Respuesta :
Answer : The enthalpy of neutralization is, -55.8 KJ/mole
Explanation :
First we have to calculate the moles of [tex]H_2SO_4[/tex] and KOH.
[tex]\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.500mole/L\times 0.0227L=0.01135mole[/tex]
[tex]\text{Moles of KOH}=\text{Concentration of KOH}\times \text{Volume of solution}=1.00mole/L\times 0.0227L=0.0227mole[/tex]
Now we have to calculate the limiting reagent.
The balanced chemical reaction will be,
[tex]H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O[/tex]
From the balanced reaction we conclude that,
As, 1 mole of [tex]H_2SO_4[/tex] react to give 2 moles of [tex]H_2O[/tex]
So, 0.01135 mole of [tex]H_2SO_4[/tex] react to give [tex]2\times 0.01135=0.0227[/tex] moles of [tex]H_2O[/tex]
and,
As, 2 moles of [tex]KOH[/tex] react to give 2 moles of [tex]H_2O[/tex]
As, 0.0227 moles of [tex]KOH[/tex] react to give 0.0227 moles of [tex]H_2O[/tex]
From this we conclude that they can form the same amount of product. This means that both will consume completely in the reaction.
Thus, the number of neutralized moles = 0.0227 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]22.7ml+22.7ml=45.4ml[/tex]
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 45.4ml=45.4g[/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 = 45.4 g
[tex]T_{final}[/tex] = final temperature of water = [tex]30.17^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature of water = [tex]23.50^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=45.4g\times 4.184J/g^oC\times (30.17-23.50)^oC[/tex]
[tex]q=1266.99J=1.267kJ[/tex]
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 = -1.267 KJ
n = number of moles used in neutralization = 0.0227 mole
[tex]\Delta H=\frac{-1.267KJ}{0.0227mole}=-55.8KJ/mole[/tex]
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, -55.8 KJ/mole
