Bond enthalpies of A_{2},B_{2} and AB are in | vedclass

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(C) Given,$(BE)_{A_{2}} : (BE)_{B_{2}} : (BE)_{AB} = 2 : 1 : 2$,where $BE$ is bond enthalpy.Let $(BE)_{A_{2}} = 2x$,$(BE)_{B_{2}} = x$,and $(BE)_{AB}

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$200 \ kJ \ mol^{-1}$

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(C) Given,$(BE)_{A_{2}} : (BE)_{B_{2}} : (BE)_{AB} = 2 : 1 : 2$,where $BE$ is bond enthalpy.Let $(BE)_{A_{2}} = 2x$,$(BE)_{B_{2}} = x$,and $(BE)_{AB} = 2x$.The reaction for the formation of $AB$ is: $\frac{1}{2} A_{2}(g) + \frac{1}{2} B_{2}(g) ightarrow AB(g)$.The enthalpy of formation is given by: $\Delta H^{\circ}_{f} = \Sigma (BE)_{	ext{reactants}} - \Sigma (BE)_{	ext{products}}$.Substituting the values: $-100 = [\frac{1}{2}(2x) + \frac{1}{2}(x)] - 2x$.$-100 = [x + 0.5x] - 2x$.$-100 = 1.5x - 2x$.$-100 = -0.5x$.$x = \frac{100}{0.5} = 200 \ kJ \ mol^{-1}$.Thus,the bond enthalpy of $B_{2}$ is $200 \ kJ \ mol^{-1}$.

Bond enthalpies of $A_{2}$,$B_{2}$ and $AB$ are in the ratio $2:1:2$. If the enthalpy of formation of $AB$ is $-100 \ kJ \ mol^{-1}$,the bond enthalpy of $B_{2}$ is:

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