Respuesta :
Answer:
-24.76 kJ/g; -601.8 kJ/mol
Explanation:
There are two heat flows in this experiment.
Heat from reaction + heat absorbed by calorimeter = 0
q1 + q2 = 0
mΔH + CΔT = 0
Data:
m = 0.1375 g
C = 3024 J/°C
ΔT = 1.126 °C
Calculations:
0.1375ΔH + 3024 × 1.126 = 0
0.1375ΔH + 3405 = 0
0.1375ΔH = -3405
ΔH = -24 760 J/g = -24.76 kJ/g
ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.
When a sample of magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity (C) of 3024 J/°C, the temperature increases by 1.126°C (ΔT). We can calculate the heat absorbed by the calorimeter (Qc) using the following expression.
[tex]Qc = C \times \Delta T = \frac{3024J}{\° C} \times 1.126 \° C \times \frac{1kJ}{1000J} = 3.405 kJ[/tex]
According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the reaction (Qr) is zero.
[tex]Qc + Qr = 0\\\\Qr = -Qc = -3.405 kJ[/tex]
3.405 kJ are released by the combustion of 0.1375 g of Mg. The heat released per gram of Mg is:
[tex]\frac{-3.405kJ}{0.1375g} = -24.76 kJ/g[/tex]
Finally, we will convert -24.67 kJ/g to kJ/mol using the molar mass of Mg (24.31 g/mol).
[tex]\frac{-24.76kJ}{g} \times \frac{24.31g}{mol} = -601.9 kJ/mol[/tex]
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter (heat capacity of 3024 J/°C), causing a temperature increase of 1.126°C. The heat given off by the burning Mg is -24.76 kJ/g and -601.9 kJ/mol.
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