Respuesta :
Answer: The enthalpy change of the reaction is -27. kJ/mol
Explanation:
To calculate the mass of water, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of water = 1 g/mL
Volume of water = 25.0 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g[/tex]
To calculate the heat released by the reaction, we use the equation:
[tex]q=mc\Delta T[/tex]
where,
q = heat released
m = Total mass = [1.25 + 25] = 26.25 g
c = heat capacity of water = 4.18 J/g°C
[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(21.9-25.8)^oC=-3.9^oC[/tex]
Putting values in above equation, we get:
[tex]q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of ammonium nitrate = 1.25 g
Molar mass of ammonium nitrate = 80 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol[/tex]
To calculate the enthalpy change of the reaction, we use the equation:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
q = amount of heat released = -0.428 kJ
n = number of moles = 0.0156 moles
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction
Putting values in above equation, we get:
[tex]\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol[/tex]
Hence, the enthalpy change of the reaction is -27. kJ/mol
