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Bond enthalpies of H2, X2 and HX are in the ratio 2 : 1 : 2. If enthalpy of formation of HX is –50 kJ mol–1, the bond enthalpy of X2 is
  • a)
    100 kJ mol–1
  • b)
    300 kJ mol–1
  • c)
    200 kJ mol–1
  • d)
    400 kJ mol–1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Bond enthalpies of H2, X2 and HX are in the ratio 2 : 1 : 2. If enthal...

Let the bond enthalpy of X – X  bond be x.
ΔHf (HX) = – 50


∴ x = 50 × 2 = 100 kJ mol–1
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Most Upvoted Answer
Bond enthalpies of H2, X2 and HX are in the ratio 2 : 1 : 2. If enthal...
Let's assume the bond enthalpy of H2 is 2x, the bond enthalpy of X2 is x, and the bond enthalpy of HX is 2x.

The enthalpy of formation of HX can be calculated using the bond enthalpy equation:
ΔHf = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products)

Since HX is the only product, the equation becomes:
ΔHf = Σ(bond enthalpies of reactants) - 2x

The bond enthalpies of reactants are H2 and X2, which are 2x and x respectively.

ΔHf = 2x + x - 2x
ΔHf = x

Therefore, the enthalpy of formation of HX is equal to x.
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Bond enthalpies of H2, X2 and HX are in the ratio 2 : 1 : 2. If enthalpy of formation of HX is –50 kJ mol–1, the bond enthalpy of X2 isa)100 kJ mol–1b)300 kJ mol–1c)200 kJ mol–1d)400 kJ mol–1Correct answer is option 'A'. Can you explain this answer?
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