Given the following information:
¹/₂ N₂ (g) + ¹/₂ O₂ (g) →NO (g) D H = 90.3 kJ NO (g) + ¹/₂ Cl₂ (g) →NOCl (g) D H = - 38.6 kJ What is the enthalpy of reaction D H _rxn for the reaction that follows?
2 NOCl (g) →N₂ (g) + O₂ (g) + Cl₂ (g) D H _{net rxn} = ??

Question

Quizplus · Accepted Answer

To find the enthalpy of the given reaction, we need to reverse both of the provided reactions and then add them together. Reversing a reaction changes the sign of ΔH. So, for the first reaction, reversing it gives ΔH = -90.3 kJ, and for the second reaction, ΔH = 38.6 kJ (since it's also reversed). Adding these together gives -90.3 kJ + 38.6 kJ = -51.7 kJ for the formation of 1 mole of NOCl. The given reaction involves the decomposition of 2 moles of NOCl, so the total ΔH for the reaction is 2 * -51.7 kJ = -103.4 kJ.

Given the Following Information