Chemistry Thermochemistry MDCAT 2011
PMDC Verified Question 78 of 81
In standard enthalpy of atomization heat of surrounding:
A
Increases
B
Increases then decreases
C
Decreases
D
Remains same
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Decreases
Concept:

The standard enthalpy of atomization (\( \Delta H_{at}^\circ \)) is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

Formula:

$$ \text{Element}_{\text{(standard state)}} \rightarrow \text{Gaseous Atoms} \quad (\Delta H = +\text{ve}) $$

Solution:

  • To turn a solid (like Na) or a diatomic gas (like \( Cl_2 \)) into separated gaseous atoms, chemical bonds or strong intermolecular forces must be broken.


  • Bond breaking strictly requires an input of energy, making atomization an endothermic process (\( \Delta H = +\text{ve} \)).


  • In an endothermic process, the reacting system actively absorbs heat from the immediate surroundings, which causes the heat of the surroundings to decrease.


Why other options are incorrect:

The heat of the surroundings would only increase if the reaction were exothermic (releasing heat). It does not remain the same because energy transfer is required to break the bonds.

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