 1592 J K-1, 5. Since △Stotal > 0, the reaction is spontaneous at 25 °C., 6. It follows that to decide spontaneity of reactions, we need to consider the entropy of, system and its surroundings., Exercises | Q 4.05 | Page 88

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Answer the following question., Obtain the relation between ΔG and △Stotal. Comment on the spontaneity of the, reaction., Solution:

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Exercises | Q 4.06 | Page 88, Answer the following question., One mole of an ideal gas is compressed from 500 cm3 against a constant pressure of, 1.2 × 105 Pa. The work involved in the process is 36.0 J. Calculate the final volume., Solution:, Given:, Initial volume (V1) = 500 cm3, External pressure (Pext) = 1.2 × 105 Pa, Work (W) = 36.0 J, To find: Final volume (V2), Formula: W = - Pext Δ V = - Pext (V2 - V1), Calculation: Initial volume (V1) = 500 cm3 = 0.5 dm3, External pressure (Pext) = 1.2 × 105 Pa = 1.2 bar

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∴ 0.3 dm3 = - V2 + 0.5 dm3, ∴ V2 = 0.2 dm3 = 200 cm3, The final volume (V2) = 200 cm3., Exercises | Q 4.07 | Page 88, Answer the following question., Calculate the maximum work when 24 g of O2 are expanded isothermally and reversibly, from the pressure of 1.6 bar to 1 bar at 298 K., Solution:, Given:, Mass of O2 = 24 g, Initial pressure = P1 = 1.6 bar, Final pressure = P2 = 1 bar, Temperature = T = 298 K, To find: Maximum work (Wmax), , Exercises | Q 4.08 | Page 88

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Answer the following question., Calculate the work done in the decomposition of 132 g of NH4NO3 at 100 °C., NH4NO3(s) → N2O(g) + 2H2O(g), State whether work is done on the system or by the system., Solution:, Given:, Decomposition of 1 mole of NH4NO3, Temperature = T = 100 °C = 373 K, To find: Work done and to determine whether work is done on the system or by the, system., Formula: W = -Δ ngRT, Calculation:, Molar mass of NH4NO3 = (2 × 14) + (3 × 16) + (4 × 1) = 80 g mol-1, , The given reaction is for 1 mole of NH4NO3. For 1.65 moles of NH4NO3, the reaction is, given as follows:, 1.65 NH4NO3(s) → 1.65 N2O(g) + 3.30 H2O(g), Now,, Δng = (moles of product gases) − (moles of reactant gases), Δng = (1.65 + 3.30) – 0 = +4.95 mol (∵ NH4NO3 is in solid state), Hence,, W = –Δng RT, = - (+ 4.95 mol) × 8.314 J K-1 mol-1 × 373 K, = - 15350 J, = - 15.35 kJ, Work is done by the system (since W < 0)., The work done is –15.35 kJ. The work is done by the system., Exercises | Q 4.09 | Page 88, Answer the following question.

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Answer the following question., Determine whether the following reaction is spontaneous under standard state, conditions., 2H2O(l) + O2(g) → 2H2O2(l), if ΔH° = 196 kJ, ΔS° = –126 J/K, does it have a cross-over temperature?, Solution:, When ∆H positive and ∆S is negative, then ∆G is positive regardless of temperature., Hence, the reaction is nonspontaneous at all temperatures. It does NOT have a crossover temperature., Exercises | Q 4.12 | Page 88, Answer the following question., Calculate ΔU at 298 K for the reaction,, C2H4(g) + HCl(g) → C2H5Cl(g), ΔH = - 72.3 kJ, How much PV work is done?, Solution:, Given:, Enthalpy change = ΔH = –72.3 kJ, Temperature = T = 298 K, To find:, PV work done and internal energy change (ΔU), Formulae:, 1. W = - ΔngRT, 2. ΔH = ΔU + ΔngRT, Calculations:, Δng = (moles of product gases) - (moles of reactant gases), Δng = 1 – 2 = –1 mol, Using formula (i), W = - ΔngRT, = - (- 1 mol) × 8.314 J K-1 mol-1 × 298 K, = 2477.57 J = 2.48 kJ, Now, using formula (ii) and rearranging,

