Respuesta :
Answer : The enthalpy of neutralization is, -113.9 KJ/mole
Explanation :
First we have to calculate the moles of [tex]H_2SO_4[/tex] and [tex]NaOH[/tex].
[tex]\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.770mole/L\times 0.070L=0.0539mole[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.420mole/L\times 0.070L=0.0294mole[/tex]
The balanced chemical reaction will be,
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
From the balanced reaction we conclude that,
As, 2 mole of [tex]NaOH[/tex] neutralizes by 1 mole of [tex]H_2SO_4[/tex]
As, 0.0294 mole of [tex]NaOH[/tex] neutralizes by [tex]\frac{0.0294}{2}=0.0147[/tex] mole of [tex]H_2SO_4[/tex]
Thus, the number of neutralized moles = 0.0147 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]70.0ml+70.0ml=140.0ml[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 140.0ml=140.0g[/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]4.184J/g^oC[/tex]
m = mass of water = 140.0 g
[tex]T_{final}[/tex] = final temperature of water = [tex]26.00^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature of metal = [tex]23.14^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=140.0g\times 4.184J/g^oC\times (26.00-23.14)^oC[/tex]
[tex]q=1675.27J=1.675kJ[/tex]
Thus, the heat released during the neutralization = -1.675 KJ
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 = -1.675 KJ
n = number of moles used in neutralization = 0.0147 mole
[tex]\Delta H=\frac{-1.675KJ}{0.0147mole}=-113.9KJ/mole[/tex]
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, -113.9 KJ/mole
