Question

1. when 0.1375 g of solid magnesium is burned in a constant - volume bomb calorimeter, the temperature increases by 1.126°c. the heat capacity of the bomb calorimeter, determined in a separate experiment, is 3024 j/°c. calculate the heat given off by the burning magnesium in kj/mol.

a. - 1.14×10³ kj/mol
b. - 82.4 kj/mol
c. - 8.52×10³ kj/mol
d. - 602 kj/mol
e. - 980 kj/mol

Explanation:

Step1: Calculate the heat absorbed by the calorimeter

The formula for heat absorbed by a calorimeter is \(q = C\Delta T\), where \(C\) is the heat capacity and \(\Delta T\) is the temperature change.
Given \(C = 3024\space J/^{\circ}C\) and \(\Delta T=1.126^{\circ}C\)
\(q = 3024\times1.126 = 3405.024\space J\)

Step2: Convert the mass of magnesium to moles

The molar mass of \(Mg\) is \(M = 24.305\space g/mol\)
The mass of \(Mg\) is \(m = 0.1375\space g\)
The number of moles \(n=\frac{m}{M}=\frac{0.1375}{24.305}\approx0.00566\space mol\)

Step3: Calculate the heat per mole

The heat given off by the reaction (\(q_{rxn}\)) is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter (\(q_{cal}\)). So \(q_{rxn}=- 3405.024\space J\)
To find the heat per mole (\(\Delta H\)), use the formula \(\Delta H=\frac{q_{rxn}}{n}\)
\(\Delta H=\frac{-3405.024}{0.00566}\approx - 602\times10^{3}\space J/mol=-602\space kJ/mol\)

Answer:

D. -602 kJ/mol