Chemistry Thermochemistry MDCAT 2016
PMDC Verified Question 71 of 81
\( Mg + \frac{1}{2} O_2 \rightarrow MgO_{(s)} \quad \Delta H = -692 \text{ kJmol}^{-1} \) at STP.
Enthalpy of the above reaction will be called:
A
\( \Delta H^\circ_{at} \)
B
\( \Delta H^\circ_n \)
C
\( \Delta H^\circ_f \)
D
\( \Delta H^\circ_{sol} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( \Delta H^\circ_f \)
Concept:

The standard enthalpy of formation (\( \Delta H_f^\circ \)) is the enthalpy change associated with the creation of exactly 1 mole of a compound directly from its constituent elements in their standard states at STP.

Formula:

$$ Mg_{(s)} + \frac{1}{2}O_{2(g)} \rightarrow MgO_{(s)} $$

Solution:

  • Magnesium (\( Mg \)) and Oxygen (\( O_2 \)) are in their standard elemental states.


  • They combine to form exactly 1 mole of the solid compound Magnesium Oxide (\( MgO \)).


  • Because it represents the formation of a compound from its raw elements, the heat change (\( -692 \text{ kJ/mol} \)) is defined as the standard enthalpy of formation, \( \Delta H_f^\circ \).


Why other options are incorrect:

It is not \( \Delta H_{at} \) (atomization) because a solid compound is formed. It is not \( \Delta H_n \) (neutralization) because there is no acid/base reaction producing water. It is not \( \Delta H_{sol} \) (solution) because no solute is dissolving in a solvent.

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