25℃时，反应N2(g) + 3H2(g) =2NH3(g)，∆rHmӨ= -92.2kJ·mol-1，若升高温度，则
举一反三
- 反应N2(g) + 3H2(g) = 2NH3(g) 的ΔrHmӨ,298 = -92.2 kJ⋅mol-1，若升高温度，则 ΔrHmӨ (T)和ΔrSmӨ (T)怎么变化
- 中国大学MOOC: 25℃时，反应N2(g) + 3H2(g) = 2NH3(g)，∆rGmӨ = -92.2kJ· mol-1，若升高温度，则
- 热化学方程式 N2(g)+3H2(g)=2NH3(g) ΔrHmӨ(298K) = -92.2 kJ·mol-1 表示
- 反应2NH3(g)= N2(g)+ H2(g), △rHmθ(298k)=92.2 kJ·mol-1则△fHmθ(NH3,g,298K)=____
- 298K时,N2(g)+3H2(g)=2NH3(g),△rHm(1)=-92.38kJ.mol-1,则1/2N2(g)+3/2H2(g)=NH3(g)△rHm(2)=()