# Determine the ΔH for the reaction in kJ/mol of ethanol?

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?

• 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. – Brinn Belyea Jul 14 '14 at 1:26
• 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? – Maher Jul 14 '14 at 1:33
• @brinnb I don't get the point of Ccal ?? – Maher Jul 14 '14 at 3:39
• The calorimeter is made of matter so it absorbs heat too. Ccal tells you how much. – Brinn Belyea Jul 14 '14 at 20:56

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