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# A chemist places 1.750 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},$ in a bomb calorimeter with a heat capacity of 12.05 kJ/K. The sample is burned and the temperature of the calorimeter increases by $4.287^{\circ} \mathrm{C}$. Calculate $\Delta E$ for the combustion of ethanol in kJ/mol.

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First of all, the heat liberated in the reaction should be determined. It occurs as follows.

$q = c \times \Delta T,$

where c stands for the heat capacity and $\Delta T$ stands for the temperature difference before and after the reaction. Now, let's see the heat.

$q = c \times \Delta T = 12.05 kJ/K \times 4.287 K = 51.66 kJ$

As the temperature increases, the heat has been given to the surroundings, so the system has lost some energy. Therefore the heat will be indicated negative.

$q = -51.66 kJ$

Subsequently, from this value, the value of $\Delta E$ can be calculated:

$\Delta E = \dfrac{q}{n}$

We also need to know the amount of substance (n) for this calculation. It can be determined my the mass (m) and the molar mass of the ethanol.

$M_{ethanol} = 46.07 g/mol$

$n_{ethanol} = \dfrac{m_{ethanol}}{M_{ethanol}} = \dfrac{1.750 g}{46.07g/mol} = 0.03799 mol$

Last, but not least, let's determine the value of $\Delta E$:

$\Delta E = \dfrac{q}{n} = \dfrac{-51.66 kJ}{0.03799 mol} = -1360 kJ/mol$

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