举一反三
- 3 、已知 298K 时25b3 f H m 0398 (Fe 3 O 4 , s) = -1118.0 kJ00b7mol -1 ,25b3 f H m 0398 (H 2 O , g) = -241.8 kJ00b7mol -1 ,则 反 应 Fe 3 O 4 (s) + 4H 2 (g) 2192 3Fe(s) + 4H 2 O(g) 的25b3 r H m 0398 = .... ......
- 已知H2O(g) = H2(g) +0.5O2(g),ΔrHm(1) = 241.8 kJ·mol–1;H2(g) = 2H(g),ΔHm(2) = 436.0 kJ·mol–1;0.5O2(g) = O(g),ΔHm(3)=247.7 kJ·mol–1则H2O中H–O键的平均键焓(单位:kJ·mol–1)为() A: 462.8 B: 925.5 C: –462.8 D: 241.8
- 【单选题】如果被测组分是 Fe 3 O 4 ,称量形式是 Fe 2 O 3 ,则换算因数是() (1.0分) A. 3Fe 2 O 3 /2Fe 3 O 4 B. 2Fe 3 O 4 /3Fe 2 O 3 C. Fe 2 O 3 /Fe 3 O 4 D. Fe 3 O 4 /Fe 2 O 3
- 若增加压力,对反应 Fe 3 O 4(S) +4H 2(g)≒ 3Fe (S) +4H 2 O (g) 的平衡无影响。
- 已知: Zn(s) + 1/2 O 2 (g) = ZnO(s) ∆ r H m q 1 = -351.5 kJ · mol -1 Hg(l) + 1/2 O 2 (g) = HgO(s , 红 ) ∆ r H m q 2 = -90.8 kJ · mol -1 则 Zn(s) + HgO(s , 红 ) = ZnO(s) + Hg(l) 的 ∆ r H m q 为( ) (kJ · mol -1 )
内容
- 0
已知反应B4C(s)()+4O2(g)()=2B2O3(s)()+CO2(g)的ΔrHmΘ=-2859()kJ·mol-1;而且ΔfHmΘ(B2O3)()=()-1273()kJ·mol-1;ΔfHmΘ(CO2)()=()-393()kJ·mol-1。则ΔfHmΘ(B4C)为()。A.()-()2859()kJ·mol()-()1B.()-()1666()kJ·mol()-()1C.()1666()kJ·mol()-()1D.()-()80.0()kJ·mol()-()1
- 1
【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
- 2
已知下面四个反应的△H°298,其中液态水的标准生成热是: A: 2H(g)+O(g)=H2O(g) △rHmθ298(1) B: H2(g)+1/2 O2(g)= H2O(1) △rHmθ298(2) C: H2(g)+1/2 O2(g)= H2O(g) △rHmθ298(3) D: H2(g)+ O(g)= H2O(1) △rHmθ298(4)
- 3
【单选题】已知反应: (1) CO(g) + H2O(g) = CO2(g) + H2(g) , (298K) = - 41.2 kJ/mol (2) CH4(g) + 2 H2O(g)= CO2(g) + 4 H2(g), (298K) = 165.0 kJ/mol 则 CH4(g) + H2O(g) = CO(g) + 3 H2(g) 的 (298K) = 。 A. - 206.2 kJ/mol B. 123.8 kJ/mol C. 206.2 kJ/mol D. - 123.8 kJ/mol
- 4
MgO的晶格能是() A: Mg(s)+1/2O(g)=MgO(s) -ΔHØm(1) B: 2Mg(s)+O(g)=2MgO(s) -ΔHØm(2) C: Mg(g)+1/2O(g)=MgO(s) -ΔHØm(3) D: Mg(g)+O(g)=MgO(s) -ΔHØm(4)