First slide

Enthalpy of neutralization

Question

Heat of neutralisation for the reaction NaOH+HCl → NaCl+H₂O is -57.1 K.J. mole^-1. The heat released when 0.25 moles of NaOH is treated with 0.25 moles of HCl is

Easy

Solution

For HCl & NaOH ; 1 mole = 1 eqivalent
given , No.of eqivalents = 1/4
1 eqivalent = 57.1
1/4 eqivalents---?
$=\frac{57.1}{4}= 14.3\;KJ/mole$

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