热化学方程式 N2(g)+3H2(g)=2NH3(g) ΔrHmӨ(298K) = -92.2 kJ·mol-1 表示
举一反三
- 反应2NH3(g)= N2(g)+ H2(g), △rHmθ(298k)=92.2 kJ·mol-1则△fHmθ(NH3,g,298K)=____
- 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
- 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)=()
- 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
- 已知298K时, ∆fGmӨ(NH3, g) = −16.45 kJ·mol−1, 则反应N2 (g) + 3H2 (g) → 2NH3 (g)在298 K、标准状态下能自发进行。