Respuesta :
Answer: ΔH for the dissolution of [tex]NH_4NO_3[/tex] is +26.0205 kJ/mol
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times C\times \Delta T[/tex]
Q = Heat released by solution = ?
C = heat capacity = [tex]4.18J/g^0C[/tex]
Initial temperature of water = [tex]T_i[/tex] = [tex]25.008^0C[/tex]
Final temperature of water = [tex]T_f[/tex] = [tex]23.348^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(23.348-25.008)^0C=-1.66^0C[/tex]
Putting in the values, we get:
[tex]Q=75.0g\times 4.18J/g^0C\times -1.66^0C=-520.41 J[/tex]
As heat released by water is equal to heat absorbed by dissolution of [tex]NH_4NO_3[/tex]
[tex]\text{Moles of}NH_4NO_3=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{1.60g}{80g/mol}=0.02mol[/tex]
Enthalpy change for 0.02 moles of [tex]NH_4NO_3[/tex] = 520.41 J
Enthalpy change for 1 mole of [tex]NH_4NO_3[/tex] = [tex]\frac{520.41}{0.02}\times 1=+26020.5J=+26.0205kJ[/tex]
ΔH for the dissolution of [tex]NH_4NO_3[/tex] is +26.0205 kJ/mol
