Answers to Problem Sheet (a) spontaneous (b) nonspontaneous (c) nonspontaneous (d) spontaneous (e) nonspontaneous

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Jordan Blankenship
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1 Answers to Problem Sheet 5 1. (a) spontaneous (b) nonspontaneous (c) nonspontaneous (d) spontaneous (e) nonspontaneous 2. (a) Heat will flow from the warmer block of iron to the colder block of iron until the maximum number of microstates of the two blocks is achieved. S universe > 0. (b) The thermal energy of the gas is constant, but the thermal energy is dispersed over a larger volume of space. S system > 0. (c) The thermal energy is increased and the number of microstates of the water is increased. S system > 0. (d) After mixing, the gas expands to twice its volume. The thermal energy is dispersed over a larger volume. S system > (a) This is an adiabatic free expansion of an ideal gas. The internal energy does not change. The temperature does not change. The final temperature is 298K. (b) q=0 (adiabatic), w = 0 (free expansion), E=0 (no temperature change), H=0 (no temperature change), S = q rev /T We need to find q rev, therefore, we need to expand the gas using a reversible path. The path is an isothermal expansion following the curve as shown in the PV graph below. w rev = nrt ln(3) = (1)(8.314)(298)ln(3)= 2722 J. q rev = 2722 J S = 2722/298 =9.13 J/K ln10 5.2

2 5. Isothermal condition, E=H=0 q= w. 43. q = 43 J S 0.14 JK 6. w = P ext V = (1) ( ) = 25.4 Latm = 2.57 kj E = C v T = 3 / 2 RT = 3 / 2 PV = ( 3 / 2 )(1)( )=38.1 Latm =3.86 kj q = E w = 3.86 ( 2.57) = 6.43 kj H = q = E+PV = =6.43 kj 7. (a) Nitrogen is an ideal gas. The heat required is p (b) The increase of the internal energy is (c) w = E q = = 9 kcal (d) If it is a process of constant volume, the required heat is q = nc v T=21 kcal (=E) 8. (a) (i) negative (ii) positive (iii) positive (iv) positive (b) (iv) 9... H vap = S T b.p. = 88 (96+273) = 3.2x10 4 Jmole H vap = (86 cal g 1 )(84 gmole 1 )=7224 cal mole

3 cal g 1 = J g 1 H vap =88426=3.75 x 10 4 J mole 1 Molar mass = 3.75 x 10 4 /384.9 = 97.4 g mole Estimate is too low since strong H bonding makes the liquid state more ordered. Therefore, requires more energy to vapourize (a) S o = S o (AgI (s) ) + S o (HCl (g) ) S o (AgCl (s) ) S o (HI (g) ) = ( ) ( )= 0.4 JK 1 (b) S o =2 S o (AgCl (s) ) 2 S o (Ag (s) ) S o (Cl 2 (g) ) = (242.6) 165.2= 58.0 JK 1 (c) S o =2 S o (Fe (s) ) + 3 S o (CaO (s) ) S o (Fe 2 O 3(s) ) 3 S o (Ca (s) ) = ( ) ( ) = 30.3 JK Devise a reversible path S H 2 O (s, 1 atm, 10 o C) H 2 O (l, 1 atm, 10 o C) warming cooling S 1 S 2 S 3 H 2 O (s, 1 atm, 0 o C) H 2 O (l, 1 atm, 0 o C) S = S 1 + S 2 + S 3 8.7ln ln 0.67 S = = 5.0 cal K Devise a reversible path S H 2 O (l, 1 atm, 125 o C) H 2 O (g, 1 atm, 125 o C) cooling warming S 1 S 2 S 3 H 2 O (l, 1 atm, 100 o C) H 2 O (g, 1 atm, 100 o C) S = S 1 + S 2 + S 3 18ln 1.17 li li

4 ln S = = 25 cal K Devise a reversible path S H 2 O (l, 3 atm, 125 o C) H 2 O (g, 1 atm, 100 o C) cooling vapourization S 1 S 2 S 3 H 2 O (l, 3 atm, 100 o C) H 2 O (l, 1 atm, 100 o C) S = S 1 + S 2 + S ln since water is a liquid 26.0 S = = 25 cal K (a) Free expansion, no heat absorbed or given off. S surroundings = 0 li (b) ln ln (a) (b) G o = H o TS o 0 = 58 x 10 3 T( 14) T = 4143 K G o < 0 spontaneous when T < 4143 K

5 21. (a) (b) G o = H o TS o 0 = 25 x 10 3 T(69.36) T = 360 K G o < 0 spontaneous when T > 360 K 22. Formation of a double bond requires input of energy. H > 0. Formation of gaseous species means S > 0. Reaction is spontaneous at high temperatures. 23. (a) The change in entropy of the water is (b) The change in entropy of the heat source is S S universe = S water +S surroundings = ln 1305 JK =1305 JK (a) Calculate G o for the equilibrium. ΔH = ( ΔH f,products) ( ΔH f,reactants) = ( 285.8) ( 230) = 55.8 kj mol 1 ΔS = ( S,products) ( S,reactants) = (70.0) ( 10.9) = J K 1 mol 1 G o = H o TS o = (0.0808)= 79.9 kj mol 1 G o < 0, spontaneous

