若已知 2NH3(g)= N2(g)+3H2(g)的△rHmθ= 92.22 kJ •mol-1，则△fHmθ(NH3,g)=( )kJ • mol-1。
A: 46.11
B: – 92.22
C: -46.11
D: 92.22
A: 46.11
B: – 92.22
C: -46.11
D: 92.22
举一反三
- 已知在298.15 K时，N2(g)+3H2(g)→2NH3(g)反应热为ΔrHθ= -23.06 kJ·mol-1 (ΔfHmθ(NH3,g,298.15K)=-46.11kJ·mol-1)，则该反应的ξ为 mol A: 1 B: 2 C: 1/2 D: 1/4
- 已知反应3H2(g)+N2(g)→2NH3(g)的△rGmӨ= -33.0kJ·mol-1，则△fGmӨ(NH3,g)=______ kJ·mol-1，△fGmӨ(N2,g)=______ kJ·mol-1。
- 反应2NH3(g)= N2(g)+ H2(g), △rHmθ(298k)=92.2 kJ·mol-1则△fHmθ(NH3,g,298K)=____
- 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
- 已知298K时下列热化学方程式： ① 2NH3(g) → N2(g) + 3H2(g) △rHmΘ = 92.2kJ·mol－1② H2(g) + 1/2O2(g) → H2O(g) △rHmΘ = -241.8kJ·mol－1③ 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) △rHmΘ = -905.6kJ·mol－1试确定△fHmΘ（NH3,g,298k）=( ) kJ·mol－1 A: -90 B: -46.1 C: 90.2 D: 46.1