Chemistry Thermochemistry MDCAT 2014
PMDC Verified Question 73 of 81
\( 2H_2 + O_2 \rightarrow 2H_2O \quad \Delta H = 205.5 \text{ kJmol}^{-1} \). What will be the enthalpy change in the above reaction?
A
\( -205.5 \text{ kJ mol}^{-1} \)
B
\( 205.5 \text{ kJ mol}^{-1} \)
C
\( 1 \text{ kJ mol}^{-1} \)
D
Zero \( \text{kJ mol}^{-1} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( -205.5 \text{ kJ mol}^{-1} \)
Concept:

Thermochemical equations must reflect the correct sign convention indicating whether heat is released or absorbed. The formation of water from hydrogen and oxygen gases is an exothermic process.

Formula:

$$ 2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)} $$

Solution:

  • The question text provides a magnitude of \( 205.5 \text{ kJ/mol} \) without a sign.


  • Because the combustion of hydrogen to form water releases energy, the enthalpy change (\( \Delta H \)) must mathematically carry a negative sign to denote an exothermic process.


  • Therefore, the correct representation of the enthalpy change for the reaction as written is \( -205.5 \text{ kJ mol}^{-1} \). (Note: While the universally accepted true value for \( \Delta H \) of this reaction is \( -571.6 \text{ kJ} \) or \( -285.8 \text{ kJ/mol} \) for formation, we must adapt to the magnitude given in the past paper).


Why other options are incorrect:

A positive value (205.5) would falsely imply the formation of water is endothermic. Zero and 1 are mathematically arbitrary distractors.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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