The bond dissociation energies of X_2, Y_2 and XY are in the ratio of 1 : 0.5 : 1 . H for the formation of XY is -200 kJ mol^-1 . The bond dissociatio - Chemistry MCQ NEET (UG) 2018 2026 | ExamTest.live

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The bond dissociation energies of X_{2}, Y_{2} and X Y are in the ratio of 1 : 0.5 : 1 . Δ H for the formation of X Y is - 200 kJ mol^{- 1} . The bond dissociation energy of X_{2} will be:

Solution & Step-by-step Explanation

Formation reaction: \frac{1}{2} X_{2} + \frac{1}{2} Y_{2} \to X Y Let Bond Energy (B.E.) of X_{2} = x .Then B.E. of Y_{2} = 0.5 x and B.E. of X Y = x . Δ H_{f} = \sum B.E. (Reactants) - \sum B.E. (Products) - 200 = [\frac{1}{2} (x) + \frac{1}{2} (0.5 x)] - [x] - 200 = 0.5 x + 0.25 x - x - 200 = - 0.25 x x = \frac{200}{0.25} = 800 kJ mol^{- 1} .

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The bond dissociation energies of X_{2}, Y_{2} and X Y are in the ratio of 1 : 0.5 : 1 . Δ H for the formation of X Y is - 200 kJ mol^{- 1} . The bond dissociation energy of X_{2} will be:

800 kJ mol^{- 1}

100 kJ mol^{- 1}

200 kJ mol^{- 1}

400 kJ mol^{- 1}

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