Question 1: N O 4 (g) NO (g) Amounts of material at the initial state: n 0 0 Amounts of material at equilibrium: (1 α)n 0 αn 0 Where α equals 0.0488 at 300 K and α equals 0.141 at 400K. Step 1: Calculate the partial pressures of N O 4 and NO P NO4 = x NO4 *P total ; P NO = x NO *P total ; This is important to know that the P total is different from the initial pressure (1bar) because of the production of extra molecules!!! The new P total can be calculated based on the perfect gas equation PV = nrt Step : Assuming that gases are perfect, α j = P j /P ө PNO ( ) θ K = P PN O4 ( ) θ P ( a) K = 1 a thus ( 0. 0488) K 300k = 1 0. 0488 = 0.01001447 ( 0. 141) K 400k = 1 0. 141 = 0.0986185 In order to calculate K 98k, one needs to go through the van t Hoff equation first to calculate the standard reaction enthalpy: lnk lnk 1 = - ( r H ө /R)(1/T 1/T 1 ) Plug in the equilibrium constants at 300K and 400K into the above equation: ln(0.0986185) ln(0.01001447) = - (/R)(1/400 1/300) r H ө =.3kJ mol -1 Because we have assumed that the standard reaction enthalpy does not change with temperature, which is also the prerequisite for the van t Hoff equation. r H ө at 98 is.3kj mol -1. lnk 98K lnk 1 = - ( r H ө /R)(1/98K 1/T 1 ) lnk 98K = -4.54398 K 98K = 0.01000 r G ө = - RTlnK = - 8.31415 JK -1 mol -1 * 98 K * ln (0.010) = 11.6 kjmol -1
Question : The standard potential for AgCl/Ag, Cl - can be calculated directly from the given equation: (note that the temperature unit is o C). Convert 98K to 5 o C first E ө /V = 0.366 4.856 x 10-4 (5 o C / o C) 3.41 x 10-6 (5 o C / o C) + 5.87 x 10-9 (5 o C / o C) 3 E ө /V = 0.31875 E ө = 0.31875 V r G ө = -vf E ө = - 1x 96485 C mol -1 x 0.31875V = -1450.76 CV mol -1 = -1450.76 J mol -1 = -1.450 kj mol -1 Note that the above value is the standard reaction Gibbs energy, which describes the following half-reaction: AgCl(s) + e - Ag(s) + Cl - (aq) r G ө = f G ө (Ag(s)+ f G ө (Cl - (aq)) - f G ө (AgCl(s) -1.450 kj mol-1 = 0 + f G ө (Cl - (aq)) (- 109.79 kj mol -1 ) f G ө (Cl - (aq)) = -131. kj mol -1 The reaction entropy can be calculated from the temperature coefficient of the potential: ϑ de r S = dt vf de θ dt = -4.856 x 10-4 6.84 x 10-6 (T / o C) + 5.87x3 x 10-9 (T/ o C) so r S ө = vf( -4.856 x 10-4 6.84 x 10-6 (5 o C / o C) + 5.87x3 x 10-9 (5 o C / o C) ) = -63.36 J mol -1 f H ө = f G ө + T x f S ө = -131. kj mol -1 + 98 (-63.36 kj mol -1 ) = 150.1 kj mol -1 Question 3 The cell reaction can be broken down into the following two half reactions: R: Au 3+ (aq) + e - Au + (aq) L: Fe 3+ (aq) + e - Fe + (aq) The cell potential E ө = E ө (Au 3+ /Au + ) - E ө (Fe 3+ /Fe + )
Here the potential for E ө (Au 3+ /Au + ) needs to be calculated from E ө (Au 3+ /Au) and E ө (Au + /Au) via their standard reaction Gibbs energy. (Note that direct subtract of their respective potential is a wrong approach) Write the reactions 1: Au 3+ (aq) + 3e - Au(s) : Au + (aq) + e - Au(s) 3: Au 3+ (aq) + e - Au + (aq) The standard reaction Gibbs energy for these reactions: r G ө (1) = - vfe ө (1) = - 3 FE ө (1) r G ө () = - vfe ө () = - 1 FE ө () r G ө (3) = - vfe ө (3) = - 3 FE ө (3) Because reaction 3 = reaction 1 - reaction r G ө(3) = r G ө(1) - r G ө() So - 3 FE ө (1) - (- 1 FE ө ()) = - FE ө (3) E ө (3) = (3*1.40 1*1.69)/ = 1.55V Thus The cell potential E ө = 1.55V 0.77V = 0.485V Because lnk =vfe ө /RT K =.54 x 10 16 Question 4: Solution: (a) The equilibrium will not shift toward either direction because both sides have the same number of molecules. Thermodynamically, K also remains to be the same since the pressure does not effect the standard reaction Gibbs energy. (b) The equilibrium shifts towards the products to absorb more heat and therefore reduces the disturbance caused by the increase of temperature. Thermodynamically, we have H > 0, so that K must increase as T increases. Therefore, the value of K rises and the equilibrium shifts toward the product side. (c) The equilibrium will shift toward the products to increase the number of moles of C(g), therefore reduces the stress produced by reducing C gas. Thermodynamically, this shift occurs in such manner to maintain a constant value for K, which would otherwise decrease. Question 5: Write the cell reaction
BaCl (s) + H (g) Ba(s) + H + (aq) + Cl - (aq) The potential for the above electrochemical cell can be calculated from E = E ө RT - ln(q) vf Q = E ө ( a + ) ( a ) H Cl a H = E ө (BaCl (s),ba) - E ө (H+/H (g)) = E ө (BaCl (s),ba) Calculate the mean activity coefficient for H + and Cl - Ionic strength I = ½ (1 x1 + 1 x1) = 1 (units mol kg -1 disappeared here due to the division by the standard molarlity) log(γ ± ) = - -1 x 1 AI 1/ = - 0.509 x 1 x(1) 1/ γ ± = 0.3097 0.341V = E ө RT ( 0. 3097 1) ( 0. 3097 1) (BaCl (s),ba) - ln( ) F 1 E ө (BaCl (s),ba) = 0.81 V Question 6: Step 1: Identify whether the mixture corresponds to the end point of the titration or in the middle of a titration process: The amount of weak acid: 0.05 x 0.001M = 5x10-5 mole The amount of strong base: 0.05 x 0.001M =.5x10-5 mole Because there are excessive amounts of weak acid in the mixture, Henderson- Hasselbalch equation should be employed here: ph = pka - log([acid]/[base]) the pka of the weak acid can be calculated from the following process: CH 3 COOH CH 3 COO- + H + r G ө = f G ө (CH3COO - ) + f G ө (H + ) - f G ө (CH3COOH) = 349.3 kj mol -1 + 0 - (-376.46)kJ mol -1 = 75.76kJ mol -1 r G ө = -RT ln(k) ln K = - 9.78 logk = - 17.15 pka = 17.15 [acid] = (5x10-5.5x10-5 )/0.075 [base] = (.5x10-5 )/0.075
ph = 17.15 log (1) = 17.15