已知△fGmɵ (N2O4，g)=9.2kJ·mol-1，则下列热化学方程式正确的是（ ）。
A: N2O4(g) → N2(g)+2O2(g) △rGmɵ = 9.2kJ·mol-1
B: N2O4(g) → N2(g)+2O2(g) △rGmɵ = -9.2kJ·mol-1
C: N2(g) +O2(l)→ N2O4(g) △rGmɵ = 9.2kJ·mol-1
D: 2N2O4(g) →2N2(g)+4O2(l) △rGmɵ = 18.4kJ·mol-1
A: N2O4(g) → N2(g)+2O2(g) △rGmɵ = 9.2kJ·mol-1
B: N2O4(g) → N2(g)+2O2(g) △rGmɵ = -9.2kJ·mol-1
C: N2(g) +O2(l)→ N2O4(g) △rGmɵ = 9.2kJ·mol-1
D: 2N2O4(g) →2N2(g)+4O2(l) △rGmɵ = 18.4kJ·mol-1
举一反三
- 【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
- 已 知 298 K 时， 反 应 N2O4(g) → N2(g) + 2O2(g) 的△rGmθ= -97.8 kJ·mol-1， 则△fGmθ(N2O4，g) = 97.8 kJ·mol-1。
- f(n)=O(g(n) 则f(n)^2=O(g(n)^2)
- 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
- f(n)=O(g(n)) 则 f(n)^2=O(g(n)^2) A: 正确 B: 错误