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Gaurav Rao Deo
Asked: 2022-11-01T21:19:30+05:30 In:

Calculate heat of dissociation for acetic acid from the following data:
`CH_(3)COOH +NaOH rarr CH_(3)COONa +H_(2)O DeltaH = – 13.2 kcal`
`H^(o+) +overset(Theta)Ohrarr H_(2)O, DeltaH =- 13.7 kcal`
Also calculate heat of dissociation for `NH_(4)OH` if
`HCI +NH_(4)OH rarr NH_(4)CI +H_(2)O, DeltaH =- 12.27 kcal`

Calculate heat of dissociation for acetic acid from the following data:
`CH_(3)COOH +NaOH rarr CH_(3)COONa +H_(2)O DeltaH = – 13.2 kcal`
`H^(o+) +overset(Theta)Ohrarr H_(2)O, DeltaH =- 13.7 kcal`
Also calculate heat of dissociation for `NH_(4)OH` if
`HCI +NH_(4)OH rarr NH_(4)CI +H_(2)O, DeltaH =- 12.27 kcal`
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  1. `CH_(3)COOH(w_(A)) +overset(Theta)OH (s_(B)) rarr CH_(3)COO^(Theta) = H_(2)O, DeltaH_(1) =- 13.2 kcal`
    `H^(o+)(S_(A)) +overset(Theta)OH(S_(B)) rarr H_(2)O DeltaH_(2) = 13.7 kcal`
    `DeltaH = DeltaH_(1) – DeltaH_(2)`
    `=- 13.2 – (-23.7) = 0.5 kcal`
    Similarly, for `NH_(4)OH:`
    `Na_(4)OH rarr NH_(4)^(o+) +overset(Theta)OH, DeltaH = ?`
    `H^(o+)(S_(A)) +NH_(4) OH (W_(B)) rarr NH_(4)^(o+) +H_(2)O, DeltaH_(1) = 12.27`
    `H^(o+)+overset(Theta)OH rarr H_(2)O`
    `{:ulbar(NH_(4)OH rarr NH_(4)^(o+)+overset(Theta)OH):}`
    `DeltaH = DeltaH_(1) – DeltaH_(2) = – 12.7 -(-13.7) = 1.43 cal`

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