02 第二章 化学热力学初步 习题参考答案P41
第二章 化学热力学初步
(习题参考答案 P42)
4、解：反应等于(1) + (2)×2- (3)
-1-1θ所以 ∆r H m =-393. 5 + (-285.9) ×2 + 890.0kJ ⋅mol = -75.3kJ ⋅mol
3θθθθθ5、解：（1）∆r H m (O 2) ] =∆f H m (PbO ) + ∆f H m (SO 2) - [∆f H m (PbS ) + ∆f H m 2
= （-215-296.8）-(-100) kJ ⋅mol -1= -411.8kJ ⋅mol -1
θθθθθ解：（2）∆r H m =4∆f H m (NO ) + ∆f H m (H 2O ) - [4∆f H m (NH 3) + 5∆f H m (O 2) ] = （4×90.4 - 285.8×6）- (46.11×4) kJ ⋅mol -1= -1168.7kJ ⋅mol -1
θθθθθ解：（3）∆r H m =∆f H m (CO 2) + ∆f H m (H 2O ) + ∆f H m (Ca 2+) - [∆f H m (CaCO 3) +
-1-1θ2∆f H m (H +) ] = （-542.7 - 393.5 - 285.8）- (-1206.9) kJ ⋅mol = -15.1kJ ⋅mol
θθθθθ解：（4）∆r H m =∆f H m (AgBr ) +∆f H m (Cl -) - [∆f H m (AgCl ) + ∆f H m (Br -) ] = （-100 – 167.2）- (-127.1-121) kJ ⋅mol -1= -19.1kJ ⋅mol -1
θθθθθ6、解： ∆r H m =3∆f H m (N 2) + 4∆f H m (H 2O ) - [2∆f H m (N 2H 4) + ∆f H m (N 2O 4) ] = -285.8×4 – [50.6×2 + 9.16] kJ ⋅mol -1 = -1253.56kJ ⋅mol -1
104kJ ∆r H θ=1000÷64×（-1253.56）= -1.96×
答:1kgN 2H 4燃烧后的∆r H θ为-1.96×104kJ 。
Q 34. 4⨯103
J ⋅K -1=31.9J ⋅K -1 11、解：（1）∆S ==T 804+273. 15
Q 2⨯6. 82⨯103
J ⋅K -1=152J ⋅K -1 （2）∆S ==T -183+273. 15
θθθθθ14、解：（1）∆r G m =∆f G m (CaCO 3) +∆f G m (H 2O ) - [∆f G m (Ca (OH ) 2) -∆f G m (CO 2) ]
= （-1128.8 - 237.2）- (896.8 - 394.4) kJ ⋅mol -1= -74.8kJ ⋅mol -1
θ
θθθθ（2）∆r G m =∆f G m (CaSO 4) +2∆f G m (H 2O ) - [∆f G m (CaSO 4⋅2H 2O ) ] = -1321.9 + (-237.2×2) –(-1797)kJ ⋅mol -1= 0.7kJ ⋅mol -1
θ > 0 反应不能自发进行 ∆r G m
θθθθθ（3）∆r G m =∆f G m (Pb ) +∆f G m (CO 2) - [∆f G m (PbO ) -∆f G m (CO ) ] = --394.1 + 0 –(-188-137.2)kJ ⋅mol -1= -69.2kJ ⋅mol -1
θ
1θθθθ18、解：∆r G m (O 2) - ∆f G m =∆f G m (C ) +∆f G m (CO ) 2
= 0 + 0 -（-137.2）kJ ⋅mol = 137.2kJ ⋅mol > 0 -1-1
1-1θθθθ(O 2) - ∆f H m ∆r H m =∆f H m (C ) + ∆f H m (CO ) =110.5kJ ⋅mol > 0 2
1θθθθ(O 2) - ∆S m = ∆S m ∆r S m (C ) + ∆S m (CO ) = -92.69J ⋅mol -1⋅K -1
答：∆r G 恒大于零，所以反应不可能自发进行。