CHEMICAL EQUILIBRIUM
129
Assuming both gases are perfect aJ = K = = (b)
pJ − p−
− )2 (pBr /p − 4α 2 p 4α 2 = = − − pBr2 /p − (1 − α 2 )p − 1 − α2
− ] [p = p −
4(0.24)2 = 0.2445 = 0.24 1 − (0.24)2
− r G − = −RT ln K = −(8.314 J K−1 mol−1 ) × (1600 K) × ln(0.2445)
= 19 kJ mol−1
(c)
− 1 1 r H − − ln K(2273 K) = ln K(1600 K) − R 2273 K 1600 K 112 × 103 J mol−1 × (−1.851 × 10−4 ) = ln(0.2445) − 8.314 J K−1 mol−1 = 1.084 K(2273 K) = e1.084 = 2.96
E9.9(b)
ν(CHCl3 ) = 1, (a)
ν(HCl) = 3,
ν(CH4 ) = −1,
ν(Cl2 ) = −3
− − − − r G− = f G − (CHCl3 , l) + 3 f G− (HCl, g) − f G− (CH4 , g) −1 −1 = (−73.66 kJ mol ) + (3) × (−95.30 kJ mol ) − (−50.72 kJ mol−1 )
= −308.84 kJ mol−1 ln K = −
− −(−308.84 × 103 J mol−1 ) r G− = 124.584 [8] = RT (8.3145 J K−1 mol−1 ) × (298.15 K)
K = 1.3 × 1054 (b)
− − − − r H − = f H − (CHCl3 , l) + 3 f H − (HCl, g) − f H − (CH4 , g) −1 −1 = (−134.47 kJ mol ) + (3) × (−92.31 kJ mol ) − (−74.81 kJ mol−1 ) = −336.59 kJ mol−1 − 1 1 r H − − [9.28] ln K(50◦ C) = ln K(25◦ C) − R 323.2 K 298.2 K −336.59 × 103 J mol−1 × (−2.594 × 10−4 K −1 ) = 114.083 = 124.584 − 8.3145 J K−1 mol−1
K(50◦ C) = 3.5 × 1049 − r G − (50◦ C) = −RT ln K(50◦ C)[18] = −(8.3145 J K −1 mol−1 )×(323.15 K)×(114.083)
= −306.52 kJ mol−1 E9.10(b)
Draw up the following table Initial amounts/mol Stated change/mol Implied change/mol Equilibrium amounts/mol Mole fractions
A 2.00 −0.79 1.21 0.1782
+
B 1.00
−0.79 0.21 0.0309
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C + 0 +0.79 +0.79 0.79 0.1163
2D 3.00 +1.58 4.58 0.6745
Total 6.00
6.79 0.9999