第五版物理化学第五章习题答案

?θΔrHmT2－T1 lgθ=(）K12.303RTT2T1θK2θΔrHm=135.78kJ?mol

∴K3θ135.781100－1000 lg=(）2.5052.303RT1100×1000K3θ＝11

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t/℃ 30 50 70 90 100 110 1/T 0.003299 0.003095 0.002914 0.002754 0.00268 0.00261 p/kPa 0.827 3.999 15.9 55.23 97.47 167 log (kPa) -0.08249 0.601951 1.201397 1.742175 1.988871 2.222716

2.52.01.5)aPK1.0/p(gl0.50.0-0.50.00260.00270.00280.00290.00300.00310.00320.00330.0034T-1/K-1

直线的斜率m =－3.345×103

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解：当温度T=473K时

lgKθ=－ΔrHθ?m2.303RT+C

∴ΔθrHm2.303R=2100

∴ΔHθrm=40.21kJ?mol－1

2C3H5OH(g)=CH3COOC2H5(g)+2H2(g)473K, ΔθfHm/kJ?mol－1 -235.34 x 0

x－2×(-235.34) = 40.21 x＝－430.47kJ·mol－1

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