Answer : The enthalpy of neutralization is, 56.012 kJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
![\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HCl%7D%3D%5Ctext%7BConcentration%20of%20HCl%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%7D%3D0.400mole%2FL%5Ctimes%200.200L%3D0.08mol)
![\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20NaOH%7D%3D%5Ctext%7BConcentration%20of%20NaOH%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%7D%3D0.400mole%2FL%5Ctimes%200.200L%3D0.08mol)
The balanced chemical reaction will be,
![HCl+NaOH\rightarrow NaCl+H_2O](https://tex.z-dn.net/?f=HCl%2BNaOH%5Crightarrow%20NaCl%2BH_2O)
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 = ![200mL+200L=400mL](https://tex.z-dn.net/?f=200mL%2B200L%3D400mL)
![\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20water%7D%3D%5Ctext%7BDensity%20of%20water%7D%5Ctimes%20%5Ctext%7BVolume%20of%20water%7D%3D1g%2Fml%5Ctimes%20400mL%3D400g)
Now we have to calculate the heat absorbed during the reaction.
![q=m\times c\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
where,
q = heat absorbed = ?
= specific heat of water = ![4.18J/g^oC](https://tex.z-dn.net/?f=4.18J%2Fg%5EoC)
m = mass of water = 400 g
= final temperature of water = ![27.78^oC=273+25.10=300.78K](https://tex.z-dn.net/?f=27.78%5EoC%3D273%2B25.10%3D300.78K)
= initial temperature of metal = ![25.10^oC=273+27.78=298.1K](https://tex.z-dn.net/?f=25.10%5EoC%3D273%2B27.78%3D298.1K)
Now put all the given values in the above formula, we get:
![q=400g\times 4.18J/g^oC\times (300.78-298.1)K](https://tex.z-dn.net/?f=q%3D400g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%28300.78-298.1%29K)
![q=4480.96J](https://tex.z-dn.net/?f=q%3D4480.96J)
Thus, the heat released during the neutralization = -4480.96 J
Now we have to calculate the enthalpy of neutralization.
![\Delta H=\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7Bq%7D%7Bn%7D)
where,
= enthalpy of neutralization = ?
q = heat released = -4480.96 J
n = number of moles used in neutralization = 0.08 mole
![\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-4480.96J%7D%7B0.08mole%7D%3D-56012J%2Fmole%3D-56.012kJ%2Fmol)
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.012 kJ/mole