Answer : The enthalpy of reaction
is, 67.716 KJ/mole
Explanation :
First we have to calculate the moles of
and
.
![\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DAgNO_3%3D%5Ctext%7BMolarity%20of%20%7DAgNO_3%5Ctimes%20%5Ctext%7BVolume%7D%3D%280.100mole%2FL%29%5Ctimes%20%280.05L%29%3D0.005mole)
![\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DHCl%3D%5Ctext%7BMolarity%20of%20%7DHCl%5Ctimes%20%5Ctext%7BVolume%7D%3D%280.100mole%2FL%29%5Ctimes%20%280.05L%29%3D0.005mole)
Now we have to calculate the moles of AgCl formed.
The balanced chemical reaction will be,
![AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)](https://tex.z-dn.net/?f=AgNO_3%28aq%29%2BHCl%28aq%29%5Crightarrow%20AgCl%28s%29%2BHNO_3%28aq%29)
As, 1 mole of
react with 1 mole of
to give 1 mole of ![AgCl](https://tex.z-dn.net/?f=AgCl)
So, 0.005 mole of
react with 0.005 mole of
to give 1 mole of ![AgCl](https://tex.z-dn.net/?f=AgCl)
The moles of AgCl formed = 0.005 mole
Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml
Now we have to calculate the mass of solution.
Mass of the solution = Density of the solution × Volume of the solution
Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g
Now we have to calculate the heat.
![q=m\times C\Delta T=m\times C \times (T_2-T_1)](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20C%5CDelta%20T%3Dm%5Ctimes%20C%20%5Ctimes%20%28T_2-T_1%29)
where,
q = heat
C = specific heat capacity = ![4.18J/g^oC](https://tex.z-dn.net/?f=4.18J%2Fg%5EoC)
m = mass = 100 g
= final temperature = ![24.21^oC](https://tex.z-dn.net/?f=24.21%5EoC)
= initial temperature = ![23.40^oC](https://tex.z-dn.net/?f=23.40%5EoC)
Now put all the given values in the above expression, we get:
![q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC](https://tex.z-dn.net/?f=q%3D100g%5Ctimes%20%284.18J%2Fg%5EoC%29%5Ctimes%20%2824.21-23.40%29%5EoC)
![q=338.58J](https://tex.z-dn.net/?f=q%3D338.58J)
Now we have to calculate the enthalpy of the reaction.
![\Delta H_{rxn}=\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Cfrac%7Bq%7D%7Bn%7D)
where,
= enthalpy of reaction = ?
q = heat of reaction = 338.58 J
n = moles of reaction = 0.005 mole
Now put all the given values in above expression, we get:
![\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Cfrac%7B338.58J%7D%7B0.005mole%7D%3D6771.6J%2Fmole%3D67.716KJ%2Fmole)
Conversion used : (1 KJ = 1000 J)
Therefore, the enthalpy of reaction
is, 67.716 KJ/mole