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. $\endgroup$ – Brinn Belyea Jul 14 '14 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 '14 at 1:33
  • $\begingroup$ @brinnb I don't get the point of Ccal ?? $\endgroup$ – Maher Jul 14 '14 at 3:39
  • $\begingroup$ The calorimeter is made of matter so it absorbs heat too. Ccal tells you how much. $\endgroup$ – Brinn Belyea Jul 14 '14 at 20:56

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


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