Calculating deltaH°f for O(g) - 142.0 kJ/mol

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To calculate the standard enthalpy of formation (deltaH°f) for atomic oxygen (O(g)), one should consider the bond energy of O2(g), which is 142.0 kJ/mol. Since the formation of one mole of O(g) from O2(g) involves breaking the O=O bond, the deltaH°f for O(g) can be determined by halving the bond energy of O2. This approach is based on the principle that breaking a bond requires energy input, thus reflecting the enthalpy change. The calculation yields a deltaH°f of 71.0 kJ/mol for O(g). This method effectively addresses the problem of finding the enthalpy for the formation of a single oxygen atom.
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Homework Statement



The bond energy of O2(g) is 142.0 kJ/mol. Calculate deltaH°f for O(g).

Homework Equations





The Attempt at a Solution



no clue
 
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Hi Nick,
You have an Oxygen molecule and it asks you for the delta Hf of Oxygen atom ONLY. So, i would suggest you to halve the bond energy of O2.

Any other suggestions or am I right?
 
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Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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