If the bond dissociation energies of XY,X_2,a | vedclass

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(B) The formation reaction of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \rightarrow XY(g)$; $\Delta_f H = -200 \ kJ \ mol^{-1}$.Let the bond dis

Vedclass · Accepted Answer

$800$

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(B) The formation reaction of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) ightarrow XY(g)$; $\Delta_f H = -200 \ kJ \ mol^{-1}$.Let the bond dissociation energies be $E(XY) = a$,$E(X_2) = a$,and $E(Y_2) = 0.5a$.The enthalpy of reaction is given by: $\Delta_f H = \sum 	ext{Bond Energy of Reactants} - \sum 	ext{Bond Energy of Products}$.$\Delta_f H = [\frac{1}{2}E(X_2) + \frac{1}{2}E(Y_2)] - E(XY)$.Substituting the values: $-200 = [\frac{1}{2}(a) + \frac{1}{2}(0.5a)] - a$.$-200 = 0.5a + 0.25a - a$.$-200 = -0.25a$.$a = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.

If the bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $\Delta_f H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$,the bond dissociation energy of $X_2$ will be in $kJ \ mol^{-1}$:

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