Question

100 ml of 1 M HCl is mixed with 200ml of 0.5 M KOH. Enthalpy change in the process is (in kJ)
a. 5.72
b. -57.2
c. -5.72
d. 11..45

Answer 1

The net equation representing the neutralization of a strong acid and a strong base,

$H^{+} (aq) + OH^{-}(aq) --> H_{2}O (l)$

Standard heat of neutralization for the titration of a strong acid and a strong base is -57.2 kJ/mol

Calculating the moles of acid :

$1 \frac{mol}{L} HCl * 100 mL * \frac{1 L}{1000 mL} = 0.1 mol$

Calculating the moles of base:

$0.5\frac{mol}{L} * 200 mL * \frac{1 L}{1000 mL} = 0.1 mol$

The enthalpy change in the above neutralization reaction will be,

$0.1 mol * \frac{-57.1 kJ}{1 mol} = -5.71 kJ$

Therefore the correct answer is c. -5.72 kJ