✔ 最佳答案
1. (b)
i.
Kp
= (PSO3)2/(PSO2)2(PO2)
= (0.32)2/(0.27)2(0.40)
= 3.51 atm-1
ii.
Initial partial pressure of O2 = 0.40 + (0.32/2) = 0.56 atm
Initial partial pressure of SO2 = 0.27 + 0.32 = 0.59 atm
(f)
B has the lowest boiling point.
A (melting point 5.5oC) and B (-16oC) have similar sizes. The hydrogen bonds between molecules of B are almost completely broken by the long alkyl groups. Therefore, the intermolecular attractions between A and B are both mainly van der Waals' attractions. As the planar molecule of A is more symmetrical than that of B, the molecules of A are packed more closely. Therefore, A has a higher melting point than B.
C is a polymer. The van der Waals' attractions between the polymer molecules are stronger than those in B, and long polymer molecules tangle each other. Therefore, C has a higher melting point.
D has a higher melting point than B, because each molecule of B is held by neighbouring molecules with a number of hydrogen bonds.
10.
(a)
(i)
No. of moles of benzene = 1.17/(12x6 + 1x5) = 0.015 mol
ΔHc[C6H6(l)] = -49020/0.015 = -3268000 J mol-1 = -3268 kJ mol-1
(ii)
C6H6(l) + 9O2(g) → 6CO2(g) + 3H2O(l)
6ΔHf[CO2(g)] + 3ΔHf[H2O(l)] - ΔHf[C6H6(l)] = ΔHc[C6H6(l)]
6(-393) + 3(-286) - ΔHf[C6H6(l)] = -3268
ΔHf[C6H6(l)] = +52 kJ mol-1
(b)
Bond enthalpy of bezene
= 6(+413) + 3(+348) + 3(+612)
= +5358 kJ mol-1
C(s) → C(g) ..... (1)
(1/2)H2(g) → H(g) ..... (2)
6C(g) + 6H(g) → C6H6(g) ..... (3)
C6H6(g) → C6H6(l) ..... (4)
6(1) + 6(2) + (3) - (4):
6C(s) + 3H2(g) → C6H6(l)
ΔHf[C6H6]
= 6(+715) + 3(+218) + (-5358) - (+31)
= +209 kJ mol-1
(c)
The actual (experimental) enthalpy change of benzene in (a)(ii) is less endothermic than the theoretical value in (b). In other words, the actual structure of benzene is more stable than that of the give structure with conjugated C=C double bonds. The extra stability of the actual structure of benzene is due to its resonance (or delocalization of π electrons.)
Besides, the bond enthalpy values used in (b) are just average values.
2009-08-18 16:19:26 補充：
In 10.(b)
Bond enthalpy change of benzene = +5358 kJ/mol
That means:
C6H6(g) → 6C(g) + 6H(g) .. ΔH = +5358 kJ/mol
as bond breaking is endothermic (ΔH = +ve)
2009-08-18 16:19:42 補充：
For Equation (3):
6C(g) + 6H(g) → C6H6(g)
This is the backward reaction of above.
Hence, ΔH = -5358 kJ/mol, because bond formation is endothermic (ΔH = -ve)
2009-08-18 16:22:02 補充：
In 10.(b)
Bond enthalpy of benzene = +5358 kJ/mol
That means:
C6H6(g) → 6C(g) + 6H(g) .. ΔH = +5358 kJ/mol
as bond breaking is endothermic (ΔH = +ve)
For Equation (3):
6C(g) + 6H(g) → C6H6(g)
This is the backward reaction of above.
Hence, ΔH = -5358 kJ/mol, because bond formation is endothermic (ΔH = -ve)
2009-08-18 16:22:41 補充：
In 10.(b)
Bond enthalpy of benzene = +5358 kJ/mol
That means:
C6H6(g) → 6C(g) + 6H(g) .. ΔH = +5358 kJ/mol
as bond breaking is endothermic (ΔH = +ve)
For Equation (3):
6C(g) + 6H(g) → C6H6(g)
This is the backward reaction of above.
Hence, ΔH = -5358 kJ/mol, because bond formation is endothermic (ΔH = -ve)
2009-08-19 18:27:17 補充：
Sorry for mistyping. The correct statements are:
Bond breaking is endothermic (ΔH = +ve)
Bond formation is exothermic (ΔH = -ve)
