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NEET CHEMISTRYThermodynamicsMedium

Question

If the bond energies of HH\text{H}-\text{H}, BrBr\text{Br}-\text{Br}, and HBr\text{H}-\text{Br} are 433433, 192192, and 364 kJ mol1364\text{ kJ mol}^{-1} respectively, the ΔH\Delta H^\circ for the reaction H2(g)+Br2(g)2HBr(g)\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g) will be:

A

+103 kJ+103\text{ kJ}

B

+261 kJ+261\text{ kJ}

C

103 kJ-103\text{ kJ}

D

261 kJ-261\text{ kJ}

Step-by-Step Solution

The standard enthalpy of reaction (ΔrH\Delta_r H^\circ) can be calculated from the bond energies of the reactants and products using the formula: ΔrH=Bond energies of reactantsBond energies of products\Delta_r H^\circ = \sum \text{Bond energies of reactants} - \sum \text{Bond energies of products} For the reaction: H2(g)+Br2(g)2HBr(g)\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g) ΔrH=[BE(HH)+BE(BrBr)][2×BE(HBr)]\Delta_r H^\circ = [\text{BE}(\text{H}-\text{H}) + \text{BE}(\text{Br}-\text{Br})] - [2 \times \text{BE}(\text{H}-\text{Br})] Given values: BE(HH)=433 kJ mol1\text{BE}(\text{H}-\text{H}) = 433\text{ kJ mol}^{-1} BE(BrBr)=192 kJ mol1\text{BE}(\text{Br}-\text{Br}) = 192\text{ kJ mol}^{-1} BE(HBr)=364 kJ mol1\text{BE}(\text{H}-\text{Br}) = 364\text{ kJ mol}^{-1} Substituting the values: ΔrH=[433+192][2×364]\Delta_r H^\circ = [433 + 192] - [2 \times 364] ΔrH=625728\Delta_r H^\circ = 625 - 728 ΔrH=103 kJ\Delta_r H^\circ = -103\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.

CHEMISTRYThermodynamicsenergiestexthtexthtextbrtextbrtexthtextbrrespectively

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