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Heat of neutralisation between HCI and N...

Heat of neutralisation between `HCI` and `NaOH` is `13.7kcal` and between `HCN` and `NaOH` is `3`kcal at `45^(@)C`. Calculate the heat of ionisation of `HCN`

Text Solution

Verified by Experts

`{:(HCI+,NaOHrarr,NaCI+,H_(2)O),(SA,SB,,):}`
`H^(o+) +overset(Theta)OH rarr H_(2)O, DeltaH_(1) =- 13.7 Kcal`
`HCN(W_(a)) +overset(Theta)OH(S_(B))rarrH_(2)O+CN^(Theta) , DeltaH_(2) = - 3kcal`
`HCN rarr H^(o+) +CN^(Theta)`,
`:. DeltaH = DeltaH_(2) - DeltaH_(1)`
`=- 3 -(-13.7) = 0.7 kcal`
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Heat of neutralization between HCl and NaOH is -13.7 k.cal . If heat of neutralization between CH_(3)COOH and NaOH is -11.7 k.cal . Calculate heat of ionization of CH_(3)COOH .

heat of neutralisation of HCl against NaOH is 13.7 kcal eq^(-1) what will be the ionisation energy of CH_(3)COOH in kcal mol^(-1) if its heat of neutralisation is 11.7 kcal eq^(-1) .

Knowledge Check

  • Heat of neutralisation is less than 13.7 kcal/mol for the reaction

    A
    `HCl + NaOH rarr NaCl + H_(2)O`
    B
    `H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O`
    C
    `HNO_(3) + NaOH rarr NaNO_(3) + H_(2)O`
    D
    `CH_(3)COOH + NaOH rarr CH_(3)COONa + H_(2)O`

  • The heat released in neutralisation of HCI and NaOH is 13.7 kcal/mol, the heat released on neutralisation of NaOH with CH_(3)COOH is 3.7 kcal/mol. The Delta H^(@) of ionsiation of CH_(3)COOH is

    A
    10.2 kcal
    B
    10 kcal
    C
    3.7 kcal
    D
    9.5 kcal

  • Heat of neutralization of HCl by NaOH is 13.7 kcal per equivalent and by NH_4OH is 12.27 kcal. The heat of dissociation of NH_4OH is

    A
    `-25.97` kcal
    B
    25.97 kcal
    C
    `-1.43` kcal
    D
    1.43 kcal

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