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NEET CHEMISTRYThermodynamicsEasy

Question

Given the following reaction: 4H(g)2H2(g)4\text{H(g)} \rightarrow 2\text{H}_2\text{(g)}. The enthalpy change for the reaction is 869.6 kJ-869.6 \text{ kJ}. The dissociation energy of the H-H bond is:

A

869.6 kJ-869.6 \text{ kJ}

B

+434.8 kJ+434.8 \text{ kJ}

C

+217.4 kJ+217.4 \text{ kJ}

D

434.8 kJ-434.8 \text{ kJ}

Step-by-Step Solution

The given reaction is the formation of 22 moles of H2\text{H}_2 gas from 44 moles of gaseous H\text{H} atoms: 4H(g)2H2(g);ΔH=869.6 kJ4\text{H(g)} \rightarrow 2\text{H}_2\text{(g)}; \Delta H = -869.6 \text{ kJ} This means the formation of 22 moles of H-H bonds releases 869.6 kJ869.6 \text{ kJ} of energy. The enthalpy change for the formation of 11 mole of H2\text{H}_2 gas from 22 moles of H\text{H} atoms is: 869.62=434.8 kJ\frac{-869.6}{2} = -434.8 \text{ kJ} 2H(g)H2(g);ΔH=434.8 kJ2\text{H(g)} \rightarrow \text{H}_2\text{(g)}; \Delta H = -434.8 \text{ kJ} The bond dissociation energy of the H-H bond is the energy required to break 11 mole of H-H bonds into gaseous H atoms. This is the reverse of the formation reaction: H2(g)2H(g)\text{H}_2\text{(g)} \rightarrow 2\text{H(g)} Therefore, the bond dissociation energy is +434.8 kJ+434.8 \text{ kJ}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsfollowingreactiontexthgrightarrowtexthtextg

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