If the bond dissociation energies of XY, X_2 and Y_2 are in the ratio 1 : 1 : 0.5 and _f H for XY is -200 kJ/mol , the bond dissociation energy of X_2 - Chemistry MCQ AIEEE 2026 | ExamTest.live

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If the bond dissociation energies of XY, X_{2} and Y_{2} are in the ratio 1 : 1 : 0.5 and Δ_{f} H for XY is - 200 kJ/mol, the bond dissociation energy of X_{2} will be:

Solution & Step-by-step Explanation

Let bond energies be x, x, 0.5 x .Reaction: \frac{1}{2} X_{2} + \frac{1}{2} Y_{2} \to XY Δ_{f} H = \frac{1}{2} BE (X_{2}) + \frac{1}{2} BE (Y_{2}) - BE (XY) - 200 = 0.5 x + 0.25 x - x - 200 = - 0.25 x ⟹ x = 800 .Thus, BE (X_{2}) = 800 kJ/mol .(Note: Source provided options may vary, but 800 is the calculated value).

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If the bond dissociation energies of XY, X_{2} and Y_{2} are in the ratio 1 : 1 : 0.5 and Δ_{f} H for XY is - 200 kJ/mol, the bond dissociation energy of X_{2} will be:

100 kJ/mol

200 kJ/mol

300 kJ/mol

800 kJ/mol

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