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Enthalpy of neutralisation of acetic aci...

Enthalpy of neutralisation of acetic acid by `NaOh` is `-50.6 kJ mol^(-1)`. Calculate `DeltaH` for ionisation of `CH_(3)COOH`. Given. The heat of neutralisation of a strong acid with a strong base is `-55.9 kJ mol^(-1)`.

Text Solution

Verified by Experts

The neutralisation of a strong acid by a strong base is represented by:
`H^(o+)(aq) +overset(Theta)OH rarr H_(2)O(l) DeltaH =- 55.9 kJ…..(i)`
We have to calculate:
`CH_(3)COOH rarr CH_(3)COO^(Theta) +H^(o+) DeltaH = ?`
Given:
`CH_(3)COOH + overset(Theta)OH rarr CH_(3)COO^(Theta) +H_(2)O, DeltaH_(2) =- 50.6 ...(ii)`
Substract equation (i) from equation (ii), we get
`DeltaH = DeltaH_(2) - DeltaH_(1)`
`=- 50.6 -(-55.9) = 5.3 kJ mol^(-1)`
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Enthalpy of neutralisation of acetic acid by NaOH is 50.6 kJ/mol. Calculate Delta H for ionisation of CH_(3)COOH . Given, the heat of neutralisation of a strong acid with a strong base is -57.3 kJ/mol.

the heat of neutralisation of acetic acid and sodium hydroxide is -50.6 kJ eq^(-1) . Find the heat of dissociation of CH_(3)COOH if the heat of neutralisation of a strong acid and a strong base is - 55.9 Kj eq^(-1) .

Knowledge Check

  • The heat of neutralisation of a strong acid by a strong base is equal to DeltaH of :

    A
    `H^(+)+OH^(-)=H_(2)O`
    B
    `H_(2)O+H^(+)=H_(3)O^(+)`
    C
    `2H_(2)+O_(2)=2H_(2)O`
    D
    `CH_(3)COOH+NaOH=CH_(3)COONa+H_(2)O`

  • Heat of neutralisation of a strong acid by a strong base is equal to DeltaH of

    A
    `H^(+)+OH^(-)=H_(2)O`
    B
    `H_(2)O+H^(+)=H_(3)O^(+)`
    C
    `2H_(2)O_(2)=2H_(2)O`
    D
    `CH_(3)COOH+NaOH=CH_(3)COONa+H_(2)O`

  • The heat of neutralisation for strong acid and strong base forming 2 moles of water is

    A
    `-2 xx57.1 kJ`
    B
    `-57.1 kJ`
    C
    `-57.1/2 kJ`
    D
    Strong acid and strong base will not undergo neutralisation

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