普通化学试卷及答案(新) 下载本文

2. ⑴ c(H)?ca??Kac?c?1.76?10-5?1?0.1?1.33?10?3(mol?dm?3)

pH=-lg1.33×10-3=2.88 , α=(1.33×10-3/0.1)×100%=1.33% ⑵ NaOH + HAc = NaAc,C(NaAc)= C(HAc)= 0.033 mol·dm-3

pH=pKa-lgc(HAc)=4.8+0=4.8 -c(Ac)⑶ C(NaAc)= 0.05 mol·dm-3,pOH=5.3 pH=14-5.3=8.7 3.⑴ △rGm(298.15K)?c(OH?)?Kbc?c?(KW/Kb)c?c?(1?10/1.76?10)?1?0.05?5.33?10(mol?dm)-14-5?6?3

??BB?fG?(—394.36)—2×86.57—2×m(B, 298.15 K)=[0+2×(—137.156)] (kJ.mol-1) = —687.548( kJ.mol-1) ⑵在25℃的标准条件下反应能自发进行。人们为此反应寻求高效催化剂具有现实意义。 ⑶?rHm(298.15K)=???B?fHm(B, 298.15 K)=[0+2×(—393.50)—2×90.25—2×

B???(—110.52)] (kJ.mol-1)= —746.46 (kJ.mol-1) ?rS???B?S?m(B, 298.15 K) m(298.15K)=??B=[191.50+2×213.64—2×210.65—2×197.56] (J·K-1·mol-1)= —197.64 (J ·K-1·mol-1)

???rG? (873.15 K) =?H (298.15 K) + T K? ?Smrmrm (298.15 K) = —746.46 kJ.mol-1 + 873.15K×(—197.64×10-3)KJ ·K-1·mol-1 = —746.46 kJ.mol-1—172.57KJ·mol-1= —919.03 KJ·mol-1

??rGm(T)-919.03?103lnK???126.6

?RT-8.314?873.15K??9.59?1054?

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