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Question

A chemist places 1.750 g of ethanol, C2H6O,\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 4.287C4.287^{\circ} \mathrm{C}. Calculate ΔE\Delta E for the combustion of ethanol in kJ/mol.

Solution

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Answered 11 months ago
Answered 11 months ago
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First of all, the heat liberated in the reaction should be determined. It occurs as follows.

q=c×ΔT,q = c \times \Delta T,

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

q=c×ΔT=12.05kJ/K×4.287K=51.66kJq = 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=51.66kJq = -51.66 kJ

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

ΔE=qn\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.

Methanol=46.07g/molM_{ethanol} = 46.07 g/mol

nethanol=methanolMethanol=1.750g46.07g/mol=0.03799moln_{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 ΔE\Delta E:

ΔE=qn=51.66kJ0.03799mol=1360kJ/mol\Delta E = \dfrac{q}{n} = \dfrac{-51.66 kJ}{0.03799 mol} = -1360 kJ/mol

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