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heat of neutralisation of HCl against Na...

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)`.

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Heat of neutralization between HCl and NaOH is -13.7 kcal "equiv"^(–1) . Heat of neutralization of H_2C_2O_4 (oxalic acid) with NaOH is -26 kcal mol^(-1) . Hence, heat of dissociation of H_(2)C_(2)O_(4) as H_2C_2O_(4) rarr 2H^(+) + C_(2)O_(4)^(2-) , is :

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 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

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

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 enthalpy of neutralisation of NH_(4)OH and CH_(3)COOH is -10.5 kcal mol^(-1) and enthalpy of neutralisation of CH_(3)COOH with strong base is -12.5 kcal mol^(-1) . The enthalpy of ionisation of NH_(4)OH will be

    A
    4.0 kcal `mol^(-1)`
    B
    3.0 kcal `mol^(-1)`
    C
    2.0 kcal `mol^(-1)`
    D
    3.2 kcal `mol^(-1)`

  • The enthalpy of neutralisation of NH_(4)OH and CH_(3)COOH is -10.5 kcal mol^(-1) and enthalpy of neutralisation of CH_(3)COOH with strong base is -12.5 kcal mol^(-1) . Enthalpy of ionistion of NH_(4)OH will

    A
    3.2 kcal `mol^-1`
    B
    2.0 kcal `mol^(-1)`
    C
    3.0 kcal `mol^(-1)`
    D
    4.0 kcal `mol^(-1)`