Chemistry Thermochemistry MDCAT 2017
PMDC Verified Question 67 of 81
Determinate the value of Enthalpy of formation of \( NH_4Cl \):
A
\( -314.55 \text{ kJmol}^{-1} \)
B
\( -788 \text{ kJmol}^{-1} \)
C
None of these
D
\( -692 \text{ kJmol}^{-1} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( -314.55 \text{ kJmol}^{-1} \)
Concept:

The enthalpy of formation (\( \Delta H_f^\circ \)) is a specific thermochemical constant for a given compound, indicating the energy change when 1 mole of it forms from its elemental constituents.

Formula:

$$ \frac{1}{2} N_{2(g)} + 2H_{2(g)} + \frac{1}{2} Cl_{2(g)} \rightarrow NH_4Cl_{(s)} $$

Solution:

  • The formation of solid ammonium chloride (\( NH_4Cl \)) from gaseous nitrogen, hydrogen, and chlorine is highly exothermic.


  • The historically accepted and experimentally determined value for the standard enthalpy of formation of \( NH_4Cl \) is exactly \( -314.5 \text{ kJ/mol} \).


Why other options are incorrect:

\( -788 \text{ kJ/mol} \) is roughly the lattice energy of \( NaCl \). \( -692 \text{ kJ/mol} \) is the enthalpy of formation for \( MgO \). These are incorrect values for ammonium chloride.

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