Respuesta :
Answer:
The enthalpy change during the reaction is -84 kJ/mole.
Explanation:
First we have to calculate the heat gained by the solution.
[tex]Q=mc\times (T_{final}-T_{initial})[/tex]
where,
Q = heat gained = ?
m = mass of the solution = 100.0 g
c = specific heat = [tex]4.2 J/^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]21.0 ^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]20.0 ^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]Q=100.0 g\times 4.2 J/^oC\times (21.0-20.0)^oC[/tex]
[tex]Q=420 J[/tex]
Now we have to calculate the enthalpy change during the reaction.
[tex]\Delta H=-\frac{Q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
Q = heat gained = 420 J
n = number of moles silver nitrate= n
[tex]AgCl(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]
Molarity of the silver nitrate solution = M = 0.100 M
Volume of the silver nitrate solution = V = 50.0 mL = 0.050 L
[tex]n=0.100M\times 0.050 L=0.005 mol[/tex]
[tex]\Delta H=-\frac{420 J}{0.005mole}=-84000J/mol=-84 kJ/mol[/tex]
(1 kJ = 1000 J)
Therefore, the enthalpy change during the reaction is -84 kJ/mole.
