Answer :
Answer : The enthalpy of neutralization is, 56.96 kJ/mole
Explanation :
First we have to calculate the moles of H₂SO₄ and NaOH.
[tex]\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.890mole/L\times 0.065L=0.0578mole[/tex]
[tex]\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.260mole/L\times 0.065L=0.0169mole[/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 NaOH neutralizes by 1 mole of H₂SO₄
So, 0.0169 mole of NaOH neutralizes by 0.00845 mole of H₂SO₄
That means, NaOH is a limiting reagent and H₂SO₄ is an excess reagent.
Now we have to calculate the moles of H₂O.
As, 2 mole of NaOH react to give 2 mole of H₂O
So, 0.0169 mole of NaOH react to give 0.0169 mole of H₂O
Now we have to calculate the mass of water.
As we know that the density of water is 1.00 g/ml. So, the mass of water will be:
The volume of water = [tex]65.0ml+65.0ml=130ml[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/ml\times 130ml=130g[/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 = 130 g
[tex]T_{final}[/tex] = final temperature of water = [tex]23.78^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature of metal = [tex]25.55^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=130g\times 4.184J/g^oC\times (25.55-23.78)^oC[/tex]
[tex]q=962.7J[/tex]
Thus, the heat released during the neutralization = -962.7 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 = -962.7 J
n = number of moles used in neutralization = 0.0169 mole
[tex]\Delta H=\frac{-962.7J}{0.0169mole}=-56964.49J/mole=-56.96kJ/mol[/tex]
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.96 kJ/mole