Respuesta :
Answer : The enthalpy of neutralization is, 56.012 kJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
[tex]\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol[/tex]
The balanced chemical reaction will be,
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH
Thus, the number of neutralized moles = 0.08 mole
Now 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]200mL+200L=400mL[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat absorbed = ?
[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
m = mass of water = 400 g
[tex]T_{final}[/tex] = final temperature of water = [tex]27.78^oC=273+25.10=300.78K[/tex]
[tex]T_{initial}[/tex] = initial temperature of metal = [tex]25.10^oC=273+27.78=298.1K[/tex]
Now put all the given values in the above formula, we get:
[tex]q=400g\times 4.18J/g^oC\times (300.78-298.1)K[/tex]
[tex]q=4480.96J[/tex]
Thus, the heat released during the neutralization = -4480.96 J
Now we have to calculate the enthalpy of neutralization.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy of neutralization = ?
q = heat released = -4480.96 J
n = number of moles used in neutralization = 0.08 mole
[tex]\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol[/tex]
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.012 kJ/mole
