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已知:(1)H2(g)+O2(g)===H2O(g) ΔH.1=a kJ/mol (2)2H2(g)+O2(g)===2H2O(g) ΔH.2=b kJ/mol (3)H2(g)+O2...

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任何中和反应生成1 mol H2O,能量变化均相同   同温同压下,H2 (g)+Cl2(g)= 2HCl(g)在光照和点燃 条件下的△H不同   已知:①2H2(g) +O2(g) =2H2O(g) △H=-a kJ·mol-1, ②2H2(g)+O2 (g)= 2H2O(1) △H=-b kJ·mol-1, 则a>b   已知:①C(s,石墨)+O2 (g)=CO2(g) △H=- 393.5kJ·mol-1, ②C(s,金刚石)+O2(g)=CO2 (g) △H=- 395.0 kJ·mol-1,则C.(s,石墨)=C(s,金刚石)△H= + 1.5 kJ·mol-1。  
a<c<0         b>d>0   2a=b<0  2c=d>0  
已知2H2(g)+O2(g)===2H2O(g) ΔH.=-483.6 kJ· mol1说明2 mol H2(g)和1 mol O2(g)的能量总和小于2 mol H2O(g)的能量   已知  (s,石墨)===C(s,金刚石) ΔH.>0,则金刚石比石墨稳定 C.已知NaOH(aq)+HCl(aq)===NaCl(aq)+H2O(l) ΔH.=-57.4 kJ·mol1, 则含20 g NaOH的稀溶液与稀盐酸完全中和,放出28.7 kJ的热量   已知2C(s)+2O2(g)===2CO2(g) ΔH.1 2C(s)+O2(g)===2CO(g) ΔH.2,则ΔH.1>ΔH.2  
241.8 kJ/mol  483.6 kJ/mol   285.8 kJ/mol  571.6 kJ/mol  
2.43 kJ  4.86 kJ  43.8 kJ  87.5 kJ  
H2O(g)=H2(g)+1/2 O2(g) ΔH=+242 kJ/mol    2 H2(g)+O2(g)=2 H2O(g) ΔH=-484 kJ/mol    H2(g)+1/2 O2(g)=H2O(g) ΔH=+242 kJ/mol    2 H2(g)+O2(g)=2 H2O(g) ΔH=+484 kJ/mol  
已知:H2(g)+O2(g)═H2O(l);△H=﹣285.8 kJ•mol﹣1 , 则H2的燃烧热为﹣285.8 kJ•mol﹣1
  
已知:S.(g)+O2(g)═SO2(g)△H1=﹣Q1;S.(s)+O2(g)═SO2(g)△H2=﹣Q2 , 则Q1<Q2
  
已知:H2SO4(浓)+NaOH(aq)═Na2SO4(aq)+H2O(l)△H1;CH3COOH(aq)+NH3•H2O(aq)═CH3COONH4(aq)+H2O(l)△H2 , 则有|△H1|<|△H2|
  
已知:Fe2O3(s)+3C(石墨)═2Fe(s)+3CO(g)△H=+489.0 kJ•mol﹣1
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C.(石墨)+O2(g)═CO2(g)△H=﹣393.5 kJ•mol﹣1
则4Fe(s)+3O2(g)═2Fe2O3(s)△H=﹣1641.0 kJ•mol﹣1  
2H2(g)+O2(g)=2H2O(l) ΔH=―142.9kJ・mol1    2H2(g)+O2(g)=2H2O(l) ΔH=―571.6kJ・mol1    2H2+O2=2H2O ΔH=―571.6kJ・mol1    H2(g)+1/2O2(g)=H2O(g) ΔH=―285.8kJ・mol1  
H2 + O2=2H2O △H= -571.6KJ/mol  H2 (g)+1/2O2(g)= H2O (l) △H= -142.9KJ/mol  H2 (g)+1/2O2(g)= H2O (l) △H= -285.8KJ/mol  2H2 (g) + O2(g) = 2H2O (g) △H= -571.6KJ/mol  
2H2(g)+O2(g)=2H2O(l) ΔH=-142.9 kJ·molˉ1   2H2(g)+O2(g)=2H2O(l) ΔH=+571.6 kJ·molˉ1   H2(g)+1/2O2(g)=2H2O(g) ΔH=-285.8 kJ·molˉ1   H2(g)+1/2O2(g)=2H2O(l) ΔH=-285.8 kJ·molˉ1  
2H2(g)+O2(g)=2H2O(l) ; ΔH=-142.9kJ·mol-1   2H2(g)+O2(g)=2H2O(l) ; ΔH=-571.6kJ·mol-1   2H2+O2=2H2O ; ΔH=-571.6kJ·mol-1   H2(g)+1/2O2(g)=H2O(g) ; ΔH=--285.8kJ·mol-1  
a<b<0  b>d>0  2a=b<0   2c=d>0  
H2O(g)==H2(g)+1/2O2(g)       △H =-485 KJ.mol-1
   
H2O(g)==H2(g)+1/2O2(g)       △H = + 485 KJ.mol-1
   
2 H2(g) + O2(g)==2 H2O(g)     △H = + 485 KJ.mol-1
   
2 H2(g) + O2(g)==2 H2O(g)     △H =-485 KJ.mol-1  
H2O(g)=H2(g)+ 1/2O2 (g)△H=+242 kJ.mol-1    2H2(g)+O2(g)=2H2O(l) △H= -484 kJ.mol-l    H2(g)+ 1/2O2(g)=H2O(g) △H=+242 kJ.mol-l    2H2(g) + O2(g) = 2 H2O(g) △H= -484 kJ  

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