Answer:
The enthalpy of the reaction is 18.616 kJ/mol.
Explanation:
Heat released by the solution = Q
Mass of the solution = 25.0 g
Heat capacity of the solution = c = 4.20 J/g K
Initial temperature of the solution ,[tex]T_1= 21.1 ^oC = 294.25 K[/tex]
Final temperature of the solution,[tex]T_2 = 17.0^oC=290.15 K[/tex]
Change in temperature of the solution = ΔT =
[tex]\Delta T=T_2-T_1=290.15 K-294.25 K=-4.1 K[/tex]
[tex]Q=mc\Delta T=25 g\times 4.20 J/g K\times (-4.1 K)=-430.5 J[/tex]
Heat gained during the reaction of ammonium nitrate = Q'
Q' = -Q (energy remained conserved)
Q'= -(-430.5 J)=430.5 J
Mass of ammonium nitrate added in water = 1.85 g
Moles of ammonium nitrate added in water =[tex]\frac{1.85 g}{80 g/mol}=0.02312 mol[/tex]
0.02312 moles of ammonium nitrate absorbs 430.5 Joules of energy. the 1 mole will absorb:
[tex]\frac{Q'}{0.02312 mol}=\frac{430.5 J}{0.02312 mol}=18,616.21 J=18.616 kJ/mol[/tex]
The enthalpy of the reaction is 18.616 kJ/mol.
