Respuesta :
Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:
[tex]Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)[/tex]
Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL
[tex]m=1.00 g/mL\times 100.0 mL = 100 g[/tex]
First we have to calculate the heat gained by the solution in coffee-cup calorimeter.
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat = [tex]4.18 J/^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]23.1^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]28.9^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC[/tex]
[tex]q=2,242.4 J=2.242 kJ [/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 = 2.242 kJ
n = number of moles fructose = [tex]\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol[/tex]
[tex]\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol[/tex]
Therefore, the enthalpy change during the reaction is -199. kJ/mol.
