Answer:
The value of Δ [tex]H_{rxn}[/tex] for the reaction = 463 [tex]\frac{KJ}{mol}[/tex]
Explanation:
[tex]q_{rxn} = - q_{sol}[/tex]
[tex]q_{rxn} = H_{rxn}[/tex]
Δ [tex]H_{rxn}[/tex] = - [tex]q_{sol}[/tex] ------ (1)
We know that
[tex]q_{sol} = m c ( T_{2} - T_{1} )[/tex]
[tex]q_{sol} =[/tex] (1)(100) × 4.18 × (32.8 - 25.6)
[tex]q_{sol} =[/tex] 3010 J = 3.01 Kilo Joule
From equation (1)
Δ [tex]H_{rxn}[/tex] = 3.01 Kilo Joule
No. of moles
[tex]N = \frac{m}{M}[/tex]
m = 0.158 gm & M = 24.31 gm Mg
No. of moles
[tex]N = \frac{0.158}{24.31}[/tex]
N = 0.0065
Therefore
Δ [tex]H_{rxn}[/tex] = [tex]\frac{3.01}{0.0065}[/tex]
Δ [tex]H_{rxn}[/tex] = 463 [tex]\frac{KJ}{mol}[/tex]
This is the value of Δ [tex]H_{rxn}[/tex] for the reaction.
