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MHT CET 202626 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
26 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

MHT CET 2026 · 26 April · Morning Shift · Chemistry · Thermodynamics (C)

What is the enthalpy of formation of HF if bond enthalpies of H-H,...

What is the enthalpy of formation of HF if bond enthalpies of H-H, F-F and H-F bonds are 434434 kJ/mol, 158158 kJ/mol and 565565 kJ/mol respectively. ?

Options

  1. A−267-267 kJ/mol
  2. B−269-269 kJ/mol
  3. C−271-271 kJ/mol
  4. D−276-276 kJ/mol

Correct answer

B. −269-269 kJ/mol

Step-by-step solution

The reaction for the formation of HFHF is:

12H2(g)+12F2(g)→HF(g)\dfrac{1}{2}H_2(g) + \dfrac{1}{2}F_2(g) \rightarrow HF(g)

The enthalpy of formation is given by:

ΔfH=∑Bond enthalpies of reactants−∑Bond enthalpies of products\Delta_f H = \sum \text{Bond enthalpies of reactants} - \sum \text{Bond enthalpies of products}

ΔfH=[12×BE(H−H)+12×BE(F−F)]−BE(H−F)\Delta_f H = \left[ \dfrac{1}{2} \times BE(H-H) + \dfrac{1}{2} \times BE(F-F) \right] - BE(H-F)

Substituting the given values:

ΔfH=[12(434)+12(158)]−565\Delta_f H = \left[ \dfrac{1}{2}(434) + \dfrac{1}{2}(158) \right] - 565

ΔfH=(217+79)−565\Delta_f H = (217 + 79) - 565

ΔfH=296−565=−269 kJ/mol\Delta_f H = 296 - 565 = -269 \text{ kJ/mol}

Answer: −269-269 kJ/mol
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