- 1 (19.57 KiB) Viewed 6 times
The standard enthalpy change for the following reaction is -314 kJ at 298 K. ΔΗ° = -314 kJ 1/2 N₂(g) + 2 H₂(g) + 1/2 Cl₂
-
answerhappygod
- Site Admin
- Posts: 899603
- Joined: Mon Aug 02, 2021 8:13 am
The standard enthalpy change for the following reaction is -314 kJ at 298 K. ΔΗ° = -314 kJ 1/2 N₂(g) + 2 H₂(g) + 1/2 Cl₂
The standard enthalpy change for the following reaction is -314 kJ at 298 K. ΔΗ° = -314 kJ 1/2 N₂(g) + 2 H₂(g) + 1/2 Cl₂(g) → NH4Cl (s) - What is the standard enthalpy change for the reaction at 298 K? N₂(g) + 4 H₂(g) + Cl₂(g) →→→ 2 NH4Cl(s) kJ