6 (b) G o = RT ln K ln K = G o /RT = ( 79.9 x 10 3 /( ))= 32.2 K = 1.0 x The reverse is the autoionization equilibrium constant for water at 25 o C. (K w =1.0 x ) 25. (a) ΔG o = 2ΔG f o (Al (s) ) + 3ΔG f o (H 2 O (l) ) [ΔG f o (Al 2 O 3(s) ) + 3ΔG f o (H 2(g) )] = kj mol 1 (Reaction is not spontaneous at 25 o C and 1 atm) (b) ΔG o = 4ΔG f o (NO (g) ) + 6ΔG f o (H 2 O (l) ) [4ΔG f o (NH 3(g) ) + 5ΔG f o (O 2(g) )] = 1009 kj mol 1 (Reaction is spontaneous at 25 o C and 1 atm) 26. (a) G o = H o TS o Evaluate S o : S o = S o ((NH 2 ) 2 CO (s) ) + S o (H 2 O (l) ) [S o (CO 2(g) ) + 2S o (NH 3(g) )] = (2192.5) = 424 JK 1 Evaluate H o : H o = H f o ((NH 2 ) 2 CO (s) ) + H f o (H 2 O (l) ) [H f o (CO 2(g) ) + 2H f o (NH 3(g) )] = ( 285.9) (246.2) = kj G o = ( 0.424) = 6.85 kj Since G o < 0 it is spontaneous under standard condition. Side note: You could find G o by this equation: G o = n prod G f o (prod) n react G f o (react) = G f o ((NH 2 ) 2 CO (s) ) + G f o (H 2 O (l) ) [G f o (CO 2(g) ) + 2G f o (NH 3(g) )] From the table, G f o (H 2 O (l)) and G f o (NH 3(g) ) are available. G f o (H 2 O (l) ) = kj/mole G f o (NH 3(g) ) = kj/mole Need to calculate: G f o ((NH 2 ) 2 CO (s) ) from: N 2 (g) + 2H 2 (g) + C (s) + 1 / 2 O 2 (g) (NH 2 ) 2 CO(s) S o (Joules mole 1 K 1 ) S o = [ (130.68) (205.14)] = JK 1 G f o ((NH 2 ) 2 CO (s) ) = H f o ((NH 2 ) 2 CO (s) ) TS o ((NH 2 ) 2 CO (s) ) = x 10 3 (298)( 456.7) = kj

7 G f o (CO 2(g) ) from: C (s) + O 2 (g) CO 2 (g) S o (Joules mole 1 K 1 ) S o = ( ) = 2.72 JK 1 G f o (CO 2(g) ) = x 10 3 (298)( 2.72) = 394 kj G f o ((NH 3(g) ) can also be calculated from: 1 / 2 N 2 (g) + 3 / 2 H 2 (g) NH 3 (g) S o (Joules mole 1 K 1 ) S o = [( 1 / 2 )(191.61) + ( 3 / 2 )(130.68) ]= = 99.3 JK 1 G f o (NH 3(g) ) = H f o (NH 3(g) ) TS o (NH 3(g) ) = x 10 3 (298)( 99.3) = 16.3 kj G o = G f o ((NH 2 ) 2 CO (s) ) + G f o (H 2 O (l) ) [G f o (CO 2(g) ) + 2G f o (NH 3(g) )] = kj [ 394 kj + 2( 16.64)] = 7.63 kj (b) Cross over temperature 0 = H o TS o T = / = K When the temperature is greater than K, the reaction is nonspontaneous. 27. ΔG o = ΔG o f (Zn 2+ (aq)) + ΔG o f (Cu (s) ) [ΔG o f (Zn (s) ) + ΔG o f (Cu 2+ (aq))] = = kj ΔG o = = nfe o E o = x 10 3 / 2 (96500) = V E o > 0; spontaneous reaction. This is a voltaic cell.

8 28. Reaction 1: Ag + + e Ag (s) E o = V ΔG o = nfe o = (1) (96500) (0.7991) = 7.71 x 10 4 J Reaction 2: Ag (s) Ag + (aq) + Cl (aq) K sp = 1.56 x ΔG o = RT lnk = ln(1.56 x ) = 5.60 x 10 4 J Reaction 1: Ag + + e Ag (s) Reaction 2: AgCl (s) Ag + (aq) + Cl (aq) AgCl (s) + e Ag (s)+ Cl (aq) ΔG o = 7.71 x x 10 4 = 2.11 x 10 4 J ΔG o = nfe o E o = 2.11 x 10 4 / = 0.22 V The low potential makes this desirable as a reference electrode. This is the silver chloride electrode that is used in the combination ph electrode. 29. ΔG o = ΔG f o (Pb (s) ) + ΔG f o (PbO 2(s) ) + 2ΔG f o (H 2 SO 4(aq) ) [ 2ΔG f o (PbSO 4(s) ) + 2ΔG f o (H 2 O (l) )] ΔG o = ( ) 2( ) 2( ) = kj ΔG o = nfe o The reaction involves 2 electrons. n=2 E o = ( x 10 3 )/2(96500) = V ΔG o > 0 and E o < 0, the charging reaction is not spontaneous.

BCIT Fall Chem Exam #2

BCIT Fall Chem Exam #2 BCIT Fall 2017 Chem 3310 Exam #2 Name: Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially