 # Chapter 6 Thermodynamics

This chapter explains the terms - system and surroundings. It also helps in discriminating between close, open and isolated systems. It also gives explanation about internal energy, work and heat. It also states first law of thermodynamics and express it mathematically. Calculation of energy changes as work and heat contributions in chemical systems are also carried out. It also explains state functions - U, H along with their correlation and experimental measures. Calculation of enthalpy changes for various types of reactions can also be carried out. It also states Hess's law of constant heat summation along with its application. It also helps in differentiating between extensive and intensive properties. Definition of spontaneous and non -spontaneous processes are also given. Explanation of entropy as a thermodynamic state function and its application for spontaneity is also given. It also explains Gibbs energy change and establishes relationship between Gibbs energy change and spontaneity.

•  Q1 Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only. Ans: In thermodynamics, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state. A state function describes the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of a path. Functions like p, V, T etc. depend only on the state of a system and not on the path. Hence, alternative (ii) is correct. Q2 For the process to occur under adiabatic conditions, the correct condition is: (i) ΔT = 0 (ii) Δp = 0 (iii) q = 0 (iv) w = 0 Ans: An adiabatic process is one that occurs without transfer of heat or matter between a system and its surroundings & it helps in explaining first law of thermodynamics Hence, under adiabatic conditions, q = 0. Therefore, alternative (iii) is correct. Q3 The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element Ans: (ii) zero Q4 ΔU0of combustion of methane is – X kJ mol–1. The value of ΔH0 is (i) = ΔU0 (ii) > ΔU0 (iii) < ΔU0 (iv) = 0 Ans: (iii) < ΔU0 Q5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1  , –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be (i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1 (iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1 Ans: According to the question, Thus, the desired equation is the one that represents the formation of CH4(g) i.e., [-393.5 + 2(-285.8) - (-890.3)] kJ Mol-1 = -74.8 kJ Mol-1 So,Enthalpy of formation of CH4(g) is -74.8 kJ Mol-1 That means answer is (i). Q6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (iv) possible at any temperature Ans: For a reaction to be spontaneous, ΔG should be negative. ΔG = ΔH- TΔS According to the question, for the given reaction, ΔS = positive ΔH= negative (since heat is evolved) ⇒ ΔG= negative Therefore, the reaction is spontaneous at any temperature. Hence, answer is (iv). Q7 In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the change in internal energy for the process? Ans: According to the first law of thermodynamics, ΔU= q + W    (i) Where, ΔU = change in internal energy for a process q = heat W = work Given, q = + 701 J (Since heat is absorbed) W = -394 J (Since work is done by the system) Substituting the values in expression (i), we get ΔU= 701 J + (-394 J) ΔU = 307 J Hence, the change in internal energy for the given process is 307 J. Q8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. NH2CN(g) + 3/2 O2(g)  →  N2(g) + CO2(g) + H2O(l) Ans: Enthalpy change for a reaction (ΔH) is given by the expression, ΔH = ΔU + ΔngRT Where, ΔU = change in internal energy Δng = change in number of moles For the given reaction, Δng = ng (products) - ng (reactants) = (2 - 1.5) moles Δng = 0.5 moles And, ΔU = -742.7 kJ mol-1 T = 298 K R = 8.314 x 10-3 kJ mol-1 K-1 Substituting the values in the expression of ΔH: ΔH = (-742.7 kJ mol-1) + (0.5 mol) (298 K) (8.314 x 10-3 kJ mol-1 K-1) = -742.7 + 1.2 ΔH = -741.5 kJ mol-1 Q9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1. Ans: From the expression of heat (q), q = m. c. ΔT Where, c = molar heat capacity m = mass of substance ΔT = change in temperature Substituting the values in the expression of q: q = (60/27 mol) (24 Jmol–1 K–1) (20K) q = 1066.7 J q = 1.07 kJ Q10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C. Cp[H2O(l)] = 75.3 J mol-1 K-1 Cp[H2O(s)] = 36.8 J mol-1 K-1 Ans: Total enthalpy change involved in the transformation is the sum of the following changes: (a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C. (b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C. (c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at -10°C. Total ΔH = Cp [H2OCI] ΔT  + ΔHfreezing + Cp[H2O(s)] ΔH = (75.3 J mol-1 K-1) (0 - 10)K + (-6.03 × 103 J mol-1) + (36.8 J mol-1 K-1) (-10 - 0)K = -753 J mol-1 - 6030 J mol-1 - 368 J mol-1 = -7151 J mol-1 = -7.151 kJ mol-1 Hence, the enthalpy change involved in the transformation is -7.151 kJ mol-1. Q11 Enthalpy of combustion of carbon to CO2 is -393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. Ans: Formation of CO2 from carbon and