Answer :
Answer : The heat of the reaction is, 1.27 kJ/mole
Explanation :
First we have to calculate the heat released.
Formula used :
[tex]Q=m\times c\times \Delta T[/tex]
or,
[tex]Q=m\times c\times (T_2-T_1)[/tex]
where,
Q = heat = ?
m = mass of sample = 1.50 g
c = specific heat of water = [tex]4.81J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]22.7^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]19.4^oC[/tex]
Now put all the given value in the above formula, we get:
[tex]Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC[/tex]
[tex]Q=-23.8095J=-0.0238kJ[/tex]
Now we have to calculate the heat of the reaction in kJ/mol.
[tex]\Delta H=\frac{Q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
Q = heat released = 0.0238 kJ
n = number of moles NH₄NO₃ = [tex]\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}=\frac{1.50g}{80g/mol}=0.01875mole[/tex]
[tex]\Delta H=\frac{0.0238kJ}{0.01875mole}=1.27kJ/mole[/tex]
Therefore, the heat of the reaction is, 1.27 kJ/mole