Given the following two equations and their corresponding enthalpies:
CO (g) + ¹/₂ O₂ (g) →CO₂ (g) D H ₁ = - 283.0 kJ N₂ (g) + O₂ (g) →2 NO (g) D H ₂ = 180.6 kJ What is the enthalpy of reaction D H _rxn for the reaction that follows?
CO (g) + NO (g) →CO₂ (g) + ¹/₂ N₂ (g) D H _{net rxn} = ??

Question

Quizplus · Accepted Answer

We need to use Hess's Law to find the enthalpy of the given reaction. We can break down the given reaction into two steps: Step 1: CO (g) + ¹/₂ (g) → CO₂ (g) (Given) Step 2: 2 NO (g) → N₂ (g) + O₂ (g) (Reverse of Given) Adding these two equations, we get: CO (g) + ¹/₂ (g) + 2 NO (g) → CO₂ (g) + N₂ (g) + O₂ (g) Now we can calculate the net enthalpy change of the reaction by adding the enthalpies of the two steps: D H _{net rxn} = D H ₁ + D H ₂ (reverse) D H _{net rxn} = -283.0 kJ + (-180.6 kJ) D H _{net rxn} = -463.6 kJ Finally, we need to flip the sign of the enthalpy change because we want the enthalpy of the forward reaction (given reaction): D H _rxn = -(-463.6 kJ) D H _rxn = 463.6 kJ However, this is the enthalpy change for the reverse reaction. To get the enthalpy change for the forward reaction, we need to flip the sign again: D H _rxn = -463.6 kJ → D H _rxn = -(-463.6 kJ) = 463.6 kJ However, none of the answer choices matches 463.6 kJ, so we need to double-check our calculations. We made an error in the calculation of the net enthalpy change: D H _{net rxn} = D H ₁ + D H ₂ (reverse) D H _{net rxn} = -283.0 kJ + (-180.6 kJ) D H _{net rxn} = -463.6 kJ This means that the correct enthalpy change for the given reaction is: D H _rxn = -(-463.6 kJ) = 463.6 kJ The only answer choice that is close to this value is C: D H _rxn = -373.3 kJ.

Given the Following Two Equations and Their Corresponding Enthalpies