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For my lab report I have to calculate the theoretical heat of reaction $\Delta H_{\text{rxn}}$ between sulfuric acid $\ce{H2SO4}$ and sodium hydroxide $\ce{NaOH}$ using their heat of formations. The reaction I have is $$\ce{H2SO4 (aq) + 2NaOH (aq) -> Na2SO4 (aq) + 2H2O (l)}$$

Now because the sulfuric acid, sodium hydroxide, and sodium sulfate are all aqueous, I thought to use the enthalpy of formation $\Delta H^\circ_f$ for their ions, especially because I couldn't find any value of $\Delta H^\circ_f$ for aqueous sulfuric acid (could only find liquid).

The problem is that my instructor said I'd have to be a little careful about sulfuric acid because it would dissociate into $\ce{H+}$ and $\ce{HSO4-}$, which I'm guessing is because hydrogen sulfate is a weak acid itself. But when I write the ionic equation, I get $$\ce{H+ (aq) + HSO4- (aq) + 2Na+ (aq) + 2OH- (aq) -> 2Na+ (aq) + H2O (l) + HSO4- (aq) + OH- (aq)}$$

Which would translate to a net ionic equation of $$\ce{H+ (aq) + OH- (aq) -> H2O (l)}$$

Now when I calculate the heat of reaction from this, I get around $-55.9\dfrac{\text{kJ}}{\text{mol}}$. But my experimental data showed that $0.02$ moles of sulfuric acid and $0.04$ moles of sodium hydroxide produced $\approx2259$ Joules of energy, which translates to 113 kJ/mol.

I don't know where I'm going wrong with this.

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  • $\begingroup$ Hydrogen sulfate is strong enough, mostly dissociated in diluted solutions. You could only ignore it, if it was very weak. $\endgroup$
    – Mithoron
    Commented May 14 at 16:22
  • $\begingroup$ You were too careful and did not think! there were two moles of H+ and OH-. $\endgroup$
    – jimchmst
    Commented Jun 16 at 2:49
  • $\begingroup$ @jimchmst I understand that there would be two moles of H+ and OH-, but the point was that the hydrogen sulfate ion (HSO4-) wouldn't completely dissociate, and so one of the OH- would be a spectator ion and the H in hydrogen sulfate would not dissociate. The commenters helpfully pointed out that the weakness of HSO4- was not enough to disregard its contribution to the total change in enthalpy. $\endgroup$ Commented Jun 16 at 3:33
  • $\begingroup$ The reaction with OH- forces the dissociation. Since HSO4- is still relatively strong its heat of neutralization is almost the same as H3O+. A weaker acid will have a lower heat of neutralization. OH- is a spectator ion only after all acids are neutralized and it is added in excess. $\endgroup$
    – jimchmst
    Commented Jun 17 at 21:02

2 Answers 2

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Arbitrary representation Species $\Delta H_f^o\;$ [kJ/mol]
A $\ce{H2SO4(aq)}$ -909.27
B $\ce{NaOH(aq)}$ -469.15
C $\ce{Na2SO4(aq)}$ -1387.10
D $\ce{H2O(l)}$ -285.83

The reaction we have is:

$$\ce{A(aq) +2B(aq)->C(aq) +2D(l)}$$

The standard molar change in enthalpy for this reaction is given by:

$$\Delta\overline{H}^o=c\;\Delta\overline{H}_{f_C}^o\;+\;d\;\Delta\overline{H}_{f_D}^o\;-\;\left(a\;\Delta\overline{H}_{f_A}^o\;+\;b\;\Delta\overline{H}_{f_B}^o\right)$$

Substituting the values from the table:

$$\Delta \overline{H}^o=-1387.10+2(-285.83)-[-909.27-2(469.15)]$$

$$\pmb{\boxed{\Delta \overline{H}^o=\pu{-111.19kJ/mol}}}$$

Calculating the standard change in enthalpy with amounts of reactants used:

$$\Delta H^o=\frac{n_A\;\Delta \overline{H}^o}{a}=\frac{\pu{(0.02mol)(\pu{-111.19kJ/mol})}}{1}=\pu{-2.22kJ}=\pu{-2220J}$$

Aside from minor, expected error related to experimental measurements, you have obtained a reasonably accurate result with a small relative error:

$$E=\frac{\left|111.19-113\right|}{111.19}=0.0163=1.63\%$$

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The problem is difficult to handle, because of the extraordinary changes in the heats of dilution of $\ce{H2SO4}$ in water. The amount of heat produced by mixing $1.000$ mole $\ce{H2SO4}$ with $n$ mole water has been determined in the Journal of Chemical Education $81, 7$, July $2004$, p. $993$. The main results are summarized here :

$n$ = $1$ mol water : $28.08$ kJ ;

$n$ = $2$ mol water : $41.93$ kJ

$n$ = $4$ mol water : $54.07$ kJ

$n$ = $10$ mol water : $67.04$ kJ

$n$ = $20$ mol water : $72.69$ kJ

$n$ = $50$ mol water : $73.36$ kJ - Gives a solution $1.00$ M.

$n$ = $500$ mol water : $76.75$ kJ - Gives a solution $0.100$ M

$n$ = $2000$ mol water : $80.90$ kJ

$n$ = $10 000$ mol water : $87.09$ kJ

$n$ = $100 000$ mol water ; $93.66$ kJ

$n$ = ∞ mol water : $96.26$ kJ

The heat of dilution is a continuous function of the amount of added water. So it is difficult to use these values to calculate precise heats of neutralization of any acidic solution with any $\ce{NaOH}$ solutions.

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  • $\begingroup$ titration usually involves a small change in total volume mitigating, but not eliminating, this effect. $\endgroup$
    – jimchmst
    Commented Jun 16 at 2:59

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