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For example, this reaction: $$\ce{2NaOH + H2SO4 -> Na2SO4 + H2O}$$

Suppose both concentrations are $1\ \mathrm{mol\ l^{-1}}$ (standard condition). There is $0.02\ \mathrm{mol}$ ($20\ \mathrm{cm^3}$) of $\ce{NaOH}$, and $0.01\ \mathrm{mol}$ ($10\ \mathrm{cm^3}$) of $\ce{H2SO4}$. So neither reactant is the limiting reactant.

Well, I know one way is to find the standard enthalpy of formation of all components and subtract the ones of products from reactants'. My question is if only experiment data is given.

For temperature change of $\Delta T$,
$\Delta H = mc\Delta T = (0.03)\times(4200)\times\Delta T$,
where $m$ is the mass of the solution. Since the density should be similar to water $1000\ \mathrm{kg\ m^{-3}}$,
$20\ \mathrm{cm^3} + 10\ \mathrm{cm^3} = 30\ \mathrm{cm^3} = 0.00003\ \mathrm{m^3}$
which translates to $0.03\ \mathrm{kg}$. $c$ is just the specific heat capacity of water, it was assumed that the specific heat capacity of the mixture is more or less the same as the water's.

Now the standard one $\Delta_\mathrm{r}H^\circ$ requires division by the amount of substance, but which one – $0.02\ \mathrm{mol}$ or $0.01\ \mathrm{mol}$? What if $\ce{NaOH}$ is $0.021\ \mathrm{mol}$? Or $\ce{H2SO4}$ is $0.011\ \mathrm{mol}$?

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  • $\begingroup$ Welcome to Chemistry.SE! Some parts of your question are a little bit puzzling to me. What do you mean by "0.02 mol (20$cm^{3}$)" - should this be a concentration? And what do you mean by "limiting reagent"? Where does "For temperature change of $\Delta T$, $\Delta H$ = $m C \Delta T$ = (0.03)*(4200)*$\Delta T$" come from, especially the values you inserted for $m$ and $C$? $\endgroup$ – Philipp Sep 10 '13 at 11:34
  • $\begingroup$ Well. Under standard condition the solution used should be 1M, right? $\endgroup$ – Ijriims Sep 10 '13 at 15:12
  • $\begingroup$ $C$ is just the specific heat capacity of water, I assumed the specific heat capacity of the mixture is more or less the same as the water's. $m$ is just the mass of the solution. Since the density should be similar to water $1000kgm^{-3}$, 20 +10 $cm^3$=30 $cm^3$ = 0.00003$m^3$ which translates to $0.03kg$ $\endgroup$ – Ijriims Sep 10 '13 at 15:20
  • $\begingroup$ Thanks for the clarification. Could you add those to your question? It is always nice if every symbol or variable is properly declared inside the question. This makes it easier to grasp where your problem lies. $\endgroup$ – Philipp Sep 10 '13 at 15:23
  • $\begingroup$ Done it. But aren't those just basic knowledge of high school chemistry? $\endgroup$ – Ijriims Sep 10 '13 at 15:30
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The difference between the Enthalpy of Reaction $\Delta_{\mathrm{r}} H$ and the Standard Enthalpy of Reaction $\Delta_{\mathrm{r}} H^{\ominus}$

The standard value is not related to a standard temperature (although Standard Enthalpies of Reaction are often given with respect to the standard temperature - they could have been given for any arbitrary temperature). The term "Standard" refers to standard pressure and the reactants and products being in their standard states. The standard states are usually those of the pure compounds. Furthermore gases are usually assumed to behave like ideal gases. This definition of standard states means that Standard Enthalpies of Reaction are often purely calculational quantities since the underlying reactions are practically not feasible. For example, the reactants usually have to be mixed in order to react with one another, meaning that they are not present in their pure form. Similarly, the products of a reaction are often formed as a mixture and not as pure substances. And real gases certainly don't show the same behavior as ideal gases. So, enthalpically the difference between the Enthalpy of Reaction and the Standard Enthalpy of Reaction is that you would have to add the Enthalpy of Mixing of the reactants to $\Delta_{\mathrm{r}} H$ and subtract the Enthalpy of Mixing of the products from it in order to get $\Delta_{\mathrm{r}} H^{\ominus}$. Furthermore you would have to account for the enthalpic difference between real and ideal gases if there are any gases involved. Furtunately, those differences are usually quite small compared to the often very big Enthalpies of Reaction so that they can be ignored without introducing a large error. But you have to be mindful of the difference between $\Delta_{\mathrm{r}} H^{\ominus}$ and $\Delta_{\mathrm{r}} H$: for example, if $\Delta_{\mathrm{r}} H^{\ominus}$ is small and the reaction is conducted under very non-ideal conditions with respect to the behavior of gases and/or including liquid mixtures with a very large Enthalpy of Mixing, then you have to expect big differences between the (tabulated) standard values and the measured values.

So, in this sense you would have quite a hard time to calculate the exact Standard Enthalpy of Reaction from your measured $\Delta_{\mathrm{R}} H$. But, under the appropriate conditions your measured value is a good approximation.

How to calculate $\Delta_{\mathrm{r}} H$?

