USNCO 2016 Local #20

So I've been trying to answer this question, and as far as I've learned it's $\Delta H^0_{\mathrm f}$(products)-$\Delta H^0_{\mathrm f}$(reactants), multiplying each by number of moles and such.

Trying to use that on this problem I came up with $\ce{0.008 OH- + 0.008 HCO3- -> 0.008 H2O + 0.008 CO3^2-}$ after (possibly incorrectly) noting that $\ce{HCO3}$ was in excess. However, this does not yield anywhere near the correct answer of C. What am I doing incorrectly?


I can obtain the correct answer using your values. Maybe double check for mistakes and typos?

Here are my quick calculations: The overall reaction is:

$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$

The equation used is: \begin{equation} \Delta H_\text{reaction}^\circ = \sum {\Delta H_{f}^\circ(\text{Products})} - \sum {\Delta H_{f}^\circ(\text{Reactants})} \end{equation}

$\Delta H_\text{products}^\circ= (\pu{0.008 mol}\times\pu{-286 kJ/mol} +\pu{0.008 mol}\times\pu{-677 kJ/mol}) = \pu{-7.704 kJ}$

$\Delta H_\text{reactants}^\circ= (\pu{0.008 mol}\times\pu{-692 kJ/mol} +\pu{0.008 mol}\times\pu{-230 kJ/mol}) = \pu{-7.376 kJ}$

$\Delta H_\text{reaction}^\circ = (-7.074 - (-7.376))~\pu{kJ/mol} = \pu{-0.328 kJ} = \pu{-328 J}$

  • $\begingroup$ Thank you! You helped me realize that I'm a buffoon and that the table is in kJ whereas the answers are in J. :P $\endgroup$ – Rohith IsMath Feb 22 '17 at 10:42
  • $\begingroup$ No problem. Could you mark the question as answered? $\endgroup$ – Bdrs Feb 22 '17 at 10:59
  • $\begingroup$ Interesting that the significant figures in the book answer seem to be wrong. I'd agree that the answer should be -328 not -330. $\endgroup$ – MaxW Feb 22 '17 at 16:45

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