Chemistry, 06.07.2019 05:30 tati7149

A0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 j/°c. the temperature increases by 1.126°c. calculate the heat given off by the burning mg, in kj/g and in kj/mol.

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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.

$Qc = C \times \Delta T = \frac{3024J}{\° C} \times 1.126 \° C \times \frac{1kJ}{1000J} = 3.405 kJ$

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.

$Qc + Qr = 0\\\\Qr = -Qc = -3.405 kJ$

3.405 kJ are released by the combustion of 0.1375 g of Mg. The heat released per gram of Mg is:

$\frac{-3.405kJ}{0.1375g} = -24.76 kJ/g$

Finally, we will convert -24.67 kJ/g to kJ/mol using the molar mass of Mg (24.31 g/mol).

$\frac{-24.76kJ}{g} \times \frac{24.31g}{mol} = -601.9 kJ/mol$

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Answer from: blacklifewed

-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

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