The key to solve your calculational problem is to look at the definition of $\Delta_{\mathrm{r}} H$:

\begin{equation} \Delta_{\mathrm{r}} H = \biggl(\frac{\partial H}{\partial \xi} \biggl)_{p, T} \ , \end{equation}

whereby the extent of reaction $\xi$ is defined as

\begin{equation} \mathrm{d} \xi = \frac{\mathrm{d}n_{i}}{\nu_{i}} \end{equation}

with $n_{i}$ and $\nu_{i}$ being the amount and the stoichiometric coefficient of substance $i$, respectively. Consider the reaction:

\begin{equation} \ce{\nu_{A} A + \nu_{B} B <=> \nu_{C} C + \nu_{D} D} \end{equation}

From its definition you see that $\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process when $\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B are converted to $\ce{\nu_{C}}$ moles of C and $\ce{\nu_{D}}$ D (= "conversion of one formula unit"). So, using the differential form of your calorimetric formula (under the assumptions you made) and "dividing" it by $\mathrm{d} \xi$ you get

\begin{equation} \mathrm{d} H = m C \mathrm{d}T \\ \frac{\partial H}{\partial \xi} = m C \frac{\partial T}{\partial \xi} \\ \Delta_{\mathrm{r}} H = m C \frac{\partial T}{\frac{\partial n_{i}}{\nu_{i}}} = m C \nu_{i} \frac{\partial T}{\partial n_{i}} \ , \end{equation}

where $m$ is the mass of the stuff you heat in the calorimeter (assumed to be constant) and $C$ its heat capacity. Considering finite instead of infinitesimal changes of $n_{i}$ leads to

\begin{equation} \Delta_{\mathrm{r}} H = m C \nu_{i} \frac{\Delta T}{\Delta n_{i}} \ . \end{equation}

So, in order to get your Enthalpy of Reaction you simply have to pick one substance from your reaction (it doesn't matter which) and substitute its stoichometric coefficient and the amount that was consumed/produced during the reaction into the equation above. Note that since $\xi$ is always positive the sign of $\Delta n_{i}$ is positive for products and negative for reactants (the same is true for $\nu_{i}$).

Side Note: Standard Enthalpy of Formation $\Delta_{\mathrm{f}} H^{\ominus}$

The Standard Enthalpy of Formation is a special Standard Enthalpy of Reaction for which further conditions are in effect:

  • the target compound is to be formed by reacting the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature.
  • The reaction equation is to be formulated in such a way that $1 \, \text{mol}$ of the compound in question is formed.
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  • $\begingroup$ Thanks for the explanation. Is there any material it is referencing? BTW, on this part "...in a isothermal and isobaric process when $ν_A$ moles of A and $ν_B$ moles of B are converted to ...", shouldn't the $ν_A$, $v_b$ be just the stoichiometric coefficients, while $n_A$, $n_B$ be instead the number of mole? Referring back to my original example, if I pick $\ce{NaOH}$, $\Delta_rH$ = $(0.03*4200*2*\Delta T\over 0.02)$ = $(0.03*4200*\Delta T)\over 0.01$. While if I pick $\ce{H2SO4}$, $\Delta_rH$ = $(0.03*4200*1*\Delta T)\over 0.01$ = $(0.03*4200*\Delta T)\over 0.01$. So both will be the same $\endgroup$ – Ijriims Sep 11 '13 at 1:49
  • $\begingroup$ @Ijriims I'm sorry but I don't know what you mean by "Is there any material it is referencing?" - do you mean where I have got the information? And for the sentence "...$\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B...": no, I mean exactly what is written there. The point about $\Delta_{\mathrm{r}} H$ is that it is the heat you get from the "conversion of one reaction formula unit", so you could say when $\Delta n_{\mathrm{A}} = \nu_{\mathrm{A}} \, \text{mol}$ and $\Delta n_{\mathrm{B}} = \nu_{\mathrm{B}} \, \text{mol}$ etc. $\endgroup$ – Philipp Sep 11 '13 at 2:33
  • $\begingroup$ @Ijriims And for the last part of your comment - "So both will be the same" - yes, that's exactly what you should expect as $\Delta_{\mathrm{r}} H$ describes the heat you get per formula unit of the reaction it should give the same result irrespective of the component as long as the reaction is complete and there are no excess of any one component. $\endgroup$ – Philipp Sep 11 '13 at 2:38
  • $\begingroup$ Yes. I mean the source. And I think the formula still works even if there is excess reactant, since it uses $\Delta n_i$, not $n_i$. You calculates how much number of mole of the excess reactant is actually being reacted, then that amount is $\Delta n_i$. $\endgroup$ – Ijriims Sep 11 '13 at 2:48
  • $\begingroup$ @Ijriims Well, as for the source: Most of it comes from memory :) Apart from that you should find most of the stuff in any undergrad text book on physical chemistry. I often use a German one by Gerd Wedler. And, of course you are right about that it still works if there is excess reactant because you only need $\Delta n$. It's nearly 5 am here in Germany and I'm a bit tired so this one slipped into my comment. I guess I thought of $\Delta n$ being easier to measure when there is no excess of the component. $\endgroup$ – Philipp Sep 11 '13 at 2:55

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