dioxygen gas can be represented as: C(s) + O2(g)  → CO2(g)           ΔfH = -393.5 kJ mol-1 (1 mole = 44 g) Heat released on formation of 44g CO2 = -393.5 kJ mol-1 Heat released on formation of 35.2 g CO2 = [ -393.5kJ mol-1  /  44g  ]  x  35.2g = -314.8 kJ mol-1 Q12 Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g) Ans: ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants. ΔrH = ΔfH (products) - ΔfH (reactants) For the given reaction, N2O4(g)+ 3CO(g)  →  N2O(g) + 3CO2(g) ΔrH = [{ΔfH (N2O) + 3ΔfH(CO2)}  -  {ΔfH(N2O4) + 3ΔfH(CO)}] Substituting the values of ΔfH for N2O, CO2, N2O4,and CO from the question, we get: ΔrH = [{81 KJ mol-1 + 3(-393) KJ mol-1} - {9.7KJ mol-1 + 3(-110)KJ mol-1}] ΔrH = -777.7 KJ mol-1 Hence, the value of ΔrH for the reaction is -777.7 KJ mol-1. Q13 Given N2(g) + 3H2(g) → 2NH3(g) ; ΔrH0 = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas? Ans: Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. Re-writing the given equation for 1 mole of NH3(g), ½ N2(g) + 3/2 H2(g) → NH3(g) Standard enthalpy of formation of NH3(g) = ½ ΔrHθ = ½ (-92.4 kJ mol-1) = - 46.2 kJ mol-1 Q14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l) ;   ΔrH0  = –726 kJ mol–1 C(g) + O2(g) → CO2(g) ;    ΔcH0 = –393 kJ mol–1 H2(g) + 1/2 O2(g) → H2O(l) ;    ΔfH0 = –286 kJ mol–1. Ans: The reaction that takes place during the formation of CH3OH(l) can be written as: C(s) + 2H2O(g) + ½O2(g)  → CH3OH(l)     (1) The reaction (1) can be obtained from the given reactions by following the algebraic calculations as: Equation (ii) + 2 × equation (iii) - equation (i) ΔfHθ [CH3OH(l)]  =  ΔcHθ  +  2ΔfHθ [H2O(l)]  - ΔrHθ = (-393 kJ mol-1) + 2 (-286 kJ mol-1) - (-726 kJ mol-1) = (-393 - 572 + 726) kJ mol-1 ΔfHθ[CH3OH(l)] = -239 kJ mol-1 Q15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). ΔvapH0(CCl4) = 30.5 kJ mol–1. ΔfH0 (CCl4) = –135.5 kJ mol–1. ΔaH0 (C) = 715.0 kJ mol–1 , where ΔaH0 is enthalpy of atomisation ΔaH0 (Cl2) = 242 kJ mol–1 Ans: The chemical equations implying to the given values of enthalpies are: (i) CCl4(l)    →    CCL4(g)      ΔvapHθ = 30.5 kJ mol-1 (ii) C(s)    →   C(g)               ΔaHθ = 715.0 kJ mol-1 (iii) Cl2(g)  →   2Cl(g)          ΔaHθ = 242 kJ mol-1 (iv) C(g)  + 4Cl(g)  →  CCl4(g)  ΔfH = -135.5 kJ mol-1 Enthalpy change for the given process  C(g)  + 4Cl(g)  →  CCl4(g)   can be calculated using the following algebraic calculations as: Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv) ΔH = ΔaHθ(C)  +  2ΔaHθ (Cl2) -  ΔvapHθ - ΔfH = (715.0 kJ mol-1) + 2(242 kJ mol-1) - (30.5 kJ mol-1) - (-135.5 kJ mol-1) ∴ΔH = 1304 kJ mol-1 Bond enthalpy of C-Cl bond in CCl4(g) = 326 kJ mol-1 Q16 For an isolated system, ΔU = 0, what will be ΔS? Ans: ΔS will be positive i.e., greater than zero Since ΔU= 0, ΔS will be positive and the reaction will be spontaneous. Q17 For the reaction at 298 K, 2A + B → C ΔH = 400 kJ mol-1and ΔS = 0.2 kJ K-1mol-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range? Ans: From the expression, ΔG = ΔH - TΔS Assuming the reaction at equilibrium, ΔTfor the reaction would be: T = (ΔH - ΔG) / ΔS T = ΔH / ΔS  (ΔG = 0 at equilibrium) = 400 kJ mol-1 /  0.2 kJ K-1mol-1 T= 2000 K For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K. Q18 For the reaction, 2Cl(g) → Cl2(g),what are the signs of ΔH and ΔS ? Ans: ΔH and ΔS are negative The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative. Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction. Q19 For the reaction 2 A(g) + B(g) → 2D(g) ΔU0 = –10.5 kJ and ΔS0 = –44.1 JK–1. Calculate ΔG0 for the reaction, and predict whether the reaction may occur spontaneously. Ans: For the given reaction, 2 A(g) + B(g) → 2D(g) Δng = 2 - (3) = -1 mole Substituting the value of ΔU0¸ in the expression of ΔH: ΔHº  = ΔUº + ΔngRT = (-10.5 kJ) - (-1) (8.314 x 10-3 kJ K-1 mol-1) (298 K) = -10.5 kJ - 2.48 kJ ΔHº  =  -8.02 kJ Substituting the values of ΔHº and ΔSº¸ in the expression of ΔH: ΔGº  =  ΔHº  - TΔSº = -8.-02 kJ - (298 K) (-44.1 J K-1) = -8.02 kJ + 13.14 kJ ΔGº = + 5.12 kJ Since ΔGº for the reaction is positive, the reaction will not occur spontaneously. Q20 The equilibrium constant for a reaction is 10. What will be the value of ΔG0 ?  R = 8.314 JK–1 mol–1, T = 300 K. Ans: From the expression, ΔG0 = -2.303 RTlogKeq ΔG0 for the reaction, = (2.303) (8.314 JK-1mol-1) (300 K) log10 = -5744.14 Jmol-1 = -5.744 kJ mol-1 Q21 Comment on the thermodynamic stability of NO(g), given 1/2 N2(g) + 1/2 O2(g) → NO(g) ; ΔrH0 = 90 kJ mol–1 NO(g) + 1/2 O2(g) → NO2(g) : ΔrH0= –74 kJ mol–1 Ans: The positive value of ΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable. The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy. Hence, unstable NO(g) changes to stable NO2(g). Q22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfH0 = –286 kJ mol–1. Ans: It is given that 286 kJ mol-1of heat is evolved on the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings. qsurr = +286 kJ mol-1 Entropy change (ΔSsurr) for the surroundings =  qsurr / 7 = 286 kJ mol-1  / 298k ΔSsurr = 959.73 J mol-1K-1