Computational heat of reation vs. using literature values [closed]








**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
 
 






---








**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/59913/edit).




Closed 3 years ago.








[Improve this question](/posts/59913/edit) 



My question is I don't understand how to explain why or what I did wrong to have such large differences between my computational heat of reaction and heat of reaction using enthalpies of formation from literature, and my experimental heat of reaction.


I performed a lab in which three methods were used to compute the heat of reaction for the reaction:


$$
\ce{Na2CO3 (s) + 2 HCl (aq) -> 2 NaCl (aq) + H2O (l) + CO2 (g)}
$$


First method I used was using experimental data from solution calorimetry and from my results I got that the average heat of reaction was $\textbf{-89.835 kJ/mol}$.


For my second method I found the heat of reaction computationally with Orca. I was instructed to use this equation instead of the original equation to calculate heat of reaction for the above reaction.


$$
\Delta U = U \left[ \ce{CO2 (g)} \right] + 3U \left[ \ce{H2O (l)} \right] - 2U \left[ \ce{H3O+ (aq)} \right] - U \left[ \ce{CO3^{2-} (aq)} \right]
$$


So basically I used Orca to find the energy for each of the molecules, and I plugged those values into their proper U(molecule) in the above formula and I calculated delta U and then I used the approximation of $\Delta H = \Delta U + R \Delta T$ to transform it into heat. My overall answer ended up being $\textbf{-378 kJ/mol}$. My work (data from Orca optimization is listed first):


\begin{align\*}
\ce{CO3^{2-} (aq)} &= -264.0667 \, \text{Ha} \\
 \ce{H3O+ (aq)} &= -76.8491 \, \text{Ha} \\
 \ce{H2O (l)} &= -76.4409 \, \text{Ha} \\
 \ce{CO2 (g)} &= -188.5873 \, \text{Ha} \\
 1 \, \text{Ha} &= 627.503 \, \text{kcal/mol} \\
 \Delta H &= \Delta U + R \Delta T \\
 \Delta U &= \left[ -188.5873 + (3\*-76.4409) \right] - \left[ (2\*-76.8491) + -264.0667 \right] = -0.1451 \, \text{Ha} \\
-0.1451 \, \text{Ha} &\rightarrow 627.503 \, \frac{\text{kcal/mol}}{\text{Ha}} \* -0.1451 \, \text{Ha} = -91.0507 \, \frac{\text{kcal}}{\text{mol}} \\
-91.0507 \, \frac{\text{kcal}}{\text{mol}} &\* 1000 \, \frac{\text{cal}}{\text{kcal}} \* 4.184 \, \frac{\text{J}}{\text{cal}} \* \frac{1}{1000} \frac{\text{kJ}}{\text{J}} = -380.956 \, \frac{\text{kcal}}{\text{mol}} \\
\Delta H\_{\text{rxn}} &= -380.956 \frac{\text{kJ}}{\text{mol}} + \left[ 8.314 \frac{\text{J}}{\text{mol K}} \* (273.15 + 25.27) \, \text{K} \* \frac{1}{1000} \frac{\text{kJ}}{\text{J}} \right] = -378.475 \, \text{kJ/mol}
\end{align\*}


My third method I used enthalpies of formation (values for looked up from various sources) and my work is shown below.


\begin{align\*}
\ce{Na2CO3 (s)} &= -1130.7 \, \text{kJ/mol} \\
 \ce{HCl (aq)} &= -167.2 \, \text{kJ/mol} \\
 \ce{NaCl (aq)} &= -407.27 \, \text{kJ/mol} \\
 \ce{H2O (l)} &= -285.8 \, \text{kJ/mol} \\
 \ce{CO2 (g)} &= -393.5 \, \text{kJ/mol} \\
\Delta H\_{\text{rxn}} &= \left[ (2\*-407.27) + (-285.8) + (-393.5) \right] - \left[ (2\*-167.2) + (-1130.7) \right] = \mathbf{-28.74 \, \textbf{kJ/mol}}
\end{align\*}


I'm not really sure how to say why the computational method did so poorly, I would have thought the computer would have done a decent job. Please let me know if something looks wrong in my calculations! Thanks for all the help! Even just a little advice will help this non-chemistry nerd!


