Answer:
[tex]\Delta H_{rxn}=-0.111 \frac{kJ}{mol}[/tex]
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:
[tex]Q_{rxn}=-C\Delta T\\\\Q_{rxn}=6.38\frac{J}{\°C}*3.2\°C=-20.416J[/tex]
Now, since this total heat flow due to the reaction is defined in terms of the heat of reaction and the total reacted moles:
[tex]Q_{rxn}=n*\Delta H_{rxn}[/tex]
Thus, we compute the moles in 8.5 g of ethanol:
[tex]n=8.5g*\frac{1mol}{46.08g}=0.185mol[/tex]
Therefore, the heat of reaction results:
[tex]\Delta H_{rxn}=\frac{Q_{rxn}}{n} =\frac{-20.416J}{0.185mol} =-110.7J/mol\\\\\Delta H_{rxn}=-0.111 \frac{kJ}{mol}[/tex]
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