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Calculate the enthalpies change for the ...

Calculate the enthalpies change for the reaction,
`H_(2)(g)+Cl_(2)(g)to2HCl(g)`
Given that, bond energies of H-H,Cl-Cl and H-Cl are 436, 243 and 432 kJ `mol^(-1)`

Text Solution

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`H_(2)(g)+Cl_(@)(g)rarr HCl(g) , Delta_(r )H=?`
Now analyses the given thermochemical equation in two parts :
(i) `{:("Bond Breaking"[Delta_(r )H_(1)],:,"Endothermic Reaction"),(H_(2)(g)rarr2H(g),,,Delta_(r )H=436 kJ),(Cl_(2)(g)rarr 2Cl(g),,,Delta_(r )H=243 kJ):}`
`rArr Delta_(r )H_(1)=436+243 = 679 kJ`
`{:("Bond Breaking"[Delta_(r )H_(2)],:,"Endothermic Reaction"),(2H(g)+2Cl(g)rarr2HCl(g),,,Delta_(r )H=-862 " kJ (Bond energy of HCl = 431 kJ/mole)"):}`
Now `Delta_(r )H = Delta_(r )H_(1)+Delta_(r )H_(2) " "` (using Hess's Law)
`rArr Delta_(r )H=+679-862=-183` kcal/mol
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