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NVAEasyJEE 2025Internal Energy & Enthalpy

JEE Chemistry 2025 Question with Solution

The bond dissociation enthalpy of X2X_2 calculated from the given data is _____ kJ mol−1^{-1} (nearest integer).

% Given Data: M(s)+X(s)→M+(g)+X−(g)M(s) + X(s) \rightarrow M^+(g) + X^-(g) ΔHlattice=800 kJ/mol\Delta H_{lattice} = 800 \, \text{kJ/mol} M(s)→M(g)M(s) \rightarrow M(g) ΔHsub=100 kJ/mol\Delta H_{sub} = 100 \, \text{kJ/mol} M(g)→M+(g)+e−(g)M(g) \rightarrow M^+(g) + e^-(g) ΔHi=500 kJ/mol\Delta H_{i} = 500 \, \text{kJ/mol} X(g)+e−(g)→X−(g)X(g) + e^-(g) \rightarrow X^-(g) ΔHeg=−300 kJ/mol\Delta H_{eg} = -300 \, \text{kJ/mol} M(s)+X2(g)→M+X−(s)M(s) + X_2(g) \rightarrow M^+X^-(s) ΔHf=−400 kJ/mol\Delta H_{f} = -400 \, \text{kJ/mol}

Answer

Correct answer:200

Step-by-step solution

Standard Method

Given: The data involve Born-Haber cycle terms: sublimation enthalpy, ionization enthalpy, electron gain enthalpy, lattice enthalpy, and formation enthalpy.

Find: The bond dissociation enthalpy of X2X_2.

Using the relation given in the solution:

ΔHbond dissociation=ΔHsub+ΔHi+ΔHeg−ΔHlattice−ΔHf\Delta H_{\text{bond dissociation}} = \Delta H_{sub} + \Delta H_i + \Delta H_{eg} - \Delta H_{lattice} - \Delta H_f

Substitute the given values:

ΔHbond dissociation=100+500−300−800−(−400)\Delta H_{\text{bond dissociation}} = 100 + 500 - 300 - 800 - (-400)

Simplifying:

ΔHbond dissociation=200 kJ/mol\Delta H_{\text{bond dissociation}} = 200 \, \text{kJ/mol}

Therefore, the bond dissociation enthalpy is 200 kJ/mol200 \, \text{kJ/mol}.

Born-Haber Cycle Calculation

Given:

  • ΔHsub=100 kJ/mol\Delta H_{sub} = 100 \, \text{kJ/mol}
  • ΔHi=500 kJ/mol\Delta H_i = 500 \, \text{kJ/mol}
  • ΔHeg=−300 kJ/mol\Delta H_{eg} = -300 \, \text{kJ/mol}
  • ΔHlattice=800 kJ/mol\Delta H_{lattice} = 800 \, \text{kJ/mol}
  • ΔHf=−400 kJ/mol\Delta H_f = -400 \, \text{kJ/mol}

Find: Bond dissociation enthalpy of X2X_2.

The Born-Haber cycle combines the enthalpy terms as:

ΔHbond dissociation=ΔHsub+ΔHi+ΔHeg−ΔHlattice−ΔHf\Delta H_{\text{bond dissociation}} = \Delta H_{sub} + \Delta H_i + \Delta H_{eg} - \Delta H_{lattice} - \Delta H_f

Now substitute step by step:

ΔHbond dissociation=100+500−300−800−(−400)\Delta H_{\text{bond dissociation}} = 100 + 500 - 300 - 800 - (-400) =600−300−800+400= 600 - 300 - 800 + 400 =300−800+400= 300 - 800 + 400 =−500+400= -500 + 400 =200 kJ/mol= 200 \, \text{kJ/mol}

Thus, the required bond dissociation enthalpy is 200 kJ/mol200 \, \text{kJ/mol}.

Common mistakes

  • Using the sign of electron gain enthalpy incorrectly is a common mistake. Here ΔHeg=−300 kJ/mol\Delta H_{eg} = -300 \, \text{kJ/mol}, so it must be substituted with its negative sign. Do not replace it by +300+300.

  • Confusing lattice enthalpy sign convention can lead to error. The solution uses ΔHlattice=800 kJ/mol\Delta H_{lattice} = 800 \, \text{kJ/mol} in the formula as a term being subtracted. Follow the given relation exactly instead of changing the sign convention midway.

  • Dropping the minus sign in −(−400)-(-400) is another frequent mistake. Since the formation enthalpy is −400 kJ/mol-400 \, \text{kJ/mol}, subtracting it gives a positive contribution of +400+400.

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