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alexira [117]
3 years ago
7

LAST QUESTION YALLL PLZZZ HELP

Chemistry
2 answers:
Law Incorporation [45]3 years ago
7 0

I've never done a reaction that involved just bond enthalpies before, so this is just a guess.

It's still products - reactants. The problem is the C=O bond. Do you count it once or twice? I'm going to choose once, but don't be surprised if it is incorrect.

Sum of the products - Sum of the reactants.

ΔH = 799 - 494

ΔH = 305

And that would be your answer to three places.

Anit [1.1K]3 years ago
7 0

Answer:

-1104 kJ

Explanation:

            C + O=O ⟶ O=C=O

Bonds:        O=O        2C=O

<em>D</em>/kJ·mol⁻¹:  494           799

The formula relating Δ<em>H</em>rxn and bond dissociation energies (<em>D</em>) is

Δ<em>H</em>rxn = Σ(<em>D</em>reactants) – Σ(<em>D</em>products)

(Note: This is an <em>exception</em> to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

<em>For the reactants</em>:

Σ(<em>D</em>reactants) = 1 × 494 = 494 kJ

<em>For the products</em>:

Σ(<em>D</em>products) = 2 × 799 = 1598 kJ

<em>For the reaction :</em>

Δ<em>H</em>rxn = 494 – 1598 = -1104 kJ

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