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
Posted: Wed Jul 06, 2022 11:09 am
by answerhappygod
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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