Respuesta :
Answer:
The enthalpy change during the reaction is -7020.09 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
[tex]q=c\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = ?
c = specific heat = [tex]994.1 J/^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]27.60^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]25.10^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=994.1 J/^oC\times (27.60-25.10)^oC[/tex]
[tex]q= 2,485.25 J[/tex]
The heat gained by water present in calorimeter. = q'
[tex]q'=mc\times (T_{final}-T_{initial})[/tex]
where,
q' = heat gained = ?
m = mass of water = [tex]1.153\times 10^3 g[/tex]
c' = specific heat of water = [tex]4.184 J/^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]27.60^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]25.10^oC[/tex]
[tex]q'=1.153\times 10^3 g\times 4.184 J/^oC\times (27.60-25.10)^oC[/tex]
q ' = 12,060.38 J
Now we have to calculate the enthalpy change during the reaction.
[tex]\Delta H=-\frac{Q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
Q = heat gained = -(q+q') = -(2,485.25 J + 12,060.38 J)= -14,545.63 J
Q = -14.54563 kJ
n = number of moles fructose = [tex]\frac{\text{Mass of benzil}}{\text{Molar mass of benzil}}=\frac{0.4351 g}{210 g/mol}=0.002072 mole[/tex]
[tex]\Delta H=-\frac{-14.54563 kJ}{0.002072 mole}=-7020.09 kJ/mole[/tex]
Therefore, the enthalpy change during the reaction is -7020.09 kJ/mole.
