Question

An 8.5 g. Of ethanol (C2H5OH) placed in a constant volume calorimeter and the temperature rose to 3.2°C. Find the heat of reaction of the ethanol (MM =46.1 g/mol) in kJ/mol. The heat capacity of the calorimeter plus water is read as 6.38 J/°C

Answer 1

`\Delta H_(rxn)=-0.111 (kJ)/(mol)`

Step-by-step explanation:

Hello!

In this case, since the total heat flow due to the reaction equals the negative of the calorimeter's heat, we can first compute the former as shown below:

`Q_(rxn)=-C\Delta T\\\\Q_(rxn)=6.38(J)/(\°C)*3.2\°C=-20.416J`

Now, since this total heat flow due to the reaction is defined in terms of the heat of reaction and the total reacted moles:

`Q_(rxn)=n*\Delta H_(rxn)`

Thus, we compute the moles in 8.5 g of ethanol:

`n=8.5g*(1mol)/(46.08g)=0.185mol`

Therefore, the heat of reaction results:

`\Delta H_(rxn)=(Q_(rxn))/(n) =(-20.416J)/(0.185mol) =-110.7J/mol\\\\\Delta H_(rxn)=-0.111 (kJ)/(mol)`

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