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ΔU = ΔH - Δ ngRT = ΔH + W = –72.3 kJ + 2.48 kJ = –69.8 kJ, ∴ The PV work done is 2.48 kJ., ∴ The internal energy change (ΔU) is –69.8 kJ., Exercises | Q 4.13 | Page 88, Answer the following question., Calculate the work done during the synthesis of NH3 in which volume changes from 8.0, dm3 to 4.0 dm3 at a constant external pressure of 43 bar. In what direction the workenergy flows?, Solution:, Given:, Initial volume (V1) = 8.0 dm 3, Final volume (V2) = 4.0 dm 3, External pressure (Pext) = 43 bar, To find:, The work done (W) and direction of the work energy flow., Formulae: W = - Pext Δ V = - Pext (V2 - V1), Calculations:, From formula,, W = - Pext Δ V = - Pext (V2 - V1), ∴ W = - 43 bar × (4.0 dm3 - 8.0 dm3) = 172 dm3 bar, Now, 1 dm3 bar = 100 J, , Since, the work is done on the system, work-energy flows into the system from, surroundings., ∴ The work done (W) = 17.2 kJ, ∴ Work energy flows into the system., Exercises | Q 4.14 | Page 88, Calculate the amount of work done in the, 1) Oxidation of 1 mole HCl(g) at 200 °C according to reaction.

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4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g), 2) Decomposition of one mole of NO at 300 °C for the reaction, 2NO(g) → N2(g) + O2(g), Solution:, Given:, 1) Oxidation of 1 mole HCl(g), Temperature = T = 200 °C = 473 K, 2) Decomposition of one mole of NO, Temperature = T = 300 °C = 573 K, To find: Work done, Formula: W = - ΔngRT, Calculation:, 1) The given reaction is for 4 moles of HCl. For 1 mole of HCl, the reaction is given as, follows:, , Hence,, W = - ΔngRT, = -(- 0.25 mol) × 8.314 J K-1 mol-1 × 473 K, = + 983 J, 2) The given reaction is for 2 moles of NO. For 1 mole of NO, the reaction is given as, follows:

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Hence,, W = - ΔngRT, = - 0 mol × 8.314 J K-1 mol-1 × 573 K, = 0 kJ, No work is done (since W = 0)., ∴ The work done is +983 J. The work is done on the system., ∴ The work done is 0 kJ. There is no work done., Exercises | Q 4.15 | Page 89, Answer the following question., When 6.0 g of O2 reacts with CIF as per, 2ClF(g) + O2(g) → Cl2O(g) + OF2(g), The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction? (Δr H° =, 205.6 kJ), Solution:, Given:, Enthalpy change for a given mass = 38.55 kJ, Mass of O2 = 6.0 g, To find: Standard enthalpy of the given reaction, Calculation:

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Exercises | Q 4.16 | Page 89, Answer the following question., Calculate the standard enthalpy of formation of CH3OH(l) from the following data:, , Solution:

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To find:, The total heat required to carry out the given reaction using 180 g of ice., Calculation:, , ∴ 10 mol of H2O requires = 60.1 kJ, ∴ Heat required = 60.1 kJ …(i), b) H2O(l) → H2O(s), 0 °C, , 100 °C, , Heat required = Mass × Specific heat × ΔT, = 180 g × 4.18 J g-1 K-1 × 100 K, = 75240 J, = 75.240 kJ, , ....(ii), , c) H2O(l) → H2O(g), 100 °C, , 100 °C, , Heat required = Latent heat of vaporization, 1 mol of H2O requires = 40.7 kJ, ∴ 1 mol of H2O = 18 g, ∴ 180 g of H2O = 10 moles of H2O

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= - (+ 6 mol) × 8.314 J K-1 mol-1 × 373 K, = - 18606.75 J, = - 18.61 kJ, Work is done by the system (since W < 0)., The work done is - 18.61 kJ. The work is done by the system.