A $1.000~\mathrm{g}$ sample of ethanol, $\ce{C2H5OH}$, was burned in a bomb calorimeter whose heat capacity had been determined to be $2.71~\mathrm{kJ/^\circ C}$. The temperature of $3.000~\mathrm{kg}$ of water rose from $24.284~^\circ\mathrm{C}$ to $26.225~^\circ\mathrm{C}$. Determine the $\Delta H$ for the reaction in $\mathrm{kJ/mol}$ of ethanol?

I said :

\begin{aligned} q_\mathrm{cal} &= C_\mathrm{cal}\cdot T\\ &= (2.71\ \mathrm{kJ/^\circ C})(1.94\ \mathrm{^\circ C})\\ &= 5.26~\mathrm{kJ}\\ \end{aligned}

Then the $q_\mathrm{r}=-5.26\ \mathrm{kJ}$

Then calculated the number of mol by dividing the mass g over molar mass and I got $0.02171~\mathrm{mol}$.

Finally I divided $-5.26\ \mathrm{kJ}$ by $0.02171\ \mathrm{mol}$ and I got $= -242.3~\mathrm{kJ/mol}$.

But the correct value is $-1365~\mathrm{kJ/mol}$. What's wrong?

  • 1
    $\begingroup$ What about the 3kg of water? The equation you chose only works if you factor the water's heat capacity into the Ccal, which did not happen here. $\endgroup$ Jul 14, 2014 at 1:26
  • $\begingroup$ So,I should get the q of water by the formula s.m.T and then take the negative value of it .. and then divide it by the number of mol. Right? $\endgroup$
    – Maher
    Jul 14, 2014 at 1:33
  • $\begingroup$ @brinnb I don't get the point of Ccal ?? $\endgroup$
    – Maher
    Jul 14, 2014 at 3:39
  • $\begingroup$ The calorimeter is made of matter so it absorbs heat too. Ccal tells you how much. $\endgroup$ Jul 14, 2014 at 20:56

1 Answer 1


$$|Q|=|(ms+C)\Delta T|=(3\times4.18+2.71)\times1.941\;\mathrm{kJ}=29.6\;\mathrm J$$ $$\Delta H_n=\Delta H/n=-29.6/(1/46)\;\mathrm{kJ}=-1361\;\mathrm{kJ}$$


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.