Answer : The molar enthalpy of reaction is, 33.3 KJ/mole
Explanation :
First we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water = [tex]60ml+60ml=120ml[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]q=m\times c\times (\Delta T)[/tex]
where,
q = heat absorbed = ?
[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
m = mass of water = 120 g
[tex]\Delta T[/tex] = change in temperature = [tex]5.18^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=120g\times 4.18J/g^oC\times (5.18)^oC[/tex]
[tex]q=2598.288J=2.60KJ[/tex]
Now we have to calculate the molar enthalpy of reaction.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy of neutralization = ?
q = heat released = 2.60 KJ
n = number of moles =
[tex]\Delta H=\frac{2.60KJ}{0.078mole}=33.3KJ/mole[/tex]
Therefore, the molar enthalpy of reaction is, 33.3 KJ/mole