<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
Density of water = 1 g/mL
Volume of water = 25.0 mL
Putting values in above equation, we get:
![1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g](https://tex.z-dn.net/?f=1g%2FmL%3D%5Cfrac%7B%5Ctext%7BMass%20of%20water%7D%7D%7B25.0mL%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20water%7D%3D%281g%2FmL%5Ctimes%2025.0mL%29%3D25g)
To calculate the heat released by the reaction, we use the equation:
![q=mc\Delta T](https://tex.z-dn.net/?f=q%3Dmc%5CDelta%20T)
where,
q = heat released
m = Total mass = [1.25 + 25] = 26.25 g
c = heat capacity of water = 4.18 J/g°C
= change in temperature = ![T_2-T_1=(21.9-25.8)^oC=-3.9^oC](https://tex.z-dn.net/?f=T_2-T_1%3D%2821.9-25.8%29%5EoC%3D-3.9%5EoC)
Putting values in above equation, we get:
![q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ](https://tex.z-dn.net/?f=q%3D26.25g%5Ctiimes%204.18J%2Fg%5EoC%5Ctimes%20%28-3.9%5EoC%29%3D-427.9J%3D-0.428kJ)
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Given mass of ammonium nitrate = 1.25 g
Molar mass of ammonium nitrate = 80 g/mol
Putting values in above equation, we get:
![\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20ammonium%20nitrate%7D%3D%5Cfrac%7B1.25g%7D%7B80g%2Fmol%7D%3D0.0156mol)
To calculate the enthalpy change of the reaction, we use the equation:
![\Delta H_{rxn}=\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Cfrac%7Bq%7D%7Bn%7D)
where,
q = amount of heat released = -0.428 kJ
n = number of moles = 0.0156 moles
= enthalpy change of the reaction
Putting values in above equation, we get:
![\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Cfrac%7B-0.428kJ%7D%7B0.0156mol%7D%3D-27.44kJ%2Fmol)
Hence, the enthalpy change of the reaction is -27. kJ/mol