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 .
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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
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