Chemistry Thermochemistry MDCAT 2016
PMDC Verified Question 70 of 81
\( \frac{1}{2} H_{2(g)} \rightarrow H_{(g)} \quad \Delta H = 218 \text{ kJmol}^{-1} \)
In this reaction \( \Delta H \) will be called:
A
Enthalpy of decomposition
B
Enthalpy of the dissociation
C
Enthalpy of atomization
D
Enthalpy of formation
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Enthalpy of atomization
Concept:

Thermochemical definitions are strictly based on the states of reactants and products, and the stoichiometry of the reaction.

Formula:

$$ \text{Element}_{(\text{standard state})} \rightarrow 1 \text{ Mole of Gaseous Atoms} $$

Solution:

  • The equation shows gaseous molecular hydrogen (\( H_2 \)), which is its standard state, being converted into purely gaseous atomic hydrogen (\( H \)).


  • Furthermore, exactly 1 mole of the atomic product is being formed.


  • This perfectly fits the definition of standard enthalpy of atomization (\( \Delta H_{at}^\circ \)).


Why other options are incorrect:

It is not formation because formation applies to compounds formed from elements. While it involves bond dissociation, 'Enthalpy of dissociation' typically refers to breaking 1 full mole of bonds (yielding 2 moles of atoms), whereas atomization specifically scales to yielding 1 mole of atoms.

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