( ) Chapter 12. Problem Problem Problem 12.3 G = T = 0 T = K. Mg CO3 = Mg O + CO2. We know that ( ) 10 amt

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1 Chater Problem. We know that ( ) Mg 3 = Mg O + G = G + R ln Π= G + R ln ( ) r r in atm amt in atm Let us find temerature at which rg = =, and from able A- we know that rg 76 7 = 76 7 const7ln + = 76 = 565 K 7 const7 ln Problem. Ni (s) + O m G Ni (l) + O fs G G fl Ni O G = G G m fs fl = = 79. able A- mh ms At melting oint, G = 79. = = K m m m m Problem.3 Ag O = Ag+ O rg = G + R ln Π = G + R ln O = G ln + R O () of

2 Of course, O in () is the artial ressure of O in the surrounding; it is not the equilibrium ressure of O above AgO. We know from able A- that rg = rg = R ln O = 35 = 66. const7 ln O If If =, then O O =., then ln O = and 35 = 6 K = K 66. const7 ln. Problem. Cr+ 3H O ox G Cr O + 3H 3 ( ) 3 HO fg ( Cr O ) fg 3 ( Cr O ) 3 ( H O) G = G G ox f 3 f = ( ) = able A- 3H + 3 O + Cr Since =, the reaction of oxidation of chromium by water vaor is exothermic oxh 3676 = + = oxg H H oxg5 = oxg5 + R ln Π = oxg5 + R ln = oxg5 + 3R ln HO HO HO oxg 5 HO oxg 5 ln = = ex HO 3R HO 3R = = ex + ex 3R 5 const7 HO oxg5 of 3 3

3 Problem.5 Sn Cl + H Sn+ HCl () HCl rg rg R ln rg R ln H = + Π = + From able A- Sn Cl Sn Cl, rg = = + (3) + = G = () H Cl H Cl, r 88.8 Aarently, () = () (3), which gives for reaction () ( ) rg = = ? const7 9 ln =.5 No, LHS is aroximately 789, which is quite different from. he equilibrium has not been attained. We can arrive at the same conclusion a bit differently. K rg = ex = ex.8 R const7 9 K is different from.7.5, which is aroximately.98. Problem.6 Fe+ = Fe O + (5) Fe+ G Fe O Fe+ C + O FeO + C + O 87.6 G = 3 of

4 7.8 G = G + R ln Π = G + R ln = Const7 ln If = 73, then G = =.37, which is, of course, zero from a ractical viewoint. G < his means that if temerature increases, G decreases, but in our case, temerature decreases, which means that G will be greater than if < 73. his means that (5) shifts to the left, and FeO will be the first to disaear. Problem.7 It never hurts to draw a thermochemical cycle Mg O (s) + Si (s) = Mg (g) +MgSi O (s) (6) Mg O + Si G r Mg (g) + Mg Si O Mg (g) + Mg O + Si O + O ( ) ( ) Mg (g) + Si O + O Mg (g) + Si+ O ( ) rg = = = = Memory A R ln = R ln = G Mg Mg r Mg G Memory A = = R Const7 r ex ex.5 atm of

5 Problem.8 Molecular mass of Ca 3 is = g mole, which means that. moles of calcium carbonate were taken. If all Ca 3 decomoses, then. moles of form. If the volume of CaO is ignored, then these. moles of occuy liter. =..86 = 8.6 Ca 3 = Ca O +, rg = 68 rg 68 5 = ex = ex + ex R R R Solution of (7) is 66 K. 5 + = ex (7) Problem.9 CoO + SO3 = CoSO (8) For (8), = + ; at = 3 K, rg =, which means that rg Solution of (9) is ressures is atm. R ln = R ln = G SO3 SO3 r 3 rg ln SO = ex = ex.8 atm 3 R Const7 3 SO + O = SO 3, rg = = = 3 K SO 3 SO3 R ln = R ln = SO O SO SO SO =.63, which means that (9) SO =.37. he sum of all three artial 5 of

6 Problem. S + Fe G + FeS C + O + S + Fe = = 8 G = + S + Fe 8 = ex = something I recorded in Memory A Const7 S + Fe = + FeS Before. 9 After. x 9 + x 9 + x = Memory A x.396. x % of S removed is S G S G = + = = K K C+ O + S = ex 5.7 const7 3 ( ) S..396 = = = 5.7 ( ) S S S S 3 = of

