# Calculating the heat of reaction between sulfuric acid and sodium hydroxide

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.

• Hydrogen sulfate is strong enough, mostly dissociated in diluted solutions. You could only ignore it, if it was very weak. Commented May 14 at 16:22
• You were too careful and did not think! there were two moles of H+ and OH-. Commented yesterday
• @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. Commented yesterday
• 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. Commented 3 hours ago

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:

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

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.

• titration usually involves a small change in total volume mitigating, but not eliminating, this effect. Commented yesterday