Enthalpy change for the reaction,4H_{(g)} rig | vedclass

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(A) The given reaction is $4H_{(g)} \rightarrow 2H_{2_{(g)}}$ with $\Delta H = -869.6 \ kJ$.This reaction represents the formation of $2$ moles of $H_

Vedclass · Accepted Answer

$434.8$

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(A) The given reaction is $4H_{(g)} \rightarrow 2H_{2_{(g)}}$ with $\Delta H = -869.6 \ kJ$.This reaction represents the formation of $2$ moles of $H_2$ molecules from $4$ moles of $H$ atoms.The dissociation energy of a bond is defined as the energy required to break $1$ mole of bonds in the gaseous state.First,reverse the reaction to represent the dissociation of $2$ moles of $H_2$ molecules:$2H_{2_{(g)}} \rightarrow 4H_{(g)}$; $\Delta H = +869.6 \ kJ$.Now,for $1$ mole of $H_2$ dissociation:$H_{2_{(g)}} \rightarrow 2H_{(g)}$; $\Delta H = \frac{869.6}{2} = 434.8 \ kJ$.Thus,the dissociation energy of the $H-H$ bond is $434.8 \ kJ$.

Enthalpy change for the reaction,$4H_{(g)} \rightarrow 2H_{2_{(g)}}$ is $-869.6 \ kJ$. The dissociation energy of $H-H$ bond is $............ \ kJ$.

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