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(B) Let the bond dissociation energies be $E_{XY} = a$,$E_{X_2} = a$,and $E_{Y_2} = 0.5a$.The reactions are:$1$) $XY(g) 	o X(g) + Y(g)$; $\Delta H_1

Vedclass · Accepted Answer

$800$

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(B) Let the bond dissociation energies be $E_{XY} = a$,$E_{X_2} = a$,and $E_{Y_2} = 0.5a$.The reactions are:$1$) $XY(g) 	o X(g) + Y(g)$; $\Delta H_1 = a$$2$) $X_2(g) 	o 2X(g)$; $\Delta H_2 = a$$3$) $Y_2(g) 	o 2Y(g)$; $\Delta H_3 = 0.5a$The formation reaction of $XY$ is: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) 	o XY(g)$.Using Hess's Law: $\Delta_f H = [\frac{1}{2} \Delta H_2 + \frac{1}{2} \Delta H_3] - \Delta H_1$$-200 = [\frac{1}{2}a + \frac{1}{2}(0.5a)] - a$$-200 = 0.5a + 0.25a - a$$-200 = -0.25a$$a = \frac{200}{0.25} = 800 \ kJ \ mol^{-1}$.

If the ratio of bond dissociation energies of $XY$,$X_2$,and $Y_2$ is $1 : 1 : 0.5$ and the enthalpy of formation of $XY$ is $\Delta_f H = -200 \ kJ \ mol^{-1}$,then the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$ is:

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