The bond enthalpies of H_2,X_2,and HX are in | vedclass

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(A) Let the bond enthalpies of $H_2$,$X_2$,and $HX$ be $2x$,$x$,and $2x$ respectively.The chemical equation for the formation of $HX$ is: $\frac{1}{2}

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) Let the bond enthalpies of $H_2$,$X_2$,and $HX$ be $2x$,$x$,and $2x$ respectively.The chemical equation for the formation of $HX$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} X_2(g) ightarrow HX(g)$.The enthalpy of reaction is given by: $\Delta H_f = \Sigma B.E.(	ext{reactants}) - \Sigma B.E.(	ext{products})$.Substituting the values: $-50 = [\frac{1}{2} 	imes B.E.(H_2) + \frac{1}{2} 	imes B.E.(X_2)] - [B.E.(HX)]$.$-50 = [\frac{1}{2} 	imes 2x + \frac{1}{2} 	imes x] - [2x]$.$-50 = x + 0.5x - 2x = -0.5x$.$x = 100 \ kJ \ mol^{-1}$.Therefore,the bond enthalpy of $H_2$ is $2x = 2 	imes 100 = 200 \ kJ \ mol^{-1}$.

The bond enthalpies of $H_2$,$X_2$,and $HX$ are in the ratio of $2 : 1 : 2$. If the standard enthalpy of formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is .... $kJ \ mol^{-1}$.

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