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NEET Hard

The bond dissociation energies of X2X_2, Y2Y_2 and XYXY are in the ratio of 1:0.5:11 : 0.5 : 1. ΔH\Delta H for the formation of XYXY is −200 kJ mol−1-200 \text{ kJ mol}^{-1}. The bond dissociation energy of X2X_2 will be

1

800 kJ mol−1^{-1}

2

100 kJ mol−1^{-1}

3

200 kJ mol−1^{-1}

4

400 kJ mol−1^{-1}

Step-by-Step Solution

The reaction for ΔfH∘(XY)\Delta_f H^\circ(XY) is 12X2(g)+12Y2(g)→XY(g)\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \rightarrow XY(g). Given bond energies of X2,Y2,XYX_2, Y_2, XY are X,0.5X,XX, 0.5X, X. ΔH=[12(X)+12(0.5X)]−X=−200\Delta H = [\frac{1}{2}(X) + \frac{1}{2}(0.5X)] - X = -200. Solving this, X2+X4−X=−200⇒−X4=−200⇒X=800 kJ/mol\frac{X}{2} + \frac{X}{4} - X = -200 \Rightarrow -\frac{X}{4} = -200 \Rightarrow X = 800 \text{ kJ/mol}.

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