Respuesta :
Answer : The enthalpy of neutralization is, 56.96 kJ/mole
Explanation :
First we have to calculate the moles of Hâ‚‚SOâ‚„ 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 Hâ‚‚SOâ‚„
So, 0.0169 mole of NaOH neutralizes by 0.00845 mole of Hâ‚‚SOâ‚„
That means, NaOH is a limiting reagent and Hâ‚‚SOâ‚„ is an excess reagent.
Now we have to calculate the moles of Hâ‚‚O.
As, 2 mole of NaOH react to give 2 mole of Hâ‚‚O
So, 0.0169 mole of NaOH react to give 0.0169 mole of Hâ‚‚O
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
