加入惰性气体对哪一个反应能增大其平衡转化率? ( )
A: C6H5C2H5(g) = C6H5C2H3(g) + H2(g)
B: CO(g) + H2O(g) = CO2(g) + H2(g)
C: 3/2H2(g) +1/2N2(g) = NH3(g)
D: CH3COOH(l) + C2H5OH(l) = H2O(l) + C2H5COOCH3(l)
A: C6H5C2H5(g) = C6H5C2H3(g) + H2(g)
B: CO(g) + H2O(g) = CO2(g) + H2(g)
C: 3/2H2(g) +1/2N2(g) = NH3(g)
D: CH3COOH(l) + C2H5OH(l) = H2O(l) + C2H5COOCH3(l)
举一反三
- 加入惰性气体对哪一个反应能增大其平衡转化率?() A: C6H5C2H5(g)=C6H5C2H3(g)+H2(g) B: CO(g)+H2O(g)=CO2(g)+H2(g) C: 3/2H2(g)+1/2N2(g)=NH3(g) D: CH3COOH(l)+C2H5OH(l)=H2O(l)+C2H5COOH3(l)
- 下列反应中 25b3 r S m 03b8 > 0 的是 。 (A) 2H 2 (g) + O 2 (g) = 2H 2 O(g) (B) N 2 (g) + 3H 2 (g) = 2NH 3 (g) (C) NH 4 Cl(s) = NH 3 (g) + HCl(g) (D) CO 2 (g)+ 2NaOH(aq) =Na 2 CO 3 (aq)+H 2 O(l)
- 【单选题】由下列数据确定CH 4 (g)的为Δ f H m Θ 为 () C ( 石墨 ) +O 2 (g)=CO 2 (g) Δ r H m Θ =-393.5kJ·mol - 1 H 2 (g)+1/2O 2 (g)=H 2 O(l) Δ r H m Θ =-285.8kJ·mol - 1 CH 4 (g)+2O 2 (g)=CO 2 (g)+2H 2 O ( l ) Δ r H m Θ =-890.3kJ·mol - 1 A. 211 kJ·mol -1 B. -74.8kJ·mol - 1 C. 890.3 kJ·mol - 1 D. 缺条件，无法算
- 某体系存在C(s)、H2O(g)、CO(g)、CO2(g)、H2(g)五种物质,相互建立了下述三个平衡,则该体系的独立组分数C为 A: sub>2O(g) + C(s) ó H 2(g) + CO(g) B: 2(g) + H 2(g) ó H 2O(g) + CO(g) C: 2(g) + C(s) ó 2CO(g) D: 3 E: 2 F: 1 G: 4
- *下列 反应的 △ r H m θ (298K) 就是 △ f H m θ (CO 2 , g,298K) 。 (A) CO(g)+1/2O 2 (g)=CO 2 (g) (B) C( 金刚石 )+O 2 (g)=CO 2 (g) (C) C( 石墨 )+O 2 (g)=CO 2 (g) (D) CO 2 (g) = C( 石墨 )+O 2 (g)