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Simple Explanation:
Hess's Law states that the enthalpy change (ΔH) for an overall reaction equals the sum of the ΔH of its elementary steps.
ΔHrxn = ΔH₁ + ΔH₂ + ΔH₃ + ... + ΔHₙ
Examples using Hess's Law
Check out the tutorial for reaction mechanisms by clicking HERE.
When given elementary steps that don't add up to the overall reaction, you can adjust these steps and their corresponding ΔH values to determine the overall reaction's ΔH value.
Step 1: Ensure All Reactants and Products Are on the Correct Side
If any reactant in the overall reaction appears as a product in an elementary step (or vice versa), flip that equation so it matches.
When flipping an equation, you must also reverse the sign of its ΔH because the direction of energy flow has changed.
Step 2: Adjust Coefficients to Match the Overall Reaction
If the coefficients in an elementary step do not match those in the overall equation, multiply the entire equation by a factor to align.
If you multiply an equation by a factor, you must also multiply its ΔH by that same factor.
Step 3: Cancel Out Intermediates and Catalysts
After making the adjustments, cancel out any substances that appear on both sides of the reaction and then add up what is left. Intermediates and catalysts don't end up in the overall equation.
Make sure the summed-up equation matches the overall equation in order to accurately calculate the ΔH values.
Step 4: Add Up the ΔH Values
Use the Hess's Law formula. The sum (ΔHrxn) represents the enthalpy change for the overall reaction.
Example #1: 2NO(g) + O₂(g) → N₂O₄(g) (ΔHf = ?)
Step 1: N₂O₄(g) → 2NO₂(g) (ΔH₁ = 57.9 KJ)
Step 2: 2NO(g) + O₂(g) → 2NO₂(g) (ΔH₂ = -113.1 KJ)
Flip the products and reactants of N₂O₄(g) → 2NO₂(g) into 2NO₂(g) → N₂O₄(g), because N₂O₄ is a product of the overall reaction. ΔH₁ becomes -57.9 KJ.
Judging by the changes made above, 2NO₂ is cancelled out because it is a catalyst of the reaction.
Add up the new enthalpy values: -57.9 + (-113.1) = -171 KJ
Therefore, ΔHf = -171 KJ
Example #2: H₂(g) + ½O₂(g) → H₂O(l) (ΔHf = ?)
Step 1: 2Hg) → H₂(g) (ΔH₁ = -436 KJ)
Step 2: 2Og) → O₂(g) (ΔH₂ = -249 KJ)
Step 3: 2H(g) + O(g) → H₂O(g) (ΔH₃ = -803 KJ)
Step 4: H₂O(g) → H₂O(l) (ΔH₄ = -44 KJ)
Steps 2Hg) → H₂(g) and 2Og) → O₂(g) are supposed to be reversed, as H₂ and O₂ are reactants, not products. ΔH₁ becomes 436 KJ, ΔH₂ becomes 249 KJ.
What is now step O₂g) →2O(g) should be multiplied by ½ to match with the coefficient of O₂ in the overall equation. This step is now ½O₂g) →O(g), ΔH₂ becomes 124.5 KJ.
Judging by the changes made above, 2H, 2O, and H₂O(g) are cancelled out because those are intermediates of the reaction.
Add up the new enthalpy values: 436 + 124.5 + (-803) + (-44) = -286.5 KJ
Therefore, ΔHf = -286.5 KJ
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