Question

If 48.6 mL of 1.4 M NaOH react with excess HCl creating 3049.6 J of heat, what is the ΔHneutralization in kJ/mol?

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Hint: Answer includes 2 decimal places.

Answer 1

Consider reaction, NaOH (aq) + HCl (aq) $\rightarrow$ NaCl (aq) + H₂O (l)

From reaction, 1 mol NaOH $\equiv$ 1 mol HCl $\equiv$ 1 mol NaCl

In above reaction, HCl is excess reactant & NaOH is limiting reactant. Hence, amount of heat released will depend on the no. of moles of NaOH reacted with HCl. Therefore, to find out heat of neutralization per mole we need to calculate moles of NaOH.

We know that, [ NaOH ] = No. of moles of NaOH / volume of NaOH solution in L

$\therefore$ No. of moles of NaOH = [ NaOH ] $\times$ volume of NaOH solution in L

$\therefore$ No. of moles of NaOH = 1.4 mol / L $\times$ 0.0486 L = 0.06804 mol

0.06804 mol NaOH reacts with 0.06804 mol HCl & produces heat 3049.6 J.

For the reaction of 0.06804 mol NaOH, q = - 3049.6 J. Hence, we can use conversion factor ( -3049.6 J / 0.06804 mol NaOH) to calculate heat of neutralization of reaction.

Hence, H reaction = 1 mol NaOH ( - 3049.6 J / 0.06804 mol NaOH )  = - 44, 820.69 J = - 44.82 kJ

ANSWER: H neutralization = - 44.82 k J /mol