7 Problem. Ca F + H O = Ca O + HF g g g g mole mole mole mole =, which tells us that a weight loss of grams means that moles of HF were roduced mole of HO was consumed atm liter R = F Kelvin mole liter 6 min 6 liters min = ; PHOV atm.9 6 liters nho= = =.8 R 98 K F B At 9 K, the weight loss is, which translates in the roduction of.69 g 5.5 C moles of HF, which gives HF,9 C F 98 6 = = D K,9 HF,9 D = = =.36.9 HO G9 = R K,9 = = X ln 38 9 ln E J E At K, the weight loss is, which translates in the roduction of g 7.55 C moles of HF, which gives HF, C F 98 = = D K, HF, D 7 = = =.57.9 HO G = R K, = = Y ln 38 ln E J a+ 9 b= X G = a b Y + = H S E 7 of

8 Problem. Let us find aroriate reactions in able A- Since [] = [a] + [b] [c] Since [] = [a] [d] Fe+ = Fe3O + [] Fe+ HO = Fe3O + H [] 3Fe+ O = Fe O, G = [a] 3 a b C + O =, G = [b] C + O =, G = 39.8 [c] c H + O = H O, G = [d] d G = = + G 8 he solution of () is =.8653 = H = HO= G R = ex G H = ex 867. HO R + + H + HO = + = H + HO () 8 of

9 Since all carbon is from CH, a number of moles of methane consumed is Since all oxygen is from Fe3O, a number of moles of magnetite reduced is n Fe3O =.78. A number of moles of iron roduced is n Fe =.3 n CH CH n Fe n er mole of Fe = =.3 3 n CH er mole of Fe = moles of CH are needed to roduce mole of Fe this is what a technologist would like to hear from us n CH = Problem.3 = + + G a b ln c dg dg G =H S =H = H + d d dg d ( ln ) H =G = a + b ln + c a + b + c = a b d d H = i ( ) H = ii ( ) = + H 68. iii ( ) Obviously, (ii) corresonds to the oxidation of Mg(g), but what s about (i) and (ii)? ln +. = 68. ln +.8 m m m m m m ( ) ( m) m = H = i m H = + ( ) ( m) 68. iii m Aarently, (ii) corresonds to the oxidation of Mg(liquid). 9 of

10 Problem. P =.8 atm and = 3 K are constant. Initial state: two hases. Gas:. moles of O and.58 moles of N. Condensed hase: Zn(L). Final (equilibrium) state: three hases. Gas: N, O, Zn. Condensed hases: Zn(L) and ZnO. We have to find an amount of Zn(L) at equilibrium. Zn ( g) + O = Zn O For this reaction, G = , which at 3K gives 566. K Zn over ure liquid Zn is given by 566 = = ex 9.9 Zn O 3 const7 () From () and () Zn 55 = ex.55ln O = 3 = N + O + Zn =.8 N =.8 O Zn = =.63 negligible 5 () V = n R.787 = =.58.5 moles or 9.7 g of Zn in gas hase V.63 N N Zn nzn nn Zn = n ZnR N V = n R all O reacted with Zn V n R 5 N N O no = n N = O = O N.63. moles of O.8 moles of Zn reacted, which is 5.9 g = 5.38 g of Zn(L) remained in the system Problem.5 CH ( ) + O + 79 N 36 + xca 3 + x + HO + O + N + xca O Ca O + = Ca, G = 68 + = 88 = R ln = R ln 3 8 K of

11 On the other hand 88 = ex. 8 const7 From (3) and () we find that x x = 36 + x (3) () Problem.6 ( ) ( ) Hg g + O = Hg O s, G = = 78 = R ln = 6 K Hg O ( O O ) = R ln From this equation O =.53. Of course, Hg =.36 and i = Pxi xi = i P N =.95. x O = =, x Hg =.53 Problem.7 In able A-, we can find Fe O + = Fe+ [] Fe+ O = Fe O, G = [a] a C + O =, G = [b] b Since [] = [a] [b] + [c] At K C + O =, G = 39.8 [c] c G = P R ln = R ln = G of

12 Since [] = [b] [c] P G = ex = R At = K (5) C + = [] = G R ln = R ln = G P At K G = ex =.578 P R At = K he solution of the system of equations () and (3) is P total = (6) P = P = By the way, at = 97 K P total = P = P = which corresonds quite well to Gaskell s numbers. of