Answer : The enthalpy of this reaction is, 1.06 kJ/mol
Explanation :
First we have to calculate the heat produced.
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = ?
m = mass of solution = 137 g
c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]21.00^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]24.70^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=137g\times 4.18J/g^oC\times (24.70-21.00)^oC[/tex]
[tex]q=2118.842J=2.12kJ[/tex]
Now we have to calculate the enthalpy of this reaction.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat released = 2.12 kJ
n = moles of compound = 2.00 mol
Now put all the given values in the above formula, we get:
[tex]\Delta H=\frac{2.12kJ}{2.00mole}[/tex]
[tex]\Delta H=1.06kJ/mol[/tex]
Thus, the enthalpy of this reaction is, 1.06 kJ/mol
