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

Question

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?

Answer 1

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