2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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}

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

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.