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Find ΔH for the combustion of ethanol (C2H6O) to carbon dioxide and liquid water from the following data. The heat capacity of the bomb calorimeter is 34.65 kJ/K and the combustion of 1.752 g of ethanol raises the temperature of the calorimeter from 294.42 K to 295.92 K .

Respuesta :

Answer : The enthalpy of combustion of ethanol is, -1366.8 KJ/mole

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_{combustion}=-q_{calorimeter}[/tex]

First we have to calculate the heat released by calorimeter.

[tex]q_{calorimeter}=-c\times (T_2-T_1)[/tex]

where,

[tex]q_{calorimeter}[/tex] = heat released by calorimeter = ?

c = heat capacity of bomb calorimeter = [tex]34.65kJ/K[/tex]

m = mass of ethanol = 1.752 g

[tex]T_1[/tex] = initial temperature of calorimeter = 294.42 K

[tex]T_2[/tex] = final temperature of calorimeter = 295.92 K

Now put all the given values in the above formula, we get

[tex]q_{calorimeter}=-34.65kJ/K\times (295.92-294.42)K[/tex]

[tex]q_{calorimeter}=-51.98kJ[/tex]

[tex]q_{combustion}=-51.98kJ[/tex]

Now we have to calculate the moles of ethanol.

[tex]\text{Moles of ethanol}=\frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{1.752g}{46.07g/mol}=0.03803mol[/tex]

Now we have to calculate the enthalpy of combustion of ethanol.

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy of combustion = ?

q = heat released = -51.98 KJ

n = number of moles used in combustion = 0.03803 mole

[tex]\Delta H=\frac{-51.98kJ}{0.03803mole}=-1366.8KJ/mole[/tex]

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of combustion of ethanol is, -1366.8 KJ/mole

pstnonsonjoku

The enthalpy change for the combustion of ethanol is -1367.76 KJ/mol.

From the question, we can see that the mass of ethanol used = 1.752 g

Heat capacity of the bomb calorimeter = 34.65 kJ/K

Temperature change of the calorimeter =  295.92 K - 294.42 K = 1.5 K

We know that heat lost by combustion of ethanol = Heat gained by the calorimeter

Hence;

Heat gained by calorimeter = Heat capacity × Temperature change

Heat gained by calorimeter = 34.65 kJ/K ×  1.5 K = 51.975 KJ

Number of moles of ethanol =  1.752g/46.07 g/mol

= 0.038 moles

ΔHrxn = -51.975 KJ/0.038 moles

ΔHrxn = -1367.76 KJ